PHYSICS COLLEGE

A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?

Answers

Answer 1

Answer:

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

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The main force(s) acting on the puck after receiving the kick is (are):_________.A) a downward force of gravity and an upward force exerted by the surfaceB) a downward force of gravity, and a horizontal force in the direction of motionC) a downward force of gravity, an upward force exerted by the surface, and a horizontal force in the direction of motionD) a downward force of gravityA) a downward force of gravity and an upward force exerted by the surface
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Which of the following is not a motivation behind taking performance enhancing drugs?A.Having bigger, stronger muscles B.improving recovery time fir injuries C. Pumping

Answers

Answer:

C. Pumping

Explanation:

???

Answer: Losing weight due to decreased appetite

Explanation: because you guys trust me

Singing that is off-pitch by more than about 1% sounds bad. How fast would a singer have to be moving relative to the rest of a band to make this much of a change in pitch due to the Doppler effect

Answers

Answer:

-3.396 m/s or 3.465 m/s

Explanation:

v = Speed of sound in air = 343 m/s

$v_s$ = Relative speed of the singer

f = Observed frequency

f' = Actual frequency

1% change can mean $f=1.01f'$

From the Doppler effect equation we have

$f=f'(v)/(v+v_s)\n\Rightarrow 1.01f'=f'(v)/(v+v_s)\n\Rightarrow 1.01=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(1.01)-343\n\Rightarrow v_s=-3.396\ m/s$

The velocity is -3.396 m/s

when $f=0.99f'$

$f=f'(v)/(v+v_s)\n\Rightarrow 0.99f'=f'(v)/(v+v_s)\n\Rightarrow 0.99=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(0.99)-343\n\Rightarrow v_s=3.46464646465\ m/s$

The velocity is 3.465 m/s

Two capacitors give an equivalent capacitance of 9.42 pF when connected in parallel and an equivalent capacitance of 1.68 pF when connected in series. What is the capacitance of each capacitor?

Answers

Answer:

$C_1=7.23pF\ and\ C_2=2.19pF$

Explanation:

Let the two capacitance are $C_1\ and\ C_2$

It is given that when capacitors are connected in parallel their equvilaent capacitance is 9.42 pF

So $C_1+ C_2=9.2$ --------EQN 1

And when they are connected in series their equivalent capacitance is 1.68 pF

So $(1)/(C_1)+(1)/(C_2)=(1)/(1.68)$

$(C_1+C_2)/(C_1C_2)=(1)/(1.68)$

$C_1C_2=1.68* 9.42=15.8256pF$

$C_1-C_2=√((C_1+C_2)^2-4C_1C_2)=√(9.42^2-4* 15.8256)=5.0432pF$ -----EQN

On solving eqn 1 and eqn 2

$C_1=7.23pF\ and\ C_2=2.19pF$

Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.

Answers

Answer:

Point motion will eventually stops due to action of g exactly perpendicular...

Explanation:

If ignoring the air resistance, the magnitude of gravitational acceleration is already strong enough to stops the acceleration. As we know that, the spring constant of a bungee spring cord will be F = -k/x, where x is the stretched length and k is the spring constant of bungee cord. If F = ma = w = mg, the g = -m k/x. Now we can clearly see that the value of g remains constant due to the fluctuating length of the cord as the motion progresses back and forth in SHM say from x1 to x2 and x2 to x1.

Please helpa car is driven 200 km west and then 80 km southwest. what is the displacement of the car from the point of orgin (magnitude and direction)? draw a diagram.

Answers

Let's take east and west to be positive and negative, respectively, and north and south to be positive and negative, respectively. Then in terms of vectors (using ijk notation), the car first moves 200 km west,

r = (-200 km) j

then 80 km southwest,

s = (-80/√2 km) i + (-80/√2 km) j

so that its total displacement is

r + s = (-80/√2 km) i + ((-200 - 80/√2) km) j

r + s ≈ (-56.6 km) i + (-256.6 km) j

This vector has magnitude

√((-56.6 km)² + (-256.6 km)²) ≈ 262.7 km

and direction θ such that

tan(θ) = (-256.6 km) / (-56.6 km) ==> θ ≈ -102.4º

relative to east, or about 12.4º west of south.

The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which data set represents constant acceleration?

Answers

Answer:

The first graph is showing the constant acceleration (1 m/s)

Explanation:

The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4

The last graph is showing constant velocity therefore there is no acceleration (a = 0)

Answer:

Explanation:

on edge

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