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Ivanshal [37]
3 years ago
6

What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant, and why

Physics
1 answer:
lutik1710 [3]3 years ago
4 0
It increases, because the centripetal acceleration is inversely proportional to the radius of the curvature.



Hopr it helps :)
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A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

Speed, v_{1} = 5 m/s

m_{2} = 1 kg

v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
3 years ago
If a 100 N net force acts on a 50 kg car, what will the acceleration of the car will be?
Allushta [10]
According to newton's law Force = mass * acceleration
so , 100 = 50 * a
so , a= 2 m/s^2
6 0
3 years ago
Destinations humans have traveled in space
raketka [301]
Yes humans have traveled in space before
8 0
3 years ago
What is static friction?​
givi [52]

Answer:

Plzzzzzzzzzzzzzzzz brainliest

Explanation:

In static friction, the frictional force resists force that is applied to an object, and the object remains at rest until the force of static friction is overcome. In kinetic friction, the frictional force resists the motion of an object. ... The frictional force itself is directed oppositely to the motion of the object.

7 0
2 years ago
A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force ju
stira [4]

Answer:

The coefficient is 0.90

Explanation:

Drawing a diagram makes thing easier, we will assume that the acceleration tends to zero because it start barely moving.

-F_s+mg*sin(\theta)+F=0\\F_s=57.1kg*9.8m/s^2*sin(28.3)+177N\\F_s=442N\\F_s=\µ*N\\N=m*g*cos(\theta)\\N=57.1*9.8*cos(28.3)=493N\\\\\µ=\frac{442N}{493N}=0.90

3 0
3 years ago
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