PHYSICS COLLEGE

A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The block rests on a frictionless surface. A 5.30×10?2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.Part AHow far does the putty-block system compress the spring?

Answers

Answer 1

Answer:

The distance the putty-block system compress the spring is 0.15 meter.

Given the following data:

Mass = 0.454 kg

Spring constant = 21.0 N/m.

Mass of putty = $5.30* 10^(-2)\;kg$

Speed = 8.97 m/s

To determine how far (distance) the putty-block system compress the spring:

First of all, we would solver for the initialmomentum of the putty.

$P_p = mass * velocity\n\nP_p = 5.30* 10^(-2)* 8.97\n\nP_p = 47.54 * 10^(-2) \;kgm/s$

Next, we would apply the law of conservation of momentum to find the final velocity of the putty-block system:

$P_p = (M_b + M_p)V\n\n47.54* 10^(-2) = (0.454 + 5.30* 10^(-2))V\n\n47.54* 10^(-2) = 0.507V\n\nV = (0.4754)/(0.507)$

Velocity, V = 0.94 m/s

To find the compression distance, we would apply the law of conservation of energy:

$U_E = K_E\n\n(1)/(2) kx^2 = (1)/(2) mv^2\n\nkx^2 =M_(bp)v^2\n\nx^2 = (M_(bp)v^2)/(k) \n\nx^2 = ((0.454 + 5.30* 10^(-2)) * 0.94^2)/(21)\n\nx^2 = ((0.507 * 0.8836))/(21)\n\nx^2 = ((0.4480))/(21)\n\nx=√(0.0213)$

x = 0.15 meter

Answer 2

Answer:

Answer:

Explanation:

Force constant of spring K = 21 N /m

we shall find the common velocity of putty-block system from law of conservation of momentum .

Initial momentum of putty

= 5.3 x 10⁻² x 8.97

= 47.54 x 10⁻² kg m/s

If common velocity after collision be V

47.54 x 10⁻² = ( 5.3x 10⁻² + .454) x V

V = .937 m/s

If x be compression on hitting the putty

1/2 k x² = 1/2 m V²

21 x² = ( 5.3x 10⁻² + .454) x .937²

x² = .0212

x = .1456 m

14.56 cm

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Answers

Answer:

* roller skates and ice skates.

* roller coaster

Explanation:

One of the best examples for this situation is when we are skating, in the initial part we must create work with a force, it compensates to move, after this the external force stops working and we continue movements with kinetic energy, if there are some ramps, we can going up, where the kinetic energy is transformed into potential energy and when going down again it is transformed into kinetic energy. This is true for both roller skates and ice skates.

Another example is the roller coaster, in this case the motor creates work to increase the energy of the car by raising it, when it reaches the top the motor is disconnected, and all the movement is carried out with changes in kinetic and potential energy. In the upper part the energy is almost all potential, it only has the kinetic energy necessary to continue the movement and in the lower part it is all kinetic; At the end of the tour, the brakes are applied that bring about the non-conservative forces that decrease the mechanical energy, transforming it into heat.

A 79-turn, 16.035-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 43 degrees away from vertical increases from 0.997 T to 6.683 T in 56.691 s. Determine the emf induced in the coil.

Answers

Answer:

The induced emf is 0.0888 V.

Explanation:

Given that,

Number of turns = 79

Diameter = 16.035 cm

Angle = 43

Change in magnetic field $\Delta B=(6.683-0.997)= 5.686\ T$

Time = 56.691 s

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=(NA\Delta B\cos\theta)/(\Delta T)$

Where, N = number of turns

A = area

B = magnetic field

Put the value into the formula

$\epsilon=(79*\pi*(8.0175*10^(-2))^2*5.686*\cos43)/(56.691)$

$\epsilon =0.0888\ V$

Hence, The induced emf is 0.0888 V.

An ordinary drinking glass is filled to the brim with water (268.4 mL) at 2.0 ° C and placed on the sunny pool deck for a swimmer to enjoy. If the temperature of the water rises to 32.0 ° C before the swimmer reaches for the glass, how much water will have spilled over the top of the glass? Assume the glass does not expand.

Answers

Answer:

$\Delta V=1.667*10^(-3)$

Explanation:

Given the initial temperature T_i=2° C

final temperature T_f= 32° C

The original volume of water Vo=268.8 mL= 0.2688 L

we need to calculate the change in the volume

As we know that volume expansion is given by

$(\Delta V)/(V_0)= \beta\Delta T$

ΔV= change in Volume

β= expansion coefficient = $207*10^(-6) K^(-1)$

therefore,

$\Delta V= \beta\Delta T V_0$

plugging values we get

$\Delta V=207*10^(-6) K^(-1) (32-2)*0.2688$

$\Delta V=1.667*10^(-3)$

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Answers

Answer:

300 clicks...

Explanation:

Output on 3 clicks = 10 dB

Increasing 10 by a factor of 100 equals 1000 dB so,

Its simple math, clicks will also increase in the same ratio and it shall take 300 clicks to increase the volume by a factor of 100.

Suppose a fast-pitch softball player does a windmill pitch, moving her hand through a circular arc with her arm straight. She releases the ball at a speed of 25.5 m/s (about 57.0 mph ). Just before the ball leaves her hand, the ball's radial acceleration is 1060 m/s2 . What is the length of her arm from the pivot point at her shoulder

Answers

Answer:

61.3 cm

Explanation:

Radial acceleration of the object in circular motion is given by formula

$a = (v^2)/(R)\n$

Given:

$a = 1060 m/s^2\nv = 25.5 m/s$

Plugging in the values in the formula

$1060 = (25.5^2)/(R)\nR = 0.613 m$

so length of his arm is 61.3 cm

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