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2 years ago

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A 0.15 kg baseball is hit by a baseball bat. Right before it is hit, the ball’s speed is 30 m/s, and right after it is hit, the

ball’s speed is 50 m/s in the opposite direction. What is the magnitude of the net impulse that is applied to the ball?
10.5 kg?m/sA. 4.5 kg?m/sB. 12.0 kg?m/sC. 7.5 kg?m/sD. 3.0 kg?m/s

1 answer:

Roman55 [17]2 years ago

5 0

Answer:

Impulse will be 12 kgm/sec

So option (b) will be correct option

Explanation:

We have given mass of the baseball m = 0.15 kg

Ball speed before hit $v_1=30m/sec$

Ball speed after hitting $v_2=-50m/sec$ ( negative direction due to opposite direction )

We have to find the impulse

We know that impulse is equal; to the change in momentum

So change in momentum = $m(v_1-v_2)=0.15(30-(-50))=0.15\times 80=12kgm/sec$

So option (b) will be correct option

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An open train car moves with speed 18.5 m/s on a flat frictionless railroad track, with no engine pulling the car. It begins to

olga2289 [7]

Answer:

The speed decreases.

Explanation:

This can be explained using the conservation of linear momentum.

Since there is no friction, the initial moment of the train must be equal to its linear moment after it is filled with water.

the initial linear momentum is

$m_{1}v_{1}$

where $m_{1}$ is the initial mass of the train, and $v_{1}$ the initial speed of the train.

And linear momentum after the water filled the train car is

$m_{2}v_{2}$

where $m_{2}$ is mass of the train after the rain, and $v_{2}$ the speed of the train after the rain

<u>the equality must be fulfilled:</u>

$m_{1}v_{1}=m_{2}v_{2}$

We know that if water is added to the train, $m_{2}$ that is the mass after the water is added, is greater than $m_{1}$ which is the mass of the train without the water.

Therefore, in order for the conservation of the linear momentum to be fulfilled: $m_{1}v_{1}=m_{2}v_{2}$

the speed after the water is added ( $v_{2}$ ) must be smaller than the initial train speed ( $v_{1}$ ) . So the speed of the car decreases.

3 0

2 years ago

Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down th

Margaret [11]

Answer:

300 m

Explanation:

The train accelerate from the rest so u = 0 m/sec

Final speed that is v = 80 m/sec

Time t = 30 sec

The distance traveled by first plane = 1200 m

We know the equation of motion $S=ut+\frac{1}{2}at^2$ where s is distance a is acceleration and u is initial velocity

Using this equation for first plane $1200=0\times 30+\frac{1}{2}a30^2$

$a=2.67\frac{m}{sec^2}$

As the acceleration is same for both the plane so a for second plane will be 2.67 $\frac{m}{sec^2}$

The another equation of motion is $v^2=u^2+2as$ using this equation for second plane $40^2=0+2\times 2.67\times s$

s = 300 m

5 0

2 years ago

In this graph, what is the displacement of the particle in the last two seconds?

nikklg [1K]

In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds?

<span>0.2 meters
2 meters
4 meters
6 meters</span>
In this graph, the displacement of the particle in the last two seconds is 2 meters.

7 0

2 years ago

A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i

Hoochie [10]

The acceleration of the box is approximately $1 m/s^2$

Explanation:

According to Newton's second law of motion, the net force acting on the box is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m = 12.0 kg is the mass of the box

a is the acceleration

The net force can be written as

$\sum F = F_a - F_f$

where

$F_a = 20 N$ is the applied forward force

$F_f=9.0 N$ is the friction force

Combining the two equations,

$F_a-F_f=ma$

And solving for the acceleration,

$a=\frac{F_a-F_f}{m}=\frac{20-9}{12}=0.9 m/s^2\sim 1 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

8 0

2 years ago

A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the stri

san4es73 [151]

Answer:

T = 8.55 N

Explanation:

When string makes an angle 40 degree with the vertical then it will have two forces on it

1) gravitational force (mg)

2) Tension force in string (T)

now we know that net force towards the center of the path is known as centripetal force and it is given as

$T - mg cos40 = F_c$

$T - (0.40\times 9.8)cos40 = \frac{mv^2}{L}$

$T = 3 + \frac{0.40\times 5^2}{1.8}$

$T = 3 + 5.55$

$T = 8.55 N$

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2 years ago