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2 years ago

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Physics

1 answer:

goblinko [34]2 years ago

6 0

Answer:

Explanation:

initial velocity u = 32.7 m /s

final velocity v = 50.3 m /s

displacement s = 44500 m

acceleration a = ?

v² = u² + 2 a s

50.3² = 32.7² + 2 x a x 44500

2530.09 = 1069.29 + 89000a

a .016 m /s²

time taken t = ?

v = u + at

50.3 = 32.7 + .016 t

t = 1100 s

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A 0.25 kg beach ball rolling at a speed of 7 m/s collides with a heavy exercise ball at rest. The beach ball bounces straight ba

svetlana [45]

Answer:

A 0.25 kg beach ball rolling at a speed of 7 m/s collides with a heavy exercise ball at rest. The beach ball bounces straight back with a speed of 4 m/s. That is the change in momentum of the beach ball? What is the impulse exerted on the beach ball? What is the impulse exerted on the exercise ball?

Explanation:

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3 years ago

the roque requried to turn the crank on an ice cream maker is 4.50 N.m how much work does it take to turn the crank through 300

Alexus [3.1K]

Answer:

the work required to turn the crank at the given revolutions is 8,483.4 J

Explanation:

Given;

torque required to turn the crank, T = 4.50 N.m

number of revolutions, = 300 turns

The work required to turn the crank is given as;

W = 2πT

W = 2 x 3.142 x 4.5

W = 28.278 J

1 revolution = 28.278 J

300 revlotions = ?

= 300 x 28.278 J

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Therefore, the work required to turn the crank at the given revolutions is 8,483.4 J

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2 years ago

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3 years ago

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A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

$v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}$

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

$v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}$

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0

3 years ago

A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget

mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, $m_{1}$ = 4 kg

Speed, $v_{1}$ = 5 m/s

$m_{2}$ = 1 kg

$v_{2}$ = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

$\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2$

By plugging in the values we get,

$KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\$

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

$KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)$

$KE = 40J + KE(lost)$

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

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3 years ago