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2 years ago

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A fan cart with the fan set to high rolled across the floor. The cart's speeds with the fan on high are shown below. If the fan

were set to medium instead, what might the cart's speed be at 5 seconds?

1 answer:

Vadim26 [7]2 years ago

3 0

Answer:

Speed of cart's might be less than the high speed after 5 seconds.

Explanation:

Given that,

A fan cart with the fan set to high rolled across the floor.

Let the speed of fan cart with set to high is $x\ cm$ per second.

The fan supplies a force to the cart. If a lower fan speed were used, less force would be applied. This would cause a slower change in the cart's speed. So, the cart would be rolling more slowly than $x\ cm$ per second after 5 seconds. The speed of cart's might be less than $x\ cm$ per second.

Force is needed

A. for a moving object to keep moving at the same speed and direction

B. for a moving object to change its speed

C. for a motionless object to remain still

D. to prevent a moving object from turning

Hence,

Speed of cart's might be less than the high speed after 5 seconds.

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A 25kg chair initially at rest on a horizontal floor requires 165 N force to set it in motion. Once the chair is in motion, a 12

bazaltina [42]

The coefficient of static friction between the chair and the floor is 0.67

Explanation:

Given:

Weight of the chair = 25kg

Force = 165 N (F_applied)

Force = 127 N (F_max)

To find: Coefficient of static friction

The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair

μ_s= $F_{max}/F_{n}$

The F_n is equal to the weight multiplied by its gravity

∴ $F_{n}$ =mg

Thus the coefficient of static friction changes as

μ_s= $F_{max}/mg$

μ_{s} = $=165N/((25kg)\times(9.80 m/s^2 ) )$

= 0.67

3 0

2 years ago

Which of the following is MOST useful to scientists in measuring the size of asteroids?

Alenkasestr [34]

Answer:c-The gravitational effect when spacecraft flies close to the asteriod

Explanation:

Gravitational effect on the spacecraft gives an estimate that how big is the asteroid by experiencing its gravitational pull.

The amount of extra thrust required to maintain the trajectory of the spacecraft during its motion hints at the scientist about the size of the asteroid.

Gravitational pull is directly proportional to the mass of object so greater the mass, greater will be the pull.

5 0

2 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

$V_{2}=V_{1}+at$ (6)

$V_{2}^{2}=V_{1}^{2}+2ad$ (7)

Where:

$a$ is the acceleration

$d=0.01 m$ is the length the ball was compressed

$t$ is the time

Finding $a$ from (7):

$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

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2 years ago

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KengaRu [80]

Answers:

1A) Al203

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3 years ago

Select the correct answer.

Triss [41]

Answer:

Jim's kinetic energy is 54.67 J.

Explanation:

Given that,

Mass, m = 15 kg

Velocity, V = 2.7 m/s

We need to find the Jim's kinetic energy. We know that when the object is in motion, it has kinetic energy. This energy is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 15\ kg\times (2.7\ m/s)^2$

E = 54.67 J

So, Jim's kinetic energy is 54.67 J. Hence, this is the required solution.

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2 years ago