Answer:
Explanation:
Distance between fringe or fringe width = xλ / d
where x is location of screen and d is slit separation
Given x = 4 m
λ = 694 nm
d = .085 x 10⁻³ m
distance between fringes
= 4 x 694 x 10⁻⁹ / .085 x 10⁻³
= 4 x 694 x 10⁻⁹ / 85 x 10⁻⁶
= 32.66 x 10⁻³ m
= 32.66 mm .
3.267 cm
b )
when submerged in water , wavelength in water becomes as follows
wavelength in water = wave length / refractive index
= 694 / 1.333 nm
= 520.63 nm
new distance between fringes
3.267 / 1.333
= 2.45 cm .
A solid exerts a force of 500 N. Calculate the pressure exerted to the surface where area
of contact is 2000 cm2.
Answer:
2500 N/m²
Explanation:
Pressure: This can be defined as the force acting normally on a surface per unit area.
The expression for pressure is give as
P = F/A...................... Equation 1
Where P = pressure (N/m²), F = force (N), A = Contact area (m²)
Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.
Substitute into equation 1
P = 500/0.2
P = 2500 N/m²
Hence the pressure exerted to the surface is 2500 N/m²
When solving vector addition problems you can use either the graphical
method or the
Answer :the resultant of two vectors can be found using either the parallelogram method or the triangle method. don't know if this was helpful ?
Explanation:
Answer:
Analytical method.
Mark Watney (Matt Damon in the Martian movie) and Marvin the Martian (Looney Tunes cartoon character) are having an argument on the surface of Mars (negligible air resistance). They are testing out their new potato launcher that fires projectiles at a constant speed. Mark launches his potato at an angle of 60◦ and Marvin launches his identical potato at an angle of 30◦ . Without any calculations try to answer the following questions, and justify each answer.
(A) Which potato lands farther away from the launcher (potatoes are launched from ground level)?
(B) Which potato spends more time in the air before hitting the ground
(C) Which potato has a greater speed just before it hits the ground?
Answer:
A) The two potatoes cover the same horizontal distance from the launcher.
B) Mark's potato spends more time in the air than Marvin's potato before hitting the ground.
C) Marvin's potato hits the ground with a greater speed than Mark's potato
Explanation:
A) For projectile motion, the final horizontal distance of the projectile from where it was initially launched (its range) is given as
R = (u² sin 2θ)/g
where
u = initial velocity of the projectile
θ = angle above the horizontal at which the projectile was launched = 30°, 60°
g = acceleration due to gravity on Mars
Since, u and g are the same for Mark and Marvin, sin 2θ would determine which range is higher.
Sin (2×60°) = sin 120°
Sin (2×30°) = sin 60°
Sin 120° = Sin 60°
Hence, the two potatoes cover the same horizontal distance from the launcher.
B) Time spent in the air for a projectile is given as
T = (2u sin θ)/g
Again, since u and g are the same for Mark and Marvin on Mars, sin θ will give the required idea of whose potato spends more time in the air.
Sin 60° = 0.866
Sin 30° = 0.50
Sin 60° > Sin 30°
Hence, Mark's potato spends more time in the air than Marvin's potato.
C) The horizontal velocity for projectile motion is constant all through the motion and is equal to u cos θ
u cos 60° < u cos 30°
And the initial vertical velocity is u sin θ
Final vertical velocity
= (initial vertical velocity) - gt
g = acceleration due to gravity on Mars
T = time of flight
For Mark,
initial vertical velocity = u sin 60°, greater than Marvin's u sin 30°
And Mark's potato's time of flight is greater as established in (B) above.
But for Marvin
initial vertical velocity = u sin 30°, less than Mark's u sin 60°
And Marvin's potato's time of flight is lesser as established in (B) above
So, at the end of the day, the final vertical velocity is almost the same for both Mark's and Marvin's potatoes.
Hence, the horizontal component of the final velocity edges the final speed of the potatoes just before hitting the ground in Marvin's favour.
Hope this Helps!!!
Proposed Exercise - Mass Center of a Composite Body Determine the coordinates (x, y) of the center of mass of the body illustrated in the picture below
Answer:
x = 3.76 cm
y = 3.76 cm
Explanation:
This composite shape can be modeled as a square (7.2 cm × 7.2 cm) minus a quarter circle in the lower left corner (3.6 cm radius) and a right triangle in the upper right corner (3.6 cm × 3.6 cm).
The centroid of a square (or any rectangle) is at x = b/2 and y = h/2.
The centroid of a quarter circle is at x = y = 4r/(3π).
The centroid of a right triangle is at x = b/3 and y = h/3.
Build a table listing each shape, the coordinates of its centroid (x and y), and its area (A). Use negative areas for the shapes that are being subtracted.
Next, multiply each coordinate by the area (Ax and Ay), sum the results (∑Ax and ∑Ay), then divide by the total area (∑Ax / ∑A and ∑Ay / ∑A). The result will be the x and y coordinates of the center of mass.
See attached image.
EASY! WILL REWARD BRAINLIEST!
Electrical current is defined as _____.
the capacity to store charge
the flow of electric charge per unit time
the amount of stored electric energy
the voltage of the battery
Electrical current is defined as the flow of electric charge per unit time.
You are on a train traveling east at speed of 19 m/s with respect to the ground. 1)If you walk east toward the front of the train, with a speed of 1.5 m/s with respect to the train, what is your velocity with respect to the ground
Answer:
Vbg = 20.5 m/s
your velocity with respect to the ground Vbg = 20.5 m/s
Explanation:
Relative velocity with respect to the ground is;
Vbg = velocity of train with respect to the ground + your velocity with respect to the train
Vbg = Vtg + Vbt ......1
Given;
velocity of train with respect to the ground;
Vtg = 19 m/s
your velocity with respect to the train;
Vbt = 1.5 m/s
Substituting the given values into the equation 1;
Vbg = 19 m/s + 1.5 m/s
Vbg = 20.5 m/s
your velocity with respect to the ground Vbg = 20.5 m/s
A ball is projected upward at time t= 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to:____________A. 9.0 sB. 9.4 sC. 9.7 sD. 8.7 sE. 10 s
Answer:
B. 9.4 s
Explanation:
In order to calculate the total time taken by the ball to hit the ground, we first analyze the upward motion. We will use subscript 1 for upward motion. Now, using 1st equation of motion:
Vf₁ = Vi₁ + gt₁
where,
Vf, = Final Velocity in upward motion = 0 m/s (ball stops at highest point)
Vi = Initial Velocity in upward motion = 36.2 m/s
g = - 9.8 m/s² (negative due to upward motion)
t₁ = Time taken in upward motion = ?
Therefore,
0 m/s = 36.2 m/s + (-9.8 m/s²)(t₁)
t₁ = (36.2 m/s)/(9.8 m/s²)
t₁ = 3.7 s
Now, using 2nd equation of motion:
h₁ = (Vi₁)(t₁) + (0.5)(g)(t₁)²
where,
h₁ = distance from top of building to highest point ball reaches = ?
Therefore,
h₁ = (36.2 m/s)(3.7 s) + (0.5)(-9.8 m/s²)(3.7 s)²
h₁ = 133.58 - 66.86 m
h₁ = 66.72 m
No, considering downward motion and using subscript 2, for it.
Using 2nd equation of motion:
h₂ = (Vi₂)(t₂) + (0.5)(g)(t₂)²
where,
h₂ = height of the highest point from ground = h₁ + height of building
h₂ = 66.72 m + 90 m = 156.72 m
Vi₂ = Initial Speed during downward motion = 0 m/s (ball stops for a moment at highest point)
t₂ = Time Taken in downward motion = ?
g = 9.8 m/s²
Therefore,
156.72 m = (0 m/s)(t₂) + (0.5)(9.8 m/s²)(t₂)²
t₂² = (156.72 m)/(4.9 m/s²)
t₂ = √31.98 s²
t₂ = 5.7 s
Now, the total time taken by ball to reach the ground is"
Total Time = T = t₁ + t₂
T = 3.7 s + 5.7 s
T = 9.4 s
Therefore, the correct answer is:
B. 9.4 s
If a metal rod is moved through magnetic field, the charged particles will feel a force, and if there is a complete circuit, a current will flow. We talk about the induced emf of the rod. The rod essentially acts like a battery, and the induced emf is the voltage of the battery. A magnetic field with a strength of 0.732 T is pointing into the page and a metal rod L=0.362 m in length is moved to the right at a speed v of 15.1m/s.
Required:
a. What is the induced emf in the rod?
b. Suppose the rod is sliding on conducting rails, and a complete circuit is formed. If the load resistance is 5.74Ω , what is the magnitude and direction (clockwise or counterclockwise) of the current flowing in the circuit?
Answer:
a. 4 V
b. 0.697 A
Explanation:
Magnetic field strength B = 0.732 T
length of rod l = 0.362 m
velocity of rod v = 15.1 m/s
a. EMF can be calculated as
E = Blv = 0.732 x 0.362 x 15.1 = 4 V
b. If the rod is connected to a conducting rail, with resistance R = 5.74Ω
current I = V/R = 4/5.74 = 0.697 A
the current flows in a clockwise direction
A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal
length lens. Calculate the size of the image formed on flim
Answer:
18.89cm
Explanation:
As we know that the person is standing 5m in front of the camera
[tex]d_0=5m=500cm[/tex]
The focal length of the lens =50cm
f=50 cm
By Lens formula we have:
[tex]\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm[/tex]
By the formula of magnification
[tex]\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm[/tex]
The height of the image formed is 18.89cm.
A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.90 m/s2 . At 20.0s after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.
(A) How high above the launch pad will the rocket eventually go?
(B) Find the rocket's velocity at its highest point.
(C) Find the magnitude of the rocket's acceleration at its highest point.
(D) Find the direction of the rocket's acceleration at its highest point.
(E) How long after it was launched will the rocket fall back to the launch pad?
(F) How fast will it be moving when it does so?
Answer:
A) 580m
B) 0 m/s
C) 9.8m/s^2
D) downward
E) 10.87s
F) 106.62 m/s
Explanation:
A) The distance traveled by the rocket is calculated by using the following expression:
[tex]y=\frac{1}{2}at^2[/tex]
a: acceleration of the rocket = 2.90 m/s^2
t: time of the flight = 20.0 s
[tex]y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m[/tex]
B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.
C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2
D) The acceleration points downward
E) The time the rocket takes to return to the ground is given by:
[tex]t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s[/tex]
10.87 seconds
F) The velocity just before the rocket arrives to the ground is:
[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}[/tex]
How can I show that the sphere of radius R performs a simple harmonic movement. how can i set its reference point and make the free body diagram.
I have the torque sum equation which is equal to the moment of inertia by angular acceleration
Explanation:
Draw a free body diagram of the pendulum (the combination of the sphere and the massless rod). There are three forces on the pendulum:
Weight force mg at the center of the sphere,
Reaction force in the x direction at the pivot,
Reaction force in the y direction at the pivot.
Sum the torques about the pivot O.
∑τ = I d²θ/dt²
mg (L sin θ) = I d²θ/dt²
For small θ, sin θ ≈ θ.
mg L θ = I d²θ/dt²
Since d²θ/dt² is directly proportional to θ, this fits the definition of simple harmonic motion.
If you wish, you can use parallel axis theorem to find the moment of inertia about O:
I = Icm + md²
I = ⅖ mr² + mL²
mg L θ = (⅖ mr² + mL²) d²θ/dt²
gL θ = (⅖ r² + L²) d²θ/dt²
A low C (f = 65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and its mass density is 5.0 g/m2, determine the tension of the wire.
Answer:
Tension of the wire(T) = 169 N
Explanation:
Given:
f = 65Hz
Length of the piano wire (L) = 2 m
Mass density = 5.0 g/m² = 0.005 kg/m²
Find:
Tension of the wire(T)
Computation:
f = v / λ
65 = v / 2L
65 = v /(2)(2)
v = 260 m/s
T = v² (m/l)
T = (260)²(0.005/2)
T = 169 N
Tension of the wire(T) = 169 N
A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its normal)?
A. 26
B. 35
C. 42
D. 22
E. 48
Answer:
35Explanation:
According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.
sini/sinr = n
n is the constant = refractive index
Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng
nw is the refractive index of water and ng is that of glass
sini/sinr = nw/ng
given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass
On substitution;
sin 30/sinr = 1.33/1.5
1.5sin30 = 1.33sinr
sinr = 1.5sin30/1.33
sinr = 0.75/1.33
sinr = 0.5639
r = arcsin0.5639
r ≈35°
angle the light leave the glass is 35°
To practice Problem-Solving Strategy 6.1: Circular motion A highway curve with radius R = 274 m is to be banked so that a car traveling v = 25.0 m/s will not skid sideways even in the absence of friction. At what angle should the curve be banked?
Answer:
The curve should be banked at an angle of 13 degrees.
Explanation:
We have,
Radius of a highway curve is 274 m
Speed of car on this curve is 25 m/s
Let [tex]\theta[/tex] is the banking angle. On a banked curve, the angle of safe diving is given by following expression.
[tex]\tan\theta=\dfrac{v^2}{Rg}[/tex]
g = 10 m/s²
Plugging all the values in above formula,
[tex]\tan\theta=\dfrac{(25)^2}{274\times 9.8}\\\\\theta=\tan^{-1}\left(\dfrac{(25)^{2}}{274\times9.8}\right)\\\\\theta=13^{\circ}[/tex]
So, the curve should be banked at an angle of 13 degrees.
Now consider a different electromagnetic wave, also described by: Ex(z,t) = Eocos(kz - ω t + φ) In this equation, k = 2π/λ is the wavenumber and ω = 2π f is the angular frequency. In this case, though, assume φ = +30o and Eo = 1 kV/m. What is the value of Ex(z,t) when z/λ = 0.25 and ft = 0.125?
Answer:
Explanation:
Ex(z,t) = Eocos(kz - ω t + φ)
k = 2π/λ , ω = 2π f
φ = +30° , E₀ = 10³ V .
z/λ = 0.25 , ft = 0.125
Ex(z,t) = Eocos(2πz/λ - 2πf t + φ)
Putting the values given above
Ex(z,t) = 10³ cos ( 2π / 4 - 2π x .125 + 30⁰ )
= 1000cos (90⁰ - 45+30)
= 1000 cos 75
=258.8 V .
What is the velocity of a car that travels 556km northwest in 3.2 hours
Answer:
173.75 km/hr in the NW direction.
Explanation:
Velocity is the time rate of change in displacement of a body. Mathematically:
v = d / t
where d = displacement
t = time
Therefore, the velocity of the car is:
v = 556 / 3.2 = 173.75 km/hr
The velocity of the car is 173.75 km/hr in the NW direction.
The velocity of a car will be "173.75 km/hr".
Displacement and Velocity,The velocity of something like a car moving northward on something like a prominent motorway as well as the velocity of something like a rocket launching towards spacecraft both might be determined or monitored.
Displacement, d = 556 km
Time, t = 3.2 hours
We know the relation,
→ Velocity = [tex]\frac{Displacement}{Time}[/tex]
or,
→ V = [tex]\frac{d}{t}[/tex]
By substituting the values, we get
= [tex]\frac{556}{3.2}[/tex]
= [tex]173.75[/tex] km/hr
Thus the response above is right.
Find out more information about velocity here:
https://brainly.com/question/6504879
The brakes of a car are applied to give it an acceleration of -2m/s^2. The car comes to a stop in 3s. What was its speed when the brakes were applied?
Answer:
So if its acceleration is -2m/s^2 that means every second the initial velocity would be subtracted by 2. So since it took 3 seconds 2*3=6. The initial velocity was 6 m/s
g A proton is held at rest in a uniform electric field. When it is released, the proton will lose... electrical potential energy. kinetic energy. both kinetic energy and electric potential energy. neither kinetic energy or electric potential energy.
Answer:
It will lose electrical potential energy.
Explanation:
A photon held at rest in a uniform electrical field will lose electrical potential energy when it is released this is because the electrical potential energy is the energy posses by the photon at rest or by virtue of the position is converted to kinetic energy which is energy posses by a body in motion.
Since the photon is released and set in motion , it now has kinetic energy and has lost the potential energy because it is set in motion.
The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magnitude of the magnetic field at twice the distance from the conductor
Answer:
B/4
Explanation:
The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:
The field at twice the distance is B/4.
Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?
Answer:
1.25kg
Explanation:
Simply multiply volume and density together
When the play button is pressed, a CD accelerates uniformly from rest to 430 rev/min in 4.0 revolutions. If the CD has a radius of 7.0 cm and a mass of 17 g , what is the torque exerted on it?
Answer:
The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].
Explanation:
As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot \Delta n[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]\ddot n[/tex] - Angular acceleration, measured in revolution per square minute.
[tex]\Delta n[/tex] - Change in angular position, measured in revolutions.
The angular acceleration is cleared and calculated:
[tex]\ddot n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \Delta n}[/tex]
Given that [tex]\dot n_{o} = 0\,\frac{rev}{min}[/tex], [tex]\dot n = 430\,\frac{rev}{min}[/tex] and [tex]\Delta n = 4\, rev[/tex], the angular acceleration is:
[tex]\ddot n = \frac{\left(430\,\frac{rev}{min} \right)^{2}-\left(0\,\frac{rev}{min} \right)^{2}}{2\cdot (4\,rev)}[/tex]
[tex]\ddot n = 23112.5\,\frac{rev}{min^{2}}[/tex]
The angular accelaration measured in radians per square second is:
[tex]\alpha = \left(23112.5\,\frac{rev}{min^{2}} \right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}} \right)[/tex]
[tex]\alpha \approx 40.339\,\frac{rad}{s^{2}}[/tex]
Net torque experimented by the CD during its accleration is equal to the product of its moment of inertia with respect to its axis of rotation and angular acceleration:
[tex]\tau = I \cdot \alpha[/tex]
Where:
[tex]I[/tex] - Moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
In addition, a CD has a form of a uniform disk, whose moment of inertia is:
[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]
Where:
[tex]m[/tex] - Mass of the CD, measured in kilograms.
[tex]r[/tex] - Radius of the CD, measured in meters.
If [tex]m = 0.017\,kg[/tex] and [tex]r = 0.07\,m[/tex], then:
[tex]I = \frac{1}{2}\cdot (0.017\,kg)\cdot (0.07\,m)^{2}[/tex]
[tex]I = 4.165\times 10^{-5}\,kg\cdot m^{2}[/tex]
Now, the net torque exerted on CD is:
[tex]\tau = (4.165\times 10^{-5}\,kg\cdot m^{2})\cdot \left(40.339\,\frac{rad}{s^{2}} \right)[/tex]
[tex]\tau = 1.680\times 10^{-3}\,N\cdot m[/tex]
The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].
Refer to a situation where you exert a force F on a crate of mass M, moving it at a speed v a distance d across a floor in a time interval t. The quantity F d/t is?
a.) kinetic energy of the crate
b.) potential energy of the crate
c.) linear momentum of the crate
d.) work you do on the crate
e.) power you supply to the crate
Answer:
e.) power you supply to the crate
Explanation:
According to given data, we have:
F = Force exerted on the crate
M = Mass of the crate
v = Speed of motion of the crate
d = Distance traveled by the crate across the floor
t = Time interval passed
Now, we try to analyze the given quantity:
=> F d/t
=> (Force)(Displacement)/(Time)
but, (Force)(Displacement) = Work Done
Therefore,
=> Work Done/Time
but, Work Done/Time = Power
Therefore,
=> Power
Hence, the quantity F d/t is:
e.) power you supply to the crate
Use the position function s(t) = -16t + v_0t + s_0 for free falling objects. A ball is thrown straight down from the top of a 600-foot building with an initial velocity of -30 feet per second. (a) Determine the position and velocity functions for the ball. (b) Determine the average velocity on the interval [1, 3]. (c) Find the instantaneous velocities when t=1 and t=3. (d) Find the time required for the ball to reach ground level. (e) Find the velocity of the ball at impact.
Answer:
a) v = -30 - 32 t , s (t) = 600 - 30 t -16 t² , b) v = -32 ft / s
c) v (1) = -62 ft / s, v (3) = -126 ft / s , d) t = 7.13 s , e) v = -258.16 ft / s
Explanation:
a) For this exercise they give us the function of the position of the ball
s (t) = s (o) + v_o t - 16 t²
notice that you forgot to write the super index
indicate the initial position of the ball
s (o) = 600 ft
also indicates initial speed
v_o = - 30 ft / s
let's substitute in the equation
s (t) = 600 - 30 t -16 t²
to find the speed we use
v = ds / dt
v = v_o - 32 t
v = -30 - 32 t
b) To find the average speed, look for the speed at the beginning and end of the time interval
t = 1 s
v (1) = -30 -32 1
v (1) = - 62 ft / s
t = 3 s
v (3) = -30 -32 3
v (3) = -126 ft / s
the average speed is
v = (v (3) -v (1)) / (3-1)
v = (-126 +62) / 2
v = -32 ft / s
c) instantaneous speeds, we already calculated them
v (1) = -62 ft / s
v (3) = -126 ft / s
d) the time to reach the ground
in this case s = 0
0 = 600 - 30 t -16 t²
t² + 1,875 t - 37.5 = 0
we solve the quadratic equation
t = [-1,875 ±√ (1,875² + 4 37.5)] / 2
t = [1,875 ± 12.39] / 2
t₁ = 7.13 s
t₂ = negative
Since the time must be positive, the correct answer is t = 7.13 s
e) the speed of the ball on reaching the ground
v = -30 - 32 t
v = -30 - 32 7.13
v = -258.16 ft / s
What’s the answer to this question?
Answer:
6 A
Explanation:
Parallel connected resistors needs to be calculated as one single resistor. To do that: [tex]\frac{1}{15}[/tex]+[tex]\frac{1}{15}[/tex]+[tex]\frac{1}{15}[/tex]=[tex]\frac{3}{15}[/tex]=[tex]R^{-1}[/tex]
[tex]\frac{3}{15} ^{-1}[/tex]= 5 Ω (total resistance)
U = R* I
[tex]\frac{U}{R}[/tex]=I
[tex]\frac{30}{5}[/tex]=6 A
Complete the first and second sentences, choosing the correct answer from the given ones.
1. A temperature of 100 K corresponds on a Celsius scale to 100 ° C / 0 ° C / 173 ° C / -173 ° C.
2. At 50 ° C, it corresponds to a Kelvin scale of 150 K / 323 K / 273 K / 223 K.
Explanation:
Complete the first and second sentences, choosing the correct answer from the given ones.
1. T = 100 K
[tex]^{\circ}C=K-273[/tex]
Put T = 100 K
[tex]T=100-273=-173^{\circ} C[/tex]
A temperature of 100 K corresponds on a Celsius scale to (-173 °C)
2. T = 50 °C
[tex]K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K[/tex]
So, At 50 °C, it corresponds to a Kelvin scale of 323 K.
someone please help me out thanks
Answer:
The answer is D)Theory
Explanation:
This is due because a theory is a scientific term and is a testable model that scientists seek to explain a phenomenon. You can also find out the answer by the process of elimination it can't be data because that would be something they already know and something they use to prove not explain. It can't be law because it isn't testable but can be used to explain. So that leaves you with two answers hypothesis and theory which are very similar but it isn't hypothesis because it isn't used to explain it to help the scientists come up with a theory and accumulate what might happen.
Two spectators at a soccer game see, and a moment later hear, the ball being kicked on the playing field. The time delay for the spectator A is 0.55 s, and for the spectator B it is 0.45 s. Sight lines from the two spectators to the player kicking the ball meet at an angle of 90°. The speed of sound in the air is 343 m/s.
How far are (a) spectator A and (b) spectator B from the player?
(c) How far are the spectators from each other?
Answer:
a)188.65m
b)154.35m
c)243.7m
Explanation:
Given data:
[tex]t_A=0.55s[/tex]
[tex]t_B=0.45s[/tex]
(a) The distance from the kicker to each of the 2 spectators is given by:
[tex]d_A=v \times t_A[/tex]
where,
v= speed of sound
[tex]t_A[/tex]=time taken for the sound waves to reach the ears
[tex]d_A=343\times 0.55=188.65[/tex]m
(b)[tex]d_B=v \times t_B[/tex]
where,
v= speed of sound
[tex]t_B[/tex]=time taken for the sound waves to reach the ears
[tex]d_B=343\times 0.45=154.35m[/tex]
(c)As the angle b/w slight lines from the two spectators to the player is right angle,
hypotenuse=the distance b/w 2 spectators
and, the slight lines are the other 2 lines
[tex]D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m[/tex]
If the radius of curvature of the cornea is 0.75 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina
Answer:
The distance from the cornea vertex to the retina is 2.36 cm
Explanation:
The question is incomplete.
The complete question is as follows;
A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.
If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?
Solution.
We use image-object reaction to calculate the distance from the cornea vertex to the retina.
Mathematically;
n1/s + n2/s’ = n2-n1/R
From the question, we identify the following;
n1 ; Refractive index of air = 1
n2 ; Refractive index of lens = 1.4
S ; Object Distance = 36 cm
S’ = ?
R ; Radius of curvature of the cornea = 0.65
Substituting these values into the equation above;
1/36 + 1.4/S’ = (1.4-1)/0.65
{S’+ 36(1.4)}/36S’ = 0.4/0.65
{S’ + 50.4}/36S’ = 0.62
S’ + 50.4 = 22.32S’
50.4 = 22.32S’ -S’
21.32S’ = 50.4
S’ = 50.4/21.32
S’ = 2.36 cm
A 888 kg car is driven clockwise around a flat circular track of radius 59 m. The speed of the car is a constant 7 m/s. Which factor, when doubled, would produce the greatest change in the centripetal force acting on the car? A. Radius of the track B. Weight of the car C. Mass of the car D. Velocity of the car
Answer:
D. Velocity of the car
Explanation:
The centripetal force acting on the car is given by the following formula:
[tex]F_c=ma_c=m\frac{v^2}{r}[/tex] (1)
m: mass of the car = 888 kg
v: tangential speed of the car = 7 m/s
r: radius of the flat circular track = 59 m
By the form of the equation (1) you can notice that the greatest change in the centripetal force is obtained when the velocity v is twice. In fact, you have:
[tex]F_c=m\frac{(2v)^2}{r}=4m\frac{v^2}{r}=4F_c[/tex]
Then, the greatest values of the centripetal force is:
[tex]F_c=4(888kg)\frac{(7m/s)^2}{59m}=2949.96N[/tex]
The greatest change in Fc is obtained by changing the value of the speed
answer
D. Velocity of the car
A rod of mass M = 154 g and length L = 35 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 11 g, moving with speed V = 9 m/s, strikes the rod at angle θ = 29° a distance D = L/3 from the end and sticks to the rod after the collision.Calculate the rotational kinetic energy, in joules, of the system after the collision.
Answer:
Explanation:
moment of inertia of the rod = 1/3 mL² , m is mass and L is length of rod.
1/3 x .154 x .35²
= .00629
moment of inertia of putty about the axis of rotation
= m d² , m is mass of putty and d is distance fro axis
= .011 x( .35 / 3 )²
= .00015
Total moment of inertia I = .00644 kgm²
angular momentum of putty about the axis of rotation
= mvRsinθ
m is mass , v is velocity , R is distance where it strikes the rod and θ is angle with the rod at which putty strikes
= .011 x 9 x .35 / 3 x sin 29
= .0056
Applying conservation of angular momentum
angular momentum of putty = angular momentum of system after of collision
.0056 = .00644 ω where ω is angular velocity of the rod after collision
ω = .87 rad /s .
Rotational energy
= 1/2 I ω²
I is total moment of inertia
= .5 x .00644 x .87²
= 2.44 x 10⁻³ J .
