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vekshin1
2 years ago
13

Calculate the weight of your energy car if it has a mass of 233.85 kg.

Physics
1 answer:
lianna [129]2 years ago
3 0

Explanation:

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Now assume that the frictional force f is not at its maximum value. What is the relation between the torque Ï„ applied to each w
leva [86]

Answer:

a)The direction the frictional force will acts is in the positive x direction.

Explanation:

a)The direction the frictional force will acts is in the positive x direction

b)in the horizontal direction, the total force F(total) is equal to 4times the frictional force in the wheel.

F(total)=4f

''f'' is taken as the frictional force.

c)4times the normal force on each wheel minus the acceleration equals zero i.e 4N(wheel)-a=0

=4N(wheel)-mg=0

d) torque is the force that tends to bend rotation

ζ=rf

but acceleration=4×frictional force

cross multiply

f=ζ/r

f=ma/4

ma/4=ζ/r

a=4ζ/r

5 0
3 years ago
Read 2 more answers
JUST PLZ HELP!!! Why does the lightbulb in the right electrical circuit turn on but not the one on the left?
jekas [21]

In the open circuit the current can not flow from one end of the power source to the other. Because of this there is no current flow, and therefore the light does not turn on.

8 0
3 years ago
Read 2 more answers
A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c
Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

I= \Delta p = m \Delta v = m (v-u)

where

m is the mass of the object

\Delta v is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s

(b) 155 N

The impulse can also be rewritten as

I=F \Delta t

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

\Delta t is the duration of the collision

In this situation, we have

\Delta t = 0.06 s

So we can re-arrange the equation to find the magnitude of the average force:

F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N

6 0
3 years ago
A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy
GREYUIT [131]

Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

8 0
2 years ago
Please help!! I will give brainliest!
grandymaker [24]

Answer:

Depends.

Explanation:

Whether the object is going left or right, the speed will stay the same until friction eventually stops it. <em>However, </em>if, for example, we're talking about an object going straight before veering right, then yes, speed <em>does</em> matter. An object will normally have to speed up or slow down momentarily when changing direction to keep itself sustained on the ground.

So, honestly? It really depends on what we're talking about!

Hope this helped!

Source(s) used: None.

7 0
3 years ago
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