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statuscvo [17]
1 year ago
9

Baseballs pitched by a machine have a horizontal velocity of 30 meters/second. The machine accelerates the baseball from 0 meter

s/second to 30 meters/second in 0.5 seconds. If a baseball has a mass of 0.15 kilograms, the force the machine exerts is _______ newtons. Use F = ma, where a= v-u/t.
Physics
1 answer:
miv72 [106K]1 year ago
4 0

using the formula

F = ma

Where F is the force applied by the machine

A is the acceleration which is also equal to v/t where v is the velocity and t is time

M is the mass

 

F = mv/t

F = (0.15kg) (30 – 0 m/s)/ 0.5 s

<span>F = 9 N</span>

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A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second
Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

5 0
2 years ago
A friend climbs an apple tree and drops a 0.22-kg apple from rest to you, standing 3.5 m below. When you catch the apple, you br
ella [17]

Answer: (A) velocity = 2.8m/s

(B) Average force = 1.9536 Newtons

Explanation:

Find detailed explanation in the attached picture

3 0
2 years ago
A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

    Angle from positive horizontal axis = 86.85⁰

 So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0
2 years ago
You are testing a new amusement park roller coaster with an empty car with a mass of 130 kg. One part of the track is a vertical
vlada-n [284]

Answer:

Work done by friction along the motion is given as

W_f = -5857.8 J

Explanation:

As per work energy theorem we can say

Work done by all forces = change in kinetic energy of the system

so here car is moving from bottom to top

so here the change in kinetic energy is total work done on the car

so here we will have

W_f + W_g = \frac{1}{2}m(v_f^2 - v_i^2)

W_f - mgH = \frac{1}{2}m(v_f^2 - v_i^2)

now plug in all data in it

W_f - (130)(9.81)(2\times 12) = \frac{1}{2}(130)(8^2 - 25^2)

W_f = 30607.2 - 36465

W_f = -5857.8 J

6 0
2 years ago
A muon formed high in the Earth's atmosphere is measured by an observer on the Earth's surface to travel at speed V - 0.983c for
Alex_Xolod [135]

Answer:

The moun lives 2.198*10^-6 s as measured by its own frame of reference

The Earth moved 648 m as measured by the moun's frame of reference

Explanation:

From the point of view of the observer on Earth the muon traveled 3.53 km at 0.983c

0.983 * 3*10^8 = 2.949*10^8 m/s

Δt = d/v = 3530 / 2.949*10^8 = 1.197*10^-5 s

The muon lived 1.197*10^-5 s from the point of view of the observer.

The equation for time dilation is:

\Delta t' = \Delta t * \sqrt{1 - \frac{v^2}{c^2}}

Then:

\Delta t' = 1.197*10^-5 * \sqrt{1 - \frac{(0.983c)^2}{c^2}} = 2.198*10^-6 s

From the point of view of the moun Earth moved at 0.983c (2.949*10^8 m/s) during a time of 2.198*10^-6, so it moved

d = v*t = 2.949*10^8 * 2.198*10^-6 = 648 m

7 0
1 year ago
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