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topjm [15]
3 years ago
10

Answer key says b but i think it is false please help me

Physics
1 answer:
klasskru [66]3 years ago
5 0
Energy in a spring:

E = 0.5 * k * x²

k spring constant = 800 n/m
x stretch of the spring = 5 cm = 0.05 m

E = 0.5 * 800 * 0.05² = 1
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Which type of reactions usually happens slowest?
m_a_m_a [10]

Answer:

option b is correct..................

7 0
3 years ago
A rock has 750 J of potential energy as it sits on a ledge. If the rock were pushed off the ledge, how much would it have just b
wariber [46]

Answer:

the 750 j will have potential energy is 375

Explanation:

750/2 is 375

8 0
2 years ago
Far out in space, a 100,000-kg rocket and a 200,000-kg rocket are docked at opposite ends of a motionless 90m-long tunnel. The t
postnew [5]

Answer:

a) X_{cm} = 60m

b) I = 5.4*10^8Kg*m^2

c) T= 4.5*10^6 N*m

d) a = 8.3*10^{-3}rad/s^2

e) W= 0.25 rad/s

Explanation:

a) we know that:

X_{cm} = \frac{m_1x_1+m_2x_2}{m1+m2}

where X_cm is the ubicaton of the center of mass, m_1 the mass of the first rocket, x_1 its distances with the rocket 1, m_2 the mass of the second rocket and x_2 its distance with the rocket 1. So, replacing values, we get:

X_{cm} = \frac{(100,000kg)(0)+(200,000kg)(90m)}{100,000kg+200,000kg}

X_{cm} = 60m

So, the center of mass is at 60m from the rocket 1.

b) we know that:

I = M_1R_1^2 +M_2R_2^2

where I is the moment of inertia, M_1 is the mass of the rocket 1, R_1 its distance from the center of mass, M_2 the mass of the second rocket and R_2 the distance between the rocket 2 and the center of mass. So, replacing values, we get:

I = (100,000kg)(60m)^2 +(200,000kg)(30m)^2

I = 5.4*10^8Kg*m^2

c) We know that:

T = Fr

where T is the net torque, F is the force and r is the distance between the rocket and the radius. Then:

FR_1+FR_2 = T

Replacing values, we get:

50,000N(60m)+50,000N(30m) = T

T= 4.5*10^6 N*m

d) We know that:

T = Ia

where T is the net torque, I the moment of inertia and a is the angular aceleration. So, replacing values, we get:

4.5*10^6Nm = 5.4*10^8Kg*m^2a

solving for a:

a = 8.3*10^{-3}rad/s^2

e) Finally, using:

W = at

where W is the angular velocity, a is the angular aceleration and t is the time.

Then, replacing values. we get:

W = (8.3*10^{-3}rad/s^2)(30s)

W = 0.25 rad/s

3 0
3 years ago
Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speed up to
Harman [31]
Accelration=changespeed/changetime
changetime=8
changespeed=45-5=40
40/8=5

5m/s^2 is acceleration
4 0
2 years ago
A kg object has 400 j of potential energy. Find how high off the groumd the object is
blondinia [14]
Potential energy = mgh
Energy, U = 400J
g = 10m/s^2
m = A kg
h = U ÷ mg
= 400÷10A
= 40÷A

mass is not given
6 0
3 years ago
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