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spin [16.1K]
3 years ago
13

A 50.0kg box is being pushed along a horizontal surface by a force of 250N directed 28 degrees below the horizontal. The coeffic

ient of kinetic friction between the box and the surface is 0.300. What is the acceleration of the box?
A) 0.769 m/s^2
B) 1.77m/s^2
C) 3.16m/s^2
D) 6.31m/s^2
E) 8.53m/s^2
Physics
1 answer:
tester [92]3 years ago
5 0
The answer is c. that was a tough question
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A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli
Lunna [17]

Answer:

a.  v=3.11mls, 29.4^{0}

b.   K.E =-697.8J

Explanation:

To calculate the values in the  question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\

Momentum of player 2 p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\

Hence the total momentum p_{12}=p_{1}+p_{2}\\

Note, since the direction of movement before collision is due south and  due north respectively we have to represent the velocity using the rectangular coordinate

Hence  p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\

(95+90)v=475i+270j\\

v=2.57i+1.45j\\

solving for the resultant velocity, we have

v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls

To calculate the direction of movement, we have

\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\  \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\

And the final kinetic energy after collision is

K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\  K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J

The decrease in Kinetic energy is

K.E =K.E_{final}- K.E_{initial}=894.7-1592.5

K.E =-697.8J

The negative sign indicate a decrease in Kinetic energy

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3 years ago
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Explanation:

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As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o
astraxan [27]

Answer:

3.6 x 10⁶ Pa

Explanation:

A = Area of the heel = 1.50 cm² = 1.50 x 10⁻⁴ m²

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g = acceleration due to gravity = 9.8 m/s²

Force of gravity on the heel is given as

F = mg

Inserting the values

F = (55) (9.8)

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Pressure exerted on the floor is given as

P = \frac{F}{A}

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6 0
3 years ago
As Clinton walks he pushes his shoe against the track.
Valentin [98]

Answer:A:The track pushes back on Clinton's shoe with the same force.

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When a certain amount of current flows through a resistor, it uses 3.00 W of power. If the current doubles, how much power will
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Explanation: Acellus

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