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valentinak56 [21]
3 years ago
7

A pitcher delivers a fast ball with a velocity of 43 m/s to the south. the batter hits the ball and gives it a velocity of 51 m/

s to the north. what was the average acceleration (magnitude and direction) of the ball during the 1.0 ms when it was in contact with the bat?
Physics
1 answer:
hoa [83]3 years ago
4 0
Assuming north as positive direction, the initial and final velocities of the ball are:
v_i=-43 m/s (with negative sign since it is due south)
v_f=+51 m/s
the time taken is t=1.0 ms=0.001 s, so the average acceleration of the ball is given by
a= \frac{v_f-v_i}{t}= \frac{51 m/s-(-43 m/s)}{0.001 s}=9.4 \cdot 10^4 m/s^2
And the positive sign tells us the direction of the acceleration is north.
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a train car of mass 444 kg moving at 5 m/s bounces into another car on the same tracks of mass 344 king. of the second car was m
Kamila [148]

Answer:

3.7 m/s

Explanation:

M = 444 kg

U = 5 m/s

m = 344 kg

u = - 5 m/s

Let the velocity of train is V and the car s v after the collision.

As the collision is elastic

By use of conservation of momentum

MU + mu = MV + mv

444 x 5 - 344 x 5 = 444 V + 344 v

500 = 444 V + 344 v

125 = 111 V + 86 v .... (1)

By using the formula of coefficient of restitution ( e = 1 for elastic collision)

e = \frac{V-v}{u-U}

-5 - 5 = V - v

V - v = - 10

v = V + 10

Substitute the value of v in equation (1)

125 = 111 V + 86 (V + 10)

125 = 197 V + 860

197 V = - 735

V = - 3.7 m/s

Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.

4 0
3 years ago
The sport with the fastest moving ball is jai alai, where measured speeds can be 286 km/h. If a professional jai alai player fac
cricket20 [7]

Answer:

d= 794.4 cmExplanation:

Given that

Speed ,V= 286 km/h

=286\times \dfrac{1000}{3600}\ m/s

V=79.44 m/s

Given that time ,t= 100 ms

t= 0.1 s

We know that ( if acceleration is zero)

Distance = Speed x time

d= V t

Now by putting the values in the above equation

d = 79.44 x 0.1 m

d= 7.944 m

We know that 1 m = 100 cm

d= 794.4 cm

5 0
3 years ago
Explain how birds, bats, insects, airplanes, rockets, and hot air balloons achieve their flight.
Oksi-84 [34.3K]

The birds, bats, insects, airplanes, rockets, and hot air balloons achieve their flight by creating high pressure below the aircraft and low pressure above it

<h3>What are living and non-living things?</h3>

They both consist of fundamentally simple building blocks. They are composed of substances or mass. Atmospheric and molecular building blocks make up the world.

In order to create high pressure below the aircraft and low pressure above it, airplanes employ specially built wings.

The wing receives sufficient airflow past it to counteract the weight and drag of the aircraft by utilizing a device to provide thrust, such as a propeller.

High pressure underneath the aircraft and low pressure above it is produced by specially constructed wings used by airplanes.

The wing receives enough airflow through it by the use of a thrust-generating device, like a propeller, to overcome the weight and drag of the aircraft.

The differences between how living and non-living things fly;

1. Moving both living and non-living objects consumes energy. Flying animals utilize their wings to create both lift and propulsion by moving them in relation to the body.

In contrast to most air vehicles, where the components that generate lift, wings, and thrust engines or propellers are distinct, the wings stay stationary.

Animal aviators like birds as well as natural parachuters like patagial as well as human inventions like aircraft as well as rockets that can power spacecraft and spaceplanes are just a few examples of the many things that can fly.

Hence, birds, bats, insects, airplanes, rockets, and hot air balloons achieve their flight by creating high pressure below the aircraft and low pressure above it

To learn more about living and nonliving things, refer to brainly.com/question/7807759

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6 0
2 years ago
1. A runner drops her phone as she is running at a constant speed of 3 miles per hour from point A to point B in a park. Describ
joja [24]

Answer:

Let's define the point A as our zero in the x-axis.

As the phone drops, it keeps the horizontal velocity that it had before, so the horizontal velocity is:

Vx = 3 mi/h.

Now, the only force acting on the phone is the gravitational force that acts in the vertical axis, then we have:

Ay = -g

where g = 9.8 m/s^2

It is dropped, so we do not have a vertical initial velocity, then for the vertical velocity we should integrate over time:

Vy = -g*t

And for the position again, we integrate over time, but now we have an initial position H, that is the height at which the phone is dropped.

Py = -(1/2)*g*t^2 + H

And the horizontal position can be found by integrating over time the horizontal velocity.

Px = (3mi/h)*t

This will be the two equations that describe the motion of the phone, and we can not solve it further because we do not know the initial height of the phone.

But in general, we have a linear equation in the horizontal axis and a quadratic equation with a negative leading coefficient in the vertical axis.

Position(t) = ( (3mi/h)*t,  -(1/2)*g*t^2 + H)

5 0
3 years ago
A block of mass M slides down a frictionless plane inclined at an angle (theta) with the horizontal. The normal reaction force e
WINSTONCH [101]

Answer: N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

Explanation:

See attached for a sketch.

From the attachment.

.

N = normal reaction force on block

W = weight of the block

theta = angle of the inclined plane to the horizontal

From the sketch, we can see that

N is equal in magnitude but opposite direction to Wy

N = Wy

And

Wy = Wcos(theta)

Wx = Wsin(theta)

Then,

N = Wy = Wcos(theta)

But W = mass × acceleration due to gravity = mg

N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

7 0
4 years ago
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