Answer:
a) v₀ₓ = v₀ cos θ , b) v_{oy} = v₀ sin θ , c) y = v_{oy}² / 2g, y = 24.25 m
e) R = 138.46 m
Explanation:
This is a projectile launch exercise
a) let's use trigonometry to find the components of the initial velocity
cos θ = v₀ₓ / v₀
v₀ₓ = v₀ cos θ
v₀ₓ = 38 cos 35
v₀ₓ = 31.13 m / s
b) sin θ = [tex]v_{oy}[/tex] / v₀
v_{oy} = v₀ sin θ
v_{oy} = 38 sint 35
v_{oy} = 21 80 m / s
c, d) to find the maximum height, the vertical speed is zero
v_{y}² = v_{oy}² - 2 g y
0 = [tex]v_{oy}[/tex]² - 2 gy
y = v_{oy}² / 2g
let's calculate
y = 21.80 2 / (2 9.8)
y = 24.25 m
e) They ask to find the horizontal distance
for this we can use the expression of reaches
R = v₀² sin 2θ / g
let's calculate
R = 38² sin (2 35) / 9.8
R = 138.46 m
g A top-fuel dragster starts from rest and has a constant acceleration of 44.0 m/s2. What are (a) the final velocity of the dragster at the end of 2.1 s, (b) the final velocity of the dragster at the end of of twice this time, or 4.2 s, (c) the displacement of the dragster at the end of 2.1 s, and (d) the displacement of the dragster at the end of twice this time, or 4.2 s?
The dragster's velocity v at time t with constant acceleration a is
[tex]v=at[/tex]
since it starts at rest.
After 2.1 s, it will attain a velocity of
[tex]v=\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)[/tex]
or 92.4 m/s.
Doubling the time would double the final velocity,
[tex]v=a(2t)=2at[/tex]
so the velocity would be twice the previous one, 184.8 m/s.
The dragster undergoes a displacement x after time t with acceleration a of
[tex]x=\dfrac12at^2[/tex]
if we take the starting line to be the origin.
After 2.1 s, it will have moved
[tex]x=\dfrac12\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)^2[/tex]
or 88 m.
Doubling the time has the effect of quadrupling the displacement, since
[tex]x=\dfrac12a(2t)^2=4\left(\dfrac12at^2\right)[/tex]
so after 4.2 s it will have moved 352 m.
Which symbol in a chemical equation separates the reactants from the products?
Answer:
the arrow symbol ⇒ in irreversible reactions and doble arrow symbol in reversible reactios⇔
Explanation:
i hope this will help you
Part A The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. The potential energy for a certain mass moving in one dimension is given by U(x) = (2.0 J/m3)x3 - (15 J/m2)x2 + (36 J/m)x - 23 J. Find the location(s) where the force on the mass is zero. 3.0 m, 5.0 m 4.0 m, 5.0 m 2.0 m, 3.0 m 1.0 m
Answer:
The location are [tex]x_1 = 2 \ and \ x_2 = 3[/tex]
Explanation:
From the question we are told that
The potential energy is [tex]U(x) = (2.0 \ J/m^3) * x^3 - (15 \ J/m) * x^2 + (36 \ J/m) * x - 23 \ J[/tex]
The force on the mass can be mathematically evaluated as
[tex]F = - \frac{d U(x)}{d x } = -( 6 x^2 - 30x +36)[/tex]
The negative sign shows that the force is moving in the opposite direction of the potential energy
[tex]F = - 6 x^2 + 30x - 36[/tex]
At critical point
[tex]\frac{d U(x)}{dx} = 0[/tex]
So
[tex]- 6 x^2 + 30x - 36 = 0[/tex]
[tex]- x^2 + 5x - 6 = 0[/tex]
Using quadratic equation formula to solve this we have that
[tex]x_1 = 2 \ and \ x_2 = 3[/tex]
If you secure a refrigerator magnet about 2mmfrom the metallic surface of a refrigerator door and then move the magnet sideways, you can feel a resistive force, indicating the presence of eddy currents in the surface.
A)Estimate the magnetic field strength Bof the magnet to be 5 mTand assume the magnet is rectangular with dimensions 4 cmwide by 2 cmhigh, so its area A is 8 cm2. Now estimate the magnetic flux ΦB into the refrigerator door behind the magnet.
Express your answer with the appropriate units.
B)If you move the magnet sideways at a speed of 2 cm/s, what is a corresponding estimate of the time rate at which the magnetic flux through an area A fixed on the refrigerator is changing as the magnet passes over? Use this estimate to estimate the emf induced under the rectangle on the door's surface.
Express your answer with the appropriate units.
Answer:
(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v
Explanation:
Solution
Given that:
A refrigerator magnet about = 2 mm
The estimated magnetic field strength of the magnet is = 5 m T
The Area = 8 cm²
Now,
(A) The magnetic flux ΦB = BA
Thus,
ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²
So,
ΦB = 4* 6 ^ ⁻6 T m²
(B)By applying Faraday's Law we have the following formula given below:
Ε = Bℓυ
Here,
ℓ = 2 cm the same as 2 * 10 ^⁻2 m
B = 5 m T = 5 * 10 ^ ⁻3 T
υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s
Thus,
Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v
E =2 * 10 ^ ⁻6 v
A) The magnetic flux ΦB into the refrigerator door behind the magnet :
4 * 6⁻⁶ Tm²B) The estimated emf induced under the rectangle on the door's surface ;
2 * 10⁻⁶ vGiven data :
magnetic field strength of magnet ( B ) = 5 mT
size of refrigerator magnet = 2 mm
Area of magnet ( A ) = 4 * 2 = 8 cm²
A) Determine the magnetic flux ΦBwhere ; ΦB = BA
ΦB = ( 5 * 10⁻³ ) * ( 4 * 10⁻² ) * ( 2 * 10⁻² ) Tm²
= 4 * 6⁻⁶ Tm²
B) Determine estimated emf inducedTo determine the estimated emf we will apply Faraday's law
Ε = Bℓυ ---- ( 2 )
where : B = 5 * 10⁻³ T, ℓ = 2 * 10⁻² m, υ = 2 * 10⁻² m/s
insert values into equation 2
E = ( 5 * 10⁻³ ) * ( 2 * 10⁻² ) * ( 2 * 10⁻² )
= 2 * 10⁻⁶ v
Hence we can conclude that The magnetic flux ΦB is 4 * 6⁻⁶ Tm² and The estimated emf induced is 2 * 10⁻⁶ v
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Work out the velocity v at the end of a rollercoaster ride (0). (rearrange the equation for KE to make velocity v the subject)
KE=1/2mv^2
Explanation:
If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :
[tex]K=\dfrac{1}{2}mv^2[/tex]
m is mass of the object
We can rearrange the above equation such that,
[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]
Hence, this is the velocity at the end of a rollercoaster ride.
Water, in a 100-mm-diameter jet with speed of 30 m/s to the right, is deflected by a cone that moves to the left at 14 m/s. Determine (a) the thickness of the jet sheet at a radius of 230 mm. and (b) the external horizontal force needed to m
Answer:
Explanation:
The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]
Calculate the inlet and exit velocity of water jet
[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]
The conservation of mass equation of steady flow
[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]
[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]
[tex]A_e\ \texttt {is the exit area of the jet}[/tex]
since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal
The expression for thickness of the jet
[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]
R is the radius
t is the thickness of the jet
D_j is the diameter of the inlet jet
[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]
(b)
[tex]R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)][/tex]
[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]
[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]
The negative sign indicate that the direction of the force will be in opposite direction of our assumption
Therefore, the horizontal force is -7603N
Although these quantities vary from one type of cell to another, a cell can be 2.2 micrometers in diameter with a cell wall 40 nm thick. If the density (mass divided by volume) of the wall material is the same as that of pure water, what is the mass (in mg) of the cell wall, assuming the cell to be spherical and the wall to be a very thin spherical shell?
Answer:
m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg
Explanation:
First, we find the the surface area of the cell wall. Since, the cell is spherical in shape. Therefore, surface area of cell wall will be:
A = 4πr²
where,
A = Surface Area = ?
r = Radius of Cell = Diameter/2 = 2.2 μm/2 = 1.1 μm = 1.1 x 10⁻⁶ m
Therefore,
A = 4π(1.1 x 10⁻⁶ m)²
A = 15.2 x 10⁻¹² m²
Now, we find the volume of the cell wall. For that purpose, we use formula:
V = At
where,
V = Volume of the Cell Wall = ?
t = Thickness of Wall = 40 nm = 4 x 10⁻⁸ m
Therefore,
V = (15.2 x 10⁻¹² m²)(4 x 10⁻⁸ m)
V = 60.82 x 10⁻²⁰ m³
Now, to find mass of cell wall, we use formula:
ρ = m/V
m = ρV
where,
ρ = density of water = 1000 kg/m³
m = Mass of Wall = ?
Therefore,
m = (1000 kg/m³)(60.82 x 10⁻²⁰ m³)
m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg
The mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg
Since we assume the cell to be spherical and the wall to be a thin spherical shell, the volume of the cell wall V = At where
A = surface area of cell = 4πR² where R = radius of cell = 2.2 μm/2 = 1.1 × 10⁻⁶ m and t = thickness of cell wall = 40 nm = 40 × 10⁻⁹ m.Volume of cell wallSo, V = 4πR²t
Substituting the values of the variables into the equation, we have
V = 4πR²t
V = 4π(1.1 × 10⁻⁶ m)² × 40 × 10⁻⁹ m.
V = 4π(1.21 × 10⁻¹² m²) × 40 × 10⁻⁹ m.
V = 193.6π × 10⁻²¹ m³
V = 608.21 × 10⁻²¹ m³
V = 6.0821 × 10⁻¹⁹ m³
V ≅ 6.082 × 10⁻¹⁹ m³
Mass of the cell wallWe know that density of cell wall, ρ = m/v where m = mass of cell wall and V = volume of cell wall.
Making m subject of the formula, we have
m = ρV
Since we assume the density of the cell wall to be equal to that of pure water, ρ = 1000 kg/m³
So, m = ρV
m = 1000 kg/m³ × 6.082 × 10⁻¹⁹ m³
m = 6.082 × 10⁻¹⁶ kg
Converting to mg, we have
m = 6.082 × 10⁻¹⁶ kg × 10⁶ mg/kg
m = 6.082 × 10⁻¹⁰ mg
So, the mass of the cell wall in mg is 6.082 × 10⁻¹⁰ mg
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What caused the disappearance of land bridges?
A. Volcanic outgassing
B. Shrinking of the polar ice caps
C. Beginning of an ice age
D. A mass extinction
Answer: B
Explanation:
I would say the shrinking of the polar ice caps because in order for ice caps to shrink, they would have to obviously melt. This will cause the sea level and total volume of sea water to rise and cover up the land bridges
Answer:B :)
Explanation:
A block is supported on a compressed spring, which projects the block straight up in the air at velocity VVoj The spring and ledge it sits on then retract. You can win a prize by hitting the block with a ball. When should you throw the ball and in what direction to be sure the ball hits the block?
A. At the instant when the block is at the highest point, directed at the spring.
B. At the instant when the block is at the highest point, directed at the block.
C. At the instant when the block leaves the spring, directed at the spring.
D. At the instant when the block leaves the spring, directed at the block.
E. When the block is back at the spring's original position, directed at that position.
Answer:
the correct answer is B
Explanation:
We analyze this exercise a little, the block goes into the air and is under the acceleration of gravity. The ball is fired by the hand and is describing a parabolic movement, subjected to the acceleration of gravity.
For the ball to hit the block we must have the distance the ball goes up equal to the distance the block moves, therefore we must shoot the ball at the block at its highest point.
Let's write the kinematic equation for the two bodies
The block. At the highest point of the path
y = - ½ g t2
The ball, in its vertical movement
y = vo t - ½ g t2
therefore the correct answer is B
If the velocity of a runner changes from -2 m/s to -4 m/s over a period of time, the
runner's kinetic energy will become:
(a) four times as great as it was.
(b) half the magnitude it was.
(c) energy is conserved.
(d) twice as great as it was.
(e) four times less than it was.
Answer:
It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4
A dimension is a physical nature of a quantity.
(i) give two (2) limitations of dimensional analysis..
(ii) if velocity (v), time (T) and force (F) were chosen as basic quantities, find the dimensions of mass?
Answer:
i) A dimension is the physical nature of a quantity. The two limitations of dimensional analysis is as following:
Dimesnional analysis is unable to derive relation when a physical quantity depends on more than three factors with dimensions. It is unable to derive a formula that contain exponential function, trigonometric function, and logarithmic function.ii) Given:
Velocity = v
Time = t
Force = F
Force = mass x acceleration
= mass x velocity/time
So, mass= (force x time) / velocity
[mass] = Ftv^-1
Hence, dimesnion of mass is Ftv^-1.
When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing is 273 mph relative to the airplane. (a) What suction pressure is developed on the wing at that point? (b) What is the pressure at the leading edge (a stagnation point) of the wing?
Answer:
P1 = 0 gage
P2 = 87.9 lb/ft³
Explanation:
Given data
Airplane flying = 200 mph = 293.33 ft/s
altitude height = 5000-ft
air velocity relative to the airplane = 273 mph = 400.4 ft/s
Solution
we know density at height 5000-ft is 2.04 × [tex]10^{-3}[/tex] slug/ft³
so here P1 + [tex]\frac{\rho v1^2}{2}[/tex] = P2 + [tex]\frac{\rho v2^2}{2}[/tex]
and here
P1 = 0 gage
because P1 = atmospheric pressure
and so here put here value and we get
P1 + [tex]\frac{\rho v1^2}{2}[/tex] = P2 + [tex]\frac{\rho v2^2}{2}[/tex]
0 + [tex]\frac{2.048 \times 10^{-3} \times 293.33^2}{2}[/tex] [tex]= P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}[/tex]
solve it we get
P2 = 87.9 lb/ft³
1. In 214 BC, Archimedes invented a large spherical-type mirror used to focus the sun's intense rays onto far away enemy boats, which would eventually light them on fire. If the boats were travelling in a nearby channel approximately 1,000 m from the river bank, what would the radius of curvature of his mirror need to be? Show your work.
Answer:
2000 m
Explanation:
since the boat is 1000 m from the river bank, the beam must be focused at this point. This indicates that the focal length is 1000 m
for a spherical mirror, the focal length is given by
f = R/2
where R is the radius of curvature
1000 = R/2
R = 2000 m
R = 2000 m
this means that the radius of curvature must be 2000 m
An object will sink in a liquid if the density of the object is greater than that of the liquid. The mass of a sphere is 0.723 g. If the volume of this sphere is less than ________ cm3, then the sphere will sink in liquid mercury (density
Answer:
= 0.0532 cm^3
Explanation:
The computation of volume of the sphere is shown below:-
[tex]Density = \frac{Mass}{Volume}[/tex]
Where,
Density = 13.6 g/cm^3
Mass of sphere = 0.723 g
now we will put the values into the above formula to reach volume of the sphere which is here below:-
[tex]Volume = \frac{0.723}{13.6}[/tex]
= 0.0532 cm^3
Therefore for computing the volume of the sphere we simply applied the above formula.
The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track and the train reaches it 76s later. The acceleration is -0.26s^2. Find its speed at the next signal.
Answer:
3.2 m/s
Explanation:
Given:
Δx = 1000 m
v₀ = 23 m/s
a = -0.26 m/s²
t = 76 s
Find: v
This problem is over-defined. We only need 3 pieces of information, and we're given 4. There are several equations we can use. For example:
v = at + v₀
v = (-0.26 m/s²) (76 s) + (23 m/s)
v = 3.2 m/s
Or:
Δx = ½ (v + v₀) t
(1000 m) = ½ (v + 23 m/s) (76 s)
v = 3.3 m/s
Or:
v² = v₀² + 2aΔx
v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)
v = 3.0 m/s
Or:
Δx = vt − ½ at²
(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²
v = 3.3 m/s
As you can see, you get slightly different answers depending on which variables you use. Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.
A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 45 g, moving with speed v = 4.23 m/s, strikes the rod at angle θ = 46° from the normal at a distance D = 2/3 L, where L = 0.95 m, from the point of rotation and sticks to the rod after the collision.
Required:
What is the angular speed ωf of the system immediately after the collision, in terms of system parameters and I?
Answer:
Explanation:
angular momentum of the putty about the point of rotation
= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .
= .045 x 4.23 x 2/3 x .95 cos46
= .0837 units
moment of inertia of rod = ml² / 3 , m is mass of rod and l is length
= 2.95 x .95² / 3
I₁ = .8874 units
moment of inertia of rod + putty
I₁ + mr²
m is mass of putty and r is distance where it sticks
I₂ = .8874 + .045 x (2 x .95 / 3)²
I₂ = .905
Applying conservation of angular momentum
angular momentum of putty = final angular momentum of rod+ putty
.0837 = .905 ω
ω is final angular velocity of rod + putty
ω = .092 rad /s .
The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm1cm and the IMA of the machine is 66 , what is the radius of the handle
Question: The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm and the IMA of the machine is 6, what is the radius of the handle?
Answer:
Radius of the handle = 3 cm = 0.03 m
Explanation:
Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod).......(1)
Diameter of the rod = 1 cm
Radius of the rod = Diameter/2
Radius of the rod = 1/2
Radius of the rod = 0.5 cm
Mechanical Accuracy of the machine, MA = 6
Substitute the values into equation (1)
6 = (Radius of the handle)/0.5
Radius of the handle = 6 * 0.5
Radius of the handle = 3 cm
The Radius of the handle is = 3 cm = 0.03 m
Calculation of the radius of the handle:Since
Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod)
Here,
Diameter of the rod = 1 cm
We know that
The radius of the rod = Diameter/2
So,
Radius of the rod = 1/2
So,
Radius of the rod = 0.5 cm
Now
Mechanical Accuracy of the machine, MA = 6
Now
6 = (Radius of the handle)/0.5
Radius of the handle = 6 * 0.5
Radius of the handle = 3 cm
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A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R. Assume that the initial height h is great enough so that the car never losses contact with the track.
Required:
a. Find an expression for the kinetic energy of the car at the top of the loop. Express the kinetic energy in terms of m, g, h, and R.
b. Find the minimum initial height h at which the car can be released that still allows the car to stay in contact with the track at the top of the loop.
Answer:
Explanation:
At height h , potential energy of coaster car having mass m = mgh .
The car will lose potential energy and gain kinetic energy.
height lost by car when it is at the top of loop of radius R
= h - 2R
potential energy lost = mg ( h - 2R )
kinetic energy gained = mg ( h - 2R )
kinetic energy = 0 + mg ( h - 2R )
= mg ( h - 2R )
b )
For the car to remain in contact with the track , if v be the minimum velocity
centripetal force at top = mg
m v² / R = mg
v² = gR
kinetic energy = 1/2 mv²
= 1/2 m x gR
= mgR /
If h be the minimum height that can give this kinetic energy
mg ( h - 2R ) = mgR / 2
h - 2R = R / 2
h = 2.5 R .
The Great Lakes are all part of what? The Mississippi River The St. Lawrence Seaway A large body of salt lakes The Missouri River
Answer:
St Lawrence Sea way
Explanation:
The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.
1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to
estimate the length of time a fully charged battery could deliver any particular current
before discharge. Approximately how much energy is stored by a 50 ampere-minute 12
volt battery?
Answer:
Energy Stored = 36000 J = 36 KJ
Explanation:
The power of a battery is given by the formula:
P = IV
where,
P = Power delivered by the battery
I = Current Supplied to the battery
V = Potential Difference between terminals of battery = 12 volt
Now, we multiply both sides by the time period (t):
Pt = VIt
where,
Pt = (Power)(Time) = Energy Stored = E = ?
It = Battery Current Rating = 50 A.min
Converting this to A.sec;
It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
E = 36000 J = 36 KJ
uring a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? 26.7 N 16.7 N 13.3 N 107 N 40.0 N
Answer:
107 N, option d
Explanation:
Given that
mass of the ball, m = 0.2 kg
initial velocity of the ball, u = 20 m/s
final velocity of the ball, v = -12 m/s
time taken, Δt = 60 ms
Solving this question makes us remember "Impulse Theorem"
It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"
Mathematically, it is represented as
FΔt = m(v - u), where
F = the average force
Δt = time taken
m = mass of the ball
v = final velocity of the ball
u = initial velocity of the ball
From the question we were given, if we substitute the values in it, we have
F = ?
Δt = 60 ms = 0.06s
m = 0.2 kg
v = -12 m/s
u = 20 m/s
F = 0.2(-12 - 20) / 0.06
F = (0.2 * -32) / 0.06
F = -6.4 / 0.06
F = -106.7 N
Thus, the magnitude is 107 N
How many significant figures does 0.09164500561 have?
Answer:
10 Sig Figs
Explanation:
Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.
pls what is the difference between Ac power and dc power
Answer:
The difference between AC and DC lies in the direction in which the electrons flow. In DC, the electrons flow steadily in a single direction, or "forward." In AC, electrons keep switching directions, sometimes going "forward" and then going "backward."
N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2 and third one is C/4, forth one is C/8 and so on. Namely, capacitance of a capacitor is one-half of the previous one. What is the equivalent capacitance of this parallel combination when N goes to inifinity?
Answer:
2C
Explanation:
The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.
So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.
Using the formula for the sum of the infinite terms of a geometric series, we have:
Sum = First term / (1 - rate)
Sum = C / (1 - 0.5)
Sum = C / 0.5 = 2C
So the equivalent capacitance of this parallel connection is 2C.
A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 110 kg and radius 4.0 m. Calculate the magnitude of the total angular momentum of the woman–disk system. (Assume that you can treat the woman as a point.)
Answer:
The angular momentum is [tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
The mass of the woman is [tex]m = 50 \ kg[/tex]
The angular speed of the rim is [tex]w = 0.80 \ rev/s = 0.8 * [\frac{2 \pi}{1} ] = 5.024 \ rad \cdot s^{-1}[/tex]
The mass of the disk is [tex]m_d = 110 \ kg[/tex]
The radius of the disk is [tex]r_d = 4.0 \ m[/tex]
The moment of inertia of the disk is mathematically represented as
[tex]I_D = \frac{1}{2} m_d r^2_d[/tex]
substituting values
[tex]I_D = \frac{1}{2} * 110 * 4^2[/tex]
[tex]I_D = 880 \ kg \cdot m^2[/tex]
The moment of inertia of the woman is
[tex]I_w = m * r_d^2[/tex]
substituting values
[tex]I_w = 50 * 4^2[/tex]
[tex]I_w =800\ kg[/tex]
The moment of inertia of the system (the woman + the large disk ) is
[tex]I_t = I_w + I_D[/tex]
substituting values
[tex]I_t = 880 +800[/tex]
[tex]I_t =1680 \ kg \cdot m^2[/tex]
The angular momentum of the system is
[tex]L = I_t w[/tex]
substituting values
[tex]L = 1680 * 5.024[/tex]
[tex]L = 8440.32 \ kg \cdot m^2 \cdot s^{-1}[/tex]
A train starts from rest and accelerates uniformly, until it has traveled 5.6 km and acquired a velocity of 42 m/s. The train then moves at a constant velocity of 42 m/s for 420 s. The train then slows down uniformly at 0.065 m/s^2, until it is brought to a halt. What is the acceleration during the first 5.6 km of travel?
Answer:
0.1575 m/s^2
Explanation:
Solution:-
- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).
- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.
- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).
- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:
[tex]v_f^2 = v_i^2 + 2*a*( s_f - s_o )[/tex]
- We will plug in the given parameters in the equation of motion given above:
[tex]42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}[/tex]
Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2
You drive in straight line at 20 m/s for 10 miles, then at 30m/s for an other 10 miles what is your average speed
Answer:
25 m/s
Explanation:
Data provided in the question
20 m/s for 10 minutes
And, the 30 m/s for another 10 minutes
Based on the above information, the average speed is
As we know that
[tex]Average\ speed = \frac{Total\ distance}{Total\ time}[/tex]
[tex]= \frac{20\times10\times60 + 30\times10\times60 }{20\times60}[/tex]
= 25 m/s
1 hour = 60 minutes
1 minute = 60 seconds
Hence, the average speed is 25 m/s
In the question, there are miles is given but instead of this we use the minutes as we have to find out the average speed and time should not be in miles it should be in minutes, hour or seconds
Therefore we considered the same
A freight car moves along a frictionless level railroad track at constant speed. The freight car is open on top. A large load of coal is suddenly dumped into the car. What happens to the speed of the freight car
Answer:
The speed of the freight car decreases.
Explanation:
According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received
In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.
Therefore the speed of the freight car decreased by applying the law of conservation of momentum i
A point charge is located at the center of a thin spherical conducting shell of inner and outer radii r1 and r2, respectively. A second thin spherical conducting shell of inner and outer radii R1 and R2, respectively, is concentric with the first shell. The flux is as follows for the different regions of this arrangement. Ф -10.3 103 N-m2/C for
0 for r<2 4:
-36.8 x 10נ N-m2/c
0 for r > R2
36.8 x 10נ N-m2/c
Determine the magnitude ond sign of the point chorge ond the charge on the surface of the two shels point charge inner shell outer shel.
Answer:
the magnitude is 7 and sign of the point charge on the surface shell is -13
Explanation:
One end of an insulated metal rod is maintained at 100c and the other end is maintained at 0.00 c by an ice–water mixture. The rod has a length of 75.0cm and a cross-sectional area of 1.25cm . The heat conducted by the rod melts a mass of 6.15g of ice in a time of 10.0 min .find the thermal conductivity k of the metal?k=............ W/(m.K)
Answer:
The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].
Explanation:
This is a situation of one-dimensional thermal conduction of a metal rod in a temperature gradient. The heat transfer rate through the metal rod is calculated by this expression:
[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]
Where:
[tex]\dot Q[/tex] - Heat transfer due to conduction, measured in watts.
[tex]L_{rod}[/tex] - Length of the metal rod, measured in meters.
[tex]A_{c,rod}[/tex] - Cross section area of the metal rod, measured in meters.
[tex]k_{rod}[/tex] - Thermal conductivity, measured in [tex]\frac{W}{m\cdot K}[/tex].
Let assume that heat conducted to melt some ice was transfered at constant rate, so that definition of power can be translated as:
[tex]\dot Q = \frac{Q}{\Delta t}[/tex]
Where Q is the latent heat required to melt the ice, whose formula is:
[tex]Q = m_{ice}\cdot L_{f}[/tex]
Where:
[tex]m_{ice}[/tex] - Mass of ice, measured in kilograms.
[tex]L_{f}[/tex] - Latent heat of fussion, measured in joules per gram.
The latent heat of fussion of water is equal to [tex]330000\,\frac{J}{g}[/tex]. Hence, the total heat received by the ice is:
[tex]Q = (6.15\,g)\cdot \left(330\,\frac{J}{g} \right)[/tex]
[tex]Q = 2029.5\,J[/tex]
Now, the heat transfer rate is:
[tex]\dot Q = \frac{2029.5\,J}{(10\,min)\cdot \left(60\,\frac{s}{min} \right)}[/tex]
[tex]\dot Q = 3.382\,W[/tex]
Turning to the thermal conduction equation, thermal conductivity is cleared and computed after replacing remaining variables: ([tex]L_{rod} = 0.75\,m[/tex], [tex]A_{c,rod} = 1.25\times 10^{-4}\,m^{2}[/tex], [tex]\Delta T = 100\,K[/tex], [tex]\dot Q = 3.382\,W[/tex])
[tex]\dot Q = \frac{k_{rod}\cdot A_{c, rod}}{L_{rod}}\cdot \Delta T[/tex]
[tex]k_{rod} = \frac{\dot Q \cdot L_{rod}}{A_{c,rod}\cdot \Delta T}[/tex]
[tex]k_{rod} = \frac{(3.382\,W)\cdot (0.75\,m)}{(1.25\times 10^{-4}\,m^{2})\cdot (100\,K)}[/tex]
[tex]k_{rod} = 202.92\,\frac{W}{m\cdot K}[/tex]
The thermal conductivity of the insulated metal rod is [tex]202.92\,\frac{W}{m\cdot K}[/tex].
