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Rainbow [258]
3 years ago
11

Zeb was lifting a box onto a moving truck. He lifted with a net force of 2000N and the box had a mass of 100 kg. What was the re

sulting acceleration?
Physics
1 answer:
Aleksandr [31]3 years ago
8 0

Answer:

<h2>20 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{2000}{100}  = 20 \\

We have the final answer as

<h3>20 m/s²</h3>

Hope this helps you

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The water was used to power a generator, creating _______________ energy
Nikitich [7]

Answer:

electric

<em>please give brainliest</em>

3 0
2 years ago
A railroad freight car rolls on a track at 3.50 m/s toward two identical coupled freight cars, which are rolling in the same dir
ladessa [460]

Answer:

Explanation:

Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence

The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.

The kinetic energy is calculated by the following equation.

E=0.5mV^2

m= mass of the cars=26500kg

V=speed

E=kinetic energy

taking into account the above, the following equation is inferred

1=  the cars are separated

2= the cars are togheter

E1=E2

E1=0.5mV1^2+0.5mV1^2+0.5m(Va)^2

where

m= mass of each car

V1= 1.2m/s

Va=3.5,m/S

E2=0.5(3)(m)V^2

m= mass of each car

V=speed (in m/s) of the three coupled cars after the first couples with the other two

Solving

0.5mV1^2+0.5mV1^2+0.5m(Va)^2=0.5(3)(m)V^2

V1^2+V1^2+(Va)^2=(3)V^2.\\2V1^2+(Va)^2=(3)V^2\\V^2=\frac{2V1^2+(Va)^2}{3} \\

V=\sqrt{\frac{2V1^2+(Va)^2}{3}} \\V=\sqrt{\frac{2(1.2)^2+(3.5)^2}{3}} \\\\V=2.245m/s

the speed  of the three coupled cars after the first couples with the other two is 2.245m/s

7 0
3 years ago
Read 2 more answers
A golf ball is dropped from rest from a height of 9.5m.  It hits the pavement then bounces back up rising  just 5.7 m before fal
Oksanka [162]

Answer: 3.4s

Explanation:

There are three stages in the motion of the ball, so you have to calculate the times for every stage.

1) Ball dropping from 9.5m: free fall

d = Vo + gt² / 2

Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²

⇒ t = √ (1.94 s²) = 1.39s

2) Ball rising 5.7m (vertical rise)

i) Determine the initial speed:

Vf² = Vo² - 2gd

Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²

⇒ Vo = 10.6 m/s

ii) time rising

Vf = Vo - gt

Vf = 0 ⇒ Vo = gt ⇒

t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s

3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)

i) d = 5.7m - 1.20m = 4.5m

ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²

⇒ t = √ (0.92 s²) = 0.96s

4) Total time

t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s

4 0
3 years ago
Read 2 more answers
suppose a child walks from the outer edge of a rotating Merry-Go-Round to the inside does angular velocity of the Merry-Go-Round
Akimi4 [234]

Okay here initially child walks on the outer edge of the disc and then started to move inside

So here as the child and Merry go round is an isolated system so there is no external Torque on this system from outside

As here we can see there is external force acting on this system by the hinge of the Merry go round as well as due to gravity so we can not use momentum conservation to solve such type of questions.

But as we can say that there is no external torque on this system about the hinge point so we will use conservation of angular momentum for this system

Here as we know that

\tau = \frac{dL}{dt}

where L = angular momentum

since here torque is ZERO

0 = \frac{dL}{dt}

L = constant

so here we can write initial angular momentum of the system as

L = (I_1 + I_2)*\omega

here we know that

I_1 = moment of inertia of merry go round

I_2 = moment of inertia of child

so here we can say

(I_1 + I_2)* \omega_1 = (I_1 + I_2')\omega_2

so here as the child moves from edge to inside the disc it moment of inertia will decrease because as we know that moment of inertia of child is given as

I_2 = mr^2

here m = mass of child

r = distance of child from center

Since child is moving inside so his distance from center is decreasing

so here moment of inertia of child is decreasing as he starts moving inside

so final angular speed of merry go round will increase as child go inside

\omega_2 = \frac{(I_1 + I_2)*\omega}{(I_1 + I_2')}

so here as

I_2' < I_2

final angular speed will be more than initial speed as child moves inside

3 0
3 years ago
Do the two cars ever have the same velocity at one instant of time? If so, between which two frames? Check all that apply. Check
alexgriva [62]

Answer:

Cars have the same velocity at one instant of time between dots 4 and 5.

Explanation:

after looking at the image it is visible that same distance is covered by both cars in the frame between 4 and 5

8 0
3 years ago
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