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masha68 [24]
3 years ago
11

A car with a mass of 2000 kg is moving around a circular curve at a uniform velocity of 25 m/s per second. The curve has a radiu

s of 80 m what is the centripedal force on the car
Physics
1 answer:
IrinaK [193]3 years ago
4 0
The centripetal force is:
 F = mv² / R
 Where:
 m: mass of the object
 v: object speed
 R: radius of the curve.
 We have to:
 m = 2000kg
 v = 25 m / s
 R = 80 meters.
 Then the centripetal force acting on the vehicle is:
 F = (2000kg * (25m / s) ²) / 80m
 F = 15625 N
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1. What is technology?<br>techndegy​
Naya [18.7K]

Answer:

the application of scientific knowledge for practical purposes, especially in industry. Another answer:the sum of techniques,skills,methods and processes used in the production of goods or services or in the accomplishment of objectives, such as scientific investigation

6 0
4 years ago
A bungee cord has a spring constant of 82.5 N/m. After jumping, a person oscillates with a period of 7.50 s. What is the mass of
Arada [10]

Answer:

Explanation:

T = 2π √ (m/k)

T / 2π = √ (m/k)

T√k = 2π √m

(T√k) / 2π = √m

( (T√k) / 2π)^2 = m

m = ( (7.5√82.5) / 2π)^2

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3 years ago
Un objeto se suelta desde determinada altura y emplea un tiempo t en caer al suelo. Si se cuadruplica la altura desde la cual se
blondinia [14]

When an object falls from a h height, you should work with the uniformly accelerated linear movement equations:

y=½*a*t²+Vo*t+yo

You should consider:

a=-g=-10m/s²

yo=h

If it’s a freefall, it means it starts from rest, which means it has no initial velocity:

Vo=0

Replacing that information in the equation:

y=½*(-10m/s²)*t²+0*t+h=-5m/s²*t²+0+h=-5m/s²*t²+h

So this is the

Besides, if you want to find out how long it takes for it to get to the floor, you should put the height of the floor as final height, which would be 0 (assuming the initial height has been measured from there):

y=0

0=-5m/s²*t²+h

5m/s²*t²=h

t²=h/(5m/s²)

t=√(h/(5m/s²))

t=√(hs²/(5m))

t=(√(h/(5m)))s

<span>If we <span>quadruple </span>h:</span>

t2=(√(h2/(5m)))s=(√(4*h1/(5m)))s=(√4)*(√h1/(5m)))s=2*(√h1/(5m)))s=2*t1

This 4 goes inside the square root, so then it converts to 2. So the new time is twice as much the previous time.

Concerning velocity, you have to use the other equation:

v=at+vo

As I said before, a is gravity and vo is zero.

v=-10m/s²*t+0=-10m/s²*t

Final velocity is directly related to time, so if time is doubled, so is velocity.

v2=-10m/s²*t2=-10m/s²*(2*t1)=2*(-10m/s²*t1)=2*v1

<span>So the correct answer is A, and the other ones are false.</span>

8 0
3 years ago
A pendulum at position A is released and swings through position B to position Con the other side.
zaharov [31]

Explanation:

Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),

the conditions can be summarized bellow

point A

initial velocity= 0

final velocity=0

P.E= Max

K.E= 0

point B

initial velocity= maximum

final velocity=maximum

P.E=K.E

point C

initial velocity= min

final velocity=min

P.E= 0

K.E= max

3 0
3 years ago
CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much do
lubasha [3.4K]

Answer:

The elastic potential energy of the spring change during this process is 21.6 J.    

Explanation:

Given that,

Spring constant of the spring, k=1.2\times 10^4\ N/m

It extends 6 cm away from its equilibrium position.

We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 1.2\times 10^4\times (0.06)^2\\\\E=21.6\ J

So, the elastic potential energy of the spring change during this process is 21.6 J.

4 0
3 years ago
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