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andrew-mc [135]
3 years ago
6

A car Rounds a curve while maintaining a constant speed. Is there a net force on the car as it rounds the curve?

Physics
1 answer:
astra-53 [7]3 years ago
7 0
Yes! the of the weight of the car and the person going around the curve would be the force you feel.
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Nuclear reactions, such as fusion and fission, convert
slava [35]

Answer:

D.

Explanation:

In nuclear reaction very small amount of matter is converted into significant amount of energy.

6 0
4 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8 m/s^2. The acce
jeka94

Answer:

The maximum height will be 7408.8 meters

Explanation:

final velocity = initial velocity + acceleration × time  

final velocity = 0 m/s + 58.8 m/s^2 ×6 s  

Final velocity = 352.8 m/s  

final velocity ^2 = initial velocity ^2 + 2 × acceleration × displacement  

(352.8)^2 = (0)^2 + 2×58.8 ×displacement  

Solving for displacement,

height = 1058.4 meters.  

After this, the rocket is in free fall, we can use the same equation.  

final velocity ^2 = initial velocity ^2 + 2 ×acceleration×displacement  

final velocity = 0  

0^2 = 352.8^2 + 2×(-9.8)×displacement  

displacement = 6350.4 meters  

the maximum height will be 7408.8 meters

3 0
4 years ago
If an object has an increasing positive acceleration, what can we assume about the forces on it?
Nesterboy [21]

Answer:

F = M a    is a vector equation

F has the same sign as A and is in the same direction as the vector a

If a is positive then F must also be positive

6 0
3 years ago
What is the meaning of Constant Acceleration?
lora16 [44]

Any change in speed or direction of motion is acceleration.

Constant acceleration can mean ...

-- speeding up at a constant rate . . . gaining the same amount
of speed each second.

-- slowing down at a constant rate . . . losing the same amount
of speed each second.

-- changing direction at a constant rate . . . for example, going
around a circular path at a constant speed.

4 0
3 years ago
A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va
ohaa [14]

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

U=\dfrac{1}{2}CV^2

C=\dfrac{\varepsilon _oA}{d}

U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2

If energy become double U' = 2 U then d'

U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.

7 0
3 years ago
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