A mass (m = 30 g) falls onto a spring (k = 7.3 N/m) from a height (h = 25 cm). The spring compresses an additional amount x before temporarily coming to a stop. What is the value of x?

Answer:

45°

Explanation:

Range is the horizontal distance travelled by an object undergoing projectile motion.

Range is given by the formula:

[tex]R=\frac{u^2sin(2\theta)}{g}[/tex]

where u = velocity, g = acceleration due to gravity and θ = angle above the horizontal.

For angle of 15°:

[tex]R=\frac{u^2sin(2*15)}{g}=\frac{0.5u^2}{g}[/tex]

We get a maximum range when sin(2θ) = 1

sin(2θ) = 1

2θ = sin⁻¹(1)

2θ = 90⁰

θ = 90°/2

θ = 45⁰

For angle of 45°:

[tex]R=\frac{u^2sin(2*45)}{g}=\frac{u^2}{g}[/tex]

Answer: the work done by the force is 0

Explanation:

F (x², xy)

121 = 11²

so R = x² + y² = 11²

p = x². Q = xy

Δp/Δy = 0, ΔQ/Δx

using Green's theorem

woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA

=  (x² + y² = 121)_∫∫ yΔA

now let x = rcosФ, y = rsinФ

ΔA = rΔrΔФ

so r from 0 to 11

and Ф from 0 to 2π

= 0_∫^2π   0_∫^11  rsinФ × rΔrΔФ

= 0_∫^2π SinФΔФ   0_∫^11  r²Δr

= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0

therefore the work done by the force is 0

Object will start moving in the direction where comparatively high force  is directing to move

When the resultant force acting on a body is not equal to zero , the force acting on the body are known as unbalance forces . The body acted upon by unbalance forces changes its state of motion.

According to Newton if a body is experiencing net force =0 (which means that all horizontal and vertical forces are cancelling each other ), then the body will acquire an equilibrium state and will not move because forces are balancing each other and making net force on the body zero .

But unbalance force can tend the body to move in that direction in which a comparatively high force is acting due to which cancellation will not occur ( equilibrium state will not occur )

hence correct option A)Object moves

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Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

[tex]M_{before} = M_{after}[/tex]

where:

M = momentum [kg*m/s]

M = m*v

where:

m = mass [kg]

v = velocity [m/s]

[tex](m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})[/tex]

where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

v2 = velocity of the tennis ball before the collision = - 30 [m/s]

v3 =  velocity of the basketball after the collision [m/s]

v4 = velocity of the tennis ball after the collision = 34 [m/s]

Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

1 - (0.05*34) = 0.5*v3

- 0.7 = 0.5*v

v = - 1.4 [m/s]

The negative sign means that the movement is towards left

Answer:

A. 1.4 m/s to the left

Explanation:

Answer:

T = D/S

Explanation:

Given the formula for calculating the average speed of an object as:

average speed = Distance/Time

Let average speed = S

Distance = D

Time = T

Substitute

S = D/T

Make T the subject of the formula:

From S = D/T

Cross multiply

ST = D

Divide both sides by S

ST/S = D/S

T = D/S

Hence the expression for he time T after rearrangement is T = D/S

Answer:

The Navy Seal is accelerating downwards at the rate of 3.3 m/s²

Explanation:

Given;

mass of the Navy Seal, m = 80 kg

the upward resistive force on her, F = 520 N

Her net downward force is given by;

[tex]F_{net} = F_{down} - F_{up}\\\\F_{net} = (80*9.8) - 520\\\\F_{net} = 264 \ N[/tex]

Her downward acceleration at this time is given by;

F = ma

a = F / m

a = 264 / 80

a = 3.3 m/s²

Therefore, the Navy Seal is accelerating downwards at the rate of 3.3 m/s²

Explanation:

distance from ground

mass

amount of compression

Answer:

Package 1 will land at 228.0 m, package 2 will land at 308.0 m, and the distance between them is 80.0 m.

Explanation:

To find the distance at which the first package will land we need to calculate the time:

[tex] Y_{f} = Y_{0} + V_{0y}t - \frac{1}{2}gt^{2} [/tex]

Where:

Y(f) is the final position = 0

Y(0) is the initial position = 160 m

V(0y) is initial speed in "y" direction = 0

g is the gravity = 9.81 m/s²

t is the time=?                                          

[tex] 0 = 160 m + 0t - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]

[tex] t = \sqrt{\frac{2*160 m}{9.81 m/s^{2}}} = 5.7 s [/tex]

Now we can find the distance of the first package:

[tex] X_{1} = V_{0x}*t = 40.0 m/s*5.7 s = 228.0 m [/tex]

Then, after 2 seconds the distance traveled by plane is (from the initial position):

[tex] X_{p} = V_{0x}*t = 40.0 m/s*2 s = 80.0 m [/tex]

Now, the distance of the second package is:

[tex] X _{2} = X_{1} + X_{p} = 228.0 m + 80.0 m = 308.0 m [/tex]

The distance between the packages is:

[tex] X = X_{2} - X_{1} = 308.0 - 228.0 m = 80.0 m [/tex]

Therefore, package 1 will land at 228.0 m, package 2 will land at 308.0 m and the distance between them is 80.0 m.

I hope it helps you!

Answer:

significant

Explanation:

The digits in a measurement that are considered significant are all of those digits that represent marked calibrations on the measuring device plus one additional digit to represent the estimated digit (tenths of the smallest calibration).

Answer:

34k+B plus 9 :)

Explanation:

Answer:

16

Explanation:

6+10=16

the box will go forward but it will be a little harder.

Explanation:

Given two capacitors 7.0 µF and 5.0 µF connected in series, let s calculate the effective capacitance.

[tex]\frac{1}{C_T} = \frac{1}{7.0} +\frac{1}{5.0}\\ \frac{1}{C_T} = \frac{5+7}{35} \\C_T = \frac{35}{12}\\ C_T = 2.9166\mu F[/tex]

Given

V = 5.0kV = 5000Volts

To get the total charge in the circuit, we will use the formula:

[tex]Q = C_TV\\Q = 2.9166\times 10^{-6} \times 5000\\Q = 0.014583 C\\Q = 1.4583 \times 10^{-2}C[/tex]

Since the same charge flows in a series connected capacitors, the charge on them in mC is 14.583mC

For the voltage across 7.0 µF:

[tex]V = \frac{Q}{C}\\V = \frac{0.014583}{0.000007} \\V = 2,083.28V[/tex]

Voltage across the 7.0 µF is 2,083.28V

The voltage across the  5.0 µF:

V = 5000 - 2,083.28

V = 2,916.72Volts

Voltage across the  5.0 µF is 2,916.72V

Answer:

The charge on the first capacitor is 14.59 mC

The voltage across the first capacitor is 2084 V

The charge on the second capacitor is 14.59 mC

The voltage across the second capacitor is 2918 V

Explanation:

Given;

first capacitor, C₁ = 7.0 µF

second capacitor, C₂ = 5.0 µF

potential difference, V = 5000 V

The equivalent capacitance is given by;

[tex]\frac{1}{C_{eq}}= \frac{1}{C_1} +\frac{1}{C_2}\\\\\frac{1}{C_{eq}}= \frac{C_1 + C_2}{C_1C_2}\\\\C_{eq} = \frac{C_1C_2}{C_1 +C_2}\\\\ C_{eq} = \frac{(7*10^{-6})(5*10^{-6})}{(7*10^{-6} \ +\ 5*10^{-6})}\\\\C_{eq} = 2.917*10^{-6} \ F[/tex]

The charge on the first capacitor is given by;

[tex]Q_1 = C_{eq} V\\\\Q_1 = 2.917*10^{-6} *5000\\\\Q_1 = 0.01459 \ C\\\\Q_1 = 14.59 \ mC[/tex]

The voltage across the first capacitor is given by;

[tex]V_1 = \frac{Q_1}{C_1} \\\\V_1 = \frac{0.01459}{7*10^{-6}}\\\\ V_1 = 2084.286 \ V[/tex]

V₁ = 2084 V

The charge on the second capacitor is given by;

[tex]Q_2 = C_{eq} V\\\\Q_2 = 2.917*10^{-6} *5000\\\\Q_2 = 0.01459 \ C\\\\Q_2 = 14.59 \ mC[/tex]

The voltage across the second capacitor is given by;

[tex]V_2 = \frac{Q_2}{C_2}\\\\ V_2 = \frac{0.01459}{5*10^{-6}}\\\\V_2 = 2918 \ V[/tex]

Answer:

The wavelength is [tex]\lambda  =  1.2  * 10^8 nm[/tex]

Explanation:

From the question we are told that

   The frequency of operation of the microwave is  [tex]f =  2.50 GHz  =  2.50 *10^{9} \ Hz[/tex]

     Generally the wavelength is mathematically represented as

          [tex]\lambda  =  \frac{c}{f}[/tex]

Here c is the speed of light with value [tex]c =  3.0 *10^{8} \  m/s[/tex]

So  

         [tex]\lambda  =  \frac{3.0 *10^{8}}{  2.50 *10^{9}}[/tex]

=>       [tex]\lambda  =  0.12 \  m [/tex]

converting to nanometer

           [tex]\lambda  =  1.2  * 10^8 nm[/tex]

Answer:

Explanation:

Given data

mass m= 1200kg

radius r=92m

coefficient of friction μ=0.65

We know that

F=mv2/r

F=μmg  

mv2/r = μsmg  

v^2/r = μsg  

vmax = √(rμg)

Substituting our data we have

vmax=√(92*0.65*9.81)

vmax=√586.638

vmax=24.22m/s

Answer:

liquids

Explanation

Answer:

2

Explanation:

Answer:

The speed is     v = 25 - 9.8 t

the velocity is   v = (25 - 9.8 t) j^

the position is   y = 25 t - 4.9 t²

Explanation:

This is a vertical throwing exercise.

The first thing we must do is set a coordinate system, in this case we will make the upward direction positive.

Let's write the kinematics equations

        v = v₀ - g t

        v² = v₀² - g (y-y₀)

        y = y₀ + v₀ t - ½ g t²

in our case the initial velocity is

         v₀ = 25 m / s

and we zero the system at the launch point

        y₀ = 0

the equations remain

      v = 25 - 9.8 t

        v² = 25² - 9.8 y

        y = 0 + 25 t - ½ 9.8 t²

The speed is     v = 25 - 9.8 t

the velocity is   v = (25 - 9.8 t) j^

the position is   y = 25 t - 4.9 t²

Explanation:

hope this helps, cheers!

The values for the average speed and average velocity are;

(a) [tex]Average \ speed = 1.37\overline{142857} \ m/s[/tex]

(b) [tex]The \ average \ velocity = 0.\overline{857142} \ m/s[/tex]

The reason the above values are correct is given as follows;

The given parameters are;

Distance of the nearest grocery store to the house, d = 60 m

Direction of the grocery store = East from the house

Walking speed, s = 1.2 m/s

Time of walking before the rain started, t₁ = 15.0 s

Time it takes to go back to the house, t₂ = 5.0 s

Speed at which the start walking again, s = 1.2 m/s

(a) The formula for average speed, [tex]s_{avg}[/tex], is given as follows;

[tex]s_{avg} = \dfrac{Total \ distance }{Sum \ of \ time \ taken}[/tex]

The distance walked before the rain, d₁ = 1.2 m/s × 15.0 s = 18 meters

Distance covered on returning to the house = d₂ = d₁ = 18 meters

The total distance covered, ∑Distance = d₁ + d₂ + 60

∴ ∑Distance = 18 m + 18 m + 60 m = 96 m

The time, t₃, it takes to get to the grocery store on the second attempt after getting the umbrella is given as follows;

[tex]t_3 = \dfrac{d}{v}[/tex]

[tex]t_3 = \dfrac{60 \ m}{1.2 \ m/s} = 50.0 \, s[/tex]

The total time, ∑Time = t₁ + t₂ + t₃

∴ ∑Time = 15.0 s + 5.0 s + 50.0 s = 70.0 s

[tex]Average \ speed = \dfrac{\sum Distance}{\sum Time}[/tex]

[tex]Average \ speed = \dfrac{96 \, m}{70 \, s} = \dfrac{48}{35} \ m/s= 1.37\overline{142857} \ m/s[/tex]

[tex]Average \ speed = 1.37\overline{142857} \ m/s[/tex]

(b) The average velocity, [tex]v_{avg}[/tex], is given as follows;

[tex]v_{avg}= \dfrac{Total \ displacement}{Total \ time}[/tex]

The total displacement = The total change in location = The change in location from the house to the grocery =  60 m

The total time is the same as for the average speed, ∑Time = 70 s

[tex]\therefore v_{avg}= \dfrac{60 \, m}{70 \, s} = \dfrac{6}{7} \ m/s = 0.\overline{857142} \ m/s[/tex]

[tex]The \ average \ velocity = 0.\overline{857142} \ m/s[/tex]

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Analysing the question:

Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s

We are given:

height of the tower (h) = 66 m

mass of the stone (m) = 0.5 kg

initial velocity of the stone (u) = 0 m/s

time taken by the stone to reach the ground (t) = t seconds

acceleration due to gravity = 10 m/s²

** Neglecting air resistance**

Finding the time taken by the stone to reach the ground:

from the second equation of motion

h = ut + 1/2at²

replacing the variables

66 = (0)(t) + 1/2 (10)(t)²

66 = 5t²

t² = 13.2

t = 3.6 seconds

I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds

but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved

Answer:

-9m/s²

Explanation:

Given parameters:

Initial velocity = 45m/s

Final velocity  = 0

duration  = 5s

Unknown:

acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time;

        Acceleration  = [tex]\frac{v- u}{t}[/tex]

v is the final velocity

u is the initial velocity

t is the time taken

 Input the parameters and solve;

     Acceleration  = [tex]\frac{0 - 45}{5}[/tex]   = -9m/s²

The car accelerates at a rate of -9m/s² which is a deceleration

Answer:

I believe the answer is Property taxes

Explanation:

Answer: I'm pretty sure property taxes

Explanation:

Answer:

slower speeds = larger and faster steering wheel movements

faster speeds = small and slow steering wheel movements

Explanation:

When driving at slower speeds you need to use larger and faster steering wheel movements. This is because at slow speeds the car does not have enough momentum to make certain maneuvers with small steering wheel movements in a given amount of time, therefore making large and faster steering wheel movements gives the car enough time with the momentum it has to make the desired maneuver. At faster speeds only small and slow steering wheel movements are needed and while cause the car to quickly change to the desired direction due to the increased momentum of the car.

Answer:True

Explanation:Matter is everything around you. Atoms and compounds are all made of very small parts of matter. Those atoms go on to build the things you see and touch every day. Matter is defined as anything that has mass and takes up space (it has volume).

Answer:

This question is incomplete

Explanation:

This question is incomplete but there are some characteristics that are peculiar to an object traveling in uniform circular motion.

An object moving in a uniform circular motion generally has a constant speed with which it uses to move in a circle (the object moves round/tangent to the circle). This then means the object will move in different direction although with the same speed. This change in direction means the object will accelerate (inwards) at different velocities (since velocity is a vector quantity that measures both magnitude and direction). Because the object moves with different velocities; this makes the object an accelerating object.

From the descriptions above, it can be conceived/visualized that the net force acting on an object in a uniform circular motion is a centripetal force. This is because the net force acting on the object is directed towards the center of the circle the object in rotating/moving in. Without this net force, the object would have moved in a straight line and thus not changing it's direction. The formula used to calculate this centripetal force is

Fc = mv²/r

where Fc is the centripetal force

m is the mass of the object

v is the velocity of the object

r is the radius of the curvature/curved path

The net force on an object moving in a circular path is directed inwards and it is known as centripetal force. The centripetal force increases with increase in speed and acceleration of the object.

The acceleration of an object travelling in a circular path is directed inwards and the magnitude of the acceleration is given as;

[tex]a_c = \frac{v^2}{r} = \omega^2 r[/tex]

where;

The net force acting on the object is given as follows;

[tex]F_c = ma_c[/tex]

The net force on the object is directed inwards and it is known as centripetal force.

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Answer:

I = 0.96 A

Explanation:

No of electrons, [tex]n=1.8\times 10^{16}[/tex]

Time, t = 3 ms = [tex]3\times 10^{-3}\ s[/tex]

We need to find the electric current. We know that electric charge per unit time is equal to the electric current.

[tex]I=\dfrac{q}{t}[/tex]

q = ne (Quantization of electric charge)

[tex]I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A[/tex]

So, the electric current is 0.96 A.

Analysing the Question:

We know that equilibrium is the state of a body when it has equal and opposite forces being applied on it

In this case, a net downward force of 496N is being applied and a net upward force of (106 + 106 + 142 + x) N

Finding the missing force:

Since we have to achieve equilibrium, the net upward forces have to be equal to the net downward forces

So,  (106 + 106 + 142 + x) = 496

354 + x = 496

x = 496 - 354

x = 142 N

Therefore, the missing force is 142 N

Answer:

new height of ramp must be 4 times that of the first ramp.

Explanation:

From conservation of energy, we know that;

Potential energy at the top of ramp = kinetic energy at the bottom of ramp.

Thus;

mgh_t = ½mv²

m will cancel out to give;

gh_t = ½v²

Thus means that the height of the first ramp is directly proportional to the square of the speed.

Thus;

h_t ∝ v²

Now, for the new ramp, we are told that we want to achieve a speed of 2v at the bottom.

Thus;

h'_t ∝ (2v)²

h'_t ∝ 4v²

From earlier we saw that;

h_t ∝ v²

Thus;

New height of ramp is;

h'_t ∝ 4h_t

Thus, new height of ramp must be 4 times that of the first ramp.

To increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.

The given parameters;

Apply the principle of conservation of energy;

P.E = K.E

[tex]mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\h = \frac{v^2}{2g}[/tex]

To increase the speed 2v at the bottom of the ramp, the height will be;

[tex]H = \frac{(2v)^2}{2g} \\\\H = \frac{4v^2}{2g} \\\\H = 4(\frac{v^2}{2g} )\\\\H = 4(h)[/tex]

Thus, to increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.

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