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zhannawk [14.2K]
3 years ago
15

A 1.5 kg baseball is pitched at 38 m/s and is hit by the batter. It heads directly back at the pitcher at 27 m/s.

Physics
1 answer:
lesantik [10]3 years ago
5 0

Answer:

The initial velocity of the ball is +38.0 m/s. The final velocity is -27.0 m/s.  

momentum = mass x velocity

(mass x final velocity) – (mass x initial velocity)

(1.5)(-27) – (1.5)(38) = -97.5

The impulse applied to the ball is -97.5 kg m/s

impulse = force x time interval

force = impulse/time interval

-97.5/0.45 s = - 216.6  

The force applied was 216.6 N.

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19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,
sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

t=15.2 s is the time it takes to the child to reach the bottom of the slope

V_{o}=0 is the initial velocity (the child started from rest)

\theta=15\° is the angle of the slope

d is the length of the slope

Now, the Force exerted on the sled along the ramp is:

F=ma (1)

Where m is the mass of the sled and a its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

F=mg sin \theta (2)

Where g=9.8 m/s^{2} is the acceleration due gravity

Then:

ma=mg sin \theta (3)

Finding a:

a=g sin \theta (4)

a=9.8 m/s^{2} sin(15\°) (5)

a=2.5 m/s^{2} (6)

Now, we will use the following kinematic equations to find d:

V=V_{o}+at (7)

V^{2}=V_{o}^{2}+2ad (8)

Where V is the final velocity

Finding V from (7):

V=at=(2.5 m/s^{2})(15.2 s) (9)

V=38 m/s (10)

Substituting (10) in (8):

(38 m/s)^{2}=2(2.5 m/s^{2})d (11)

Finding d:

d=288.8 m

6 0
3 years ago
What is the resultant of the two vectors shown?
lianna [129]

Explanation:

The correct is (c). cuz, the tail of A and the tip of B will join

3 0
2 years ago
A motorcycle that is slowing down uniformly covers 2.0 successive km in 80 s and 120 s, respectively. Calculate (a) the accelera
8090 [49]

Answer:

acceleration = -0.042 m/s²

velocity at beginning =  14.167 m/s

velocity at  end = 5.7183 m/s

Explanation:

given data

distance d1 = 1 km

distance d2 = 2 km

time  t1 = 80 s

time t2 = 120 s + 80s = 200 s

to find out

acceleration and velocity at beginning and end

solution

we apply here law of motion that is

d = vt + 1/2×at²    

put value

1000 = v(80) + 1/2×a(80)²           ........................1

and

2000 = v(200) + 1/2×a(200)²      ........................2

so from equation 1 and 2 we get a and v

a = -0.042 m/s² and

v = 14.167 m/s

so by kinematic final velocity will be

V² = v² + 2ad

V² = (14.167)² + 2×(-0.042)×(2000)

V²  = 32.70

V = 5.7183 m/s

so

acceleration = -0.042 m/s²

velocity at beginning =  14.167 m/s

velocity at  end = 5.7183 m/s

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Explanation:

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