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Slav-nsk [51]
3 years ago
15

A box is given a push so that it slides across the floor. how far will it go given that the How far will it go, given that the c

oefficient of kinetic friction is 0.13 and the push imparts an initial speed of 3.5 m/s
Physics
1 answer:
melisa1 [442]3 years ago
6 0

Answer:

4.808m

Explanation:

Force is given as:

F = ma

Since the force is a frictional force,

Fr = F

Fr = ma

Frictional force, Fr, is given as:

-umg = ma

a = -ug = -0.13 * 9.8 = -1.274 m/s²

Using one of the equations of motion, we can find the distance:

v² = u² + 2as

Since the box comes to a halt, v = 0m/s:

0 = u² + 2as

s = -u²/2a

s = -(-3.5²)/(2 * 1.274) = 4.808m

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93.5 if it’s wrong sorry sis I need my homework done too
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Russell bradley carried 207 kg of bricks 3.65 m up a ladder. If the amount of work required to perform that task is used to comp
MrRa [10]

Answer:

Explanation:

Work done in carrying bricks

mgh

= 207 x 9.8 x 3.65

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Work done in compressing gas

PΔV

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1.8 x 10⁶ ΔV = 7404.4

ΔV  = 7404.4  / 1.8 x 10⁶m³

= 4113.33 x 10⁻⁶ m³

= 4113.33 cc

8 0
3 years ago
Read 2 more answers
1. An object (m = 500 g) with an initial speed of 0.2 m/s collides with another object (m = 1.5 kg) which was at rest before the
Anettt [7]

Answer:

Explanation:

1 )

We shall apply conservation of momentum law to solve the problem.

mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .

.5 x .2 = (1.5 + .5)V

V = .05 m /s

2 ) We shall use formula for velocity of object after elastic collision as follows

v₁ = \frac{(m_1-m_2)}{(m_1+m_2)} u_1+\frac{2m_2u_2}{(m_1+m_2)}

m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.

Putting the values

= \frac{(200-1000)}{(1000+200)} \times 1 +\frac{2 \times1000\times0}{(1000+200)}

= - .66 m /s

Since the sign is negative so it will be in opposite direction .

4 0
3 years ago
What is the heat capacity of an object at 25.5∘C that absorbs 45 kJ of heat and is heated to 28.2∘C?
sergey [27]

Answer:

16.6 kJ/°C

Explanation:

given,

Amount of heat absorbed = 45 kJ

initial temperature, T₁ = 25.5°C

final temperature, T₂ = 28.2°C

change in temperature = T₂ - T₁

                                       = 28.2 - 25.5  = 2.7° C

Heat\ capacity = \dfrac{Heat\ absorbed}{\Delta T}

Heat\ capacity = \dfrac{45\ kJ}{2.7}

Heat\ capacity = 16.6\ kJ/^0C

Heat capacity of the object is equal to 16.6 kJ/°C

4 0
3 years ago
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A fair ground ride spins its occupants inside a flying saucer-shaped container. if the horizontal circular path the riders follo
Licemer1 [7]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below is the solution:

 <span>centripetal accel = 1.5*g 
ω²r = 1.5*9.8m/s² 
ω² * 8m = 14.7 m/s² 
ω = 1.36 rad/s * 1rev/2πrads * 60s/min = 12.9 rpm</span>

6 0
3 years ago
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