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Cloud [144]
3 years ago
15

What is initial velocity?​

Physics
1 answer:
LUCKY_DIMON [66]3 years ago
8 0

Initial velocity is when the motion starts.

Explanation:

Initial velocity is is the velocity at time interval t = 0 is represented by u.

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A tennis player hits a ball at an angle of 50 degrees above horizontal so that it has an acceleration of 12 m/s2. What is the ho
harkovskaia [24]
So we want to know what is the magnitude of the horizontal component of acceleration ah if we know that the overall acceleration a=12 m/s^2 and the angle of overall acceleration and the horizontal acceleration is α=50°. We know that ah=a*cosα. So now it isn't hard to get the horizontal component: ah=12*cos50=12*0.64=7.71 m/s^2. So the correct answer is ah=7.71 m/s^2. 
6 0
3 years ago
I need help! plz, It's about energy. its just middle school science
Anit [1.1K]

Answer:

it would be 12/A

Explanation:

if you dived 5 by 12 youll get 60 and you cant get 60 y anything else in there so therefor its 12 :)

8 0
3 years ago
A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler
Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

W=K_f - K_i

Since the cart was initially at rest, K_i = 0, so

W=K_f = \frac{1}{2}mv^2 (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, F_g = 99.5 N:

m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg

So solving eq.(1) for v, we find the final speed of the cart:

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s

2) 2.51\cdot 10^7 J

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

F=4.28 \cdot 10^5 N

d=586 m

So the work done is

W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J

3)  2.51\cdot 10^7 J

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

W=\Delta K = K_f - K_i

where

W is the work done

\Delta K is the change in kinetic energy

Therefore, the change in kinetic energy is

\Delta K = W = 2.51\cdot 10^7 J

4) 37.2 m/s

According to the work-energy theorem,

W=\Delta K = K_f - K_i

where

K_f is the final kinetic energy of the train

K_i = 0 is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

W=K_f = \frac{1}{2}mv^2

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s

7 0
4 years ago
Can anybody help me please?
vovangra [49]

Answer:

1,2,5

Explanation:

7 0
3 years ago
Mary applies a force of 70 N to push a box with an acceleration of 0.50 m/s2. When she increases the pushing force to 78 N, the
neonofarm [45]

Explanation:

Below is an attachment containing the solution.

3 0
3 years ago
Read 2 more answers
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