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likoan [24]
3 years ago
8

A 100-n force has a horizontal component of 80 n and a vertical component of 60 n. the force is applied to a box which rests on

a level frictionless floor. the cart starts from rest, and moves 2.0 m horizontally along the floor. what is the cart's final kinetic energy?
Physics
1 answer:
iragen [17]3 years ago
6 0
<span>160 Joules

   For this problem, we can ignore the vertical component of the applied force and focus on only the horizontal component of 80 N and since work is defined as force over distance, let's multiply the force by the distance: 80 N * 2.0 m = 160 Nm = 160 kg*m^2/s^2 = 160 Joules.

   So the cart has a final kinetic energy of 160 Joules.</span>
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Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a
weeeeeb [17]

Answer:

force becomes one - ninth

Explanation:

According to Coulomb's law in electrostatics, two charges can exert a force of attraction or repulsion on each other which is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

Here both the charges remains same but the distance is variable.

So, we can say that

F \alpha \frac{1}{d^{2}}    .... (1)

Where d be the distance between the tow charges

As the distance between two charges increases by factor of three, let the new force be F'.

F' \alpha \frac{1}{9d^{2}}   .... (2)

Divide equation (2) by equation (1), we get

\frac{F'}{F}=\frac{d^{2}}{9d^{2}}

F'=\frac{F}{9}

Thus, the force becomes one - ninth times the initial force.

6 0
2 years ago
A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular t
Natasha_Volkova [10]

Answer:

v=12.5 i + 12.5 j m/s

Explanation:

Given that

m₁=m₂ = m

m₃ = 2 m

Given that speed of the two pieces

u₁=- 25 j m/s

u₂ =- 25 i m/s

Lets take the speed of the third mass = v m/s

From linear momentum conservation

Pi= Pf

0 = m₁u₁+m₂u₂ + m₃ v

0 = -25 j m  - 25 i m + 2 m v

2 v=25 j   + 25 i m/s

v=12.5 i + 12.5 j m/s

Therefore the speed of the third mass will be v=12.5 i + 12.5 j m/s

4 0
3 years ago
A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.
Anestetic [448]

Answer:

\omega=0.37 [rad/s]  

Explanation:

We can use the conservation of the angular momentum.

L=mvR

I\omega=mvR

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

(I_{proffesor - stool}+mR^{2})\omega=mvR

Now, we just need to solve it for ω.

\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}

\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}      

\omega=0.37 [rad/s]  

I hope it helps you!

5 0
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Naddik [55]

Answer:

α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²

Explanation:

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In order to describe motion along a straight line, you must state the speed and direction of the motion. Those two quantities, together, comprise what's known as "velocity".

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