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densk [106]
4 years ago
6

STATE THE HOOKE'S LAW

Physics
2 answers:
victus00 [196]4 years ago
6 0
The Hooke's law is a principal of physics that states that the force needed to extend or compress a spring by some distance x scales linearly with respect to that distance.
Anvisha [2.4K]4 years ago
6 0

a law stating that the strain in a solid is proportional to the applied stress within the elastic limit of that solid.


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A 180 meter long train, travelling at a constant speed, is passing a car driving on an adjacent lane at a speed of 72 km/h. Know
Lana71 [14]

Answer:

The velocity of the train is 82.8 km/h

Explanation:

The equation for the position of the train and the car is as follows:

x = x0 + v · t  

Where:

x = position at time "t".

x0 = initial position.

v = velocity.

t = time.

First, let´s calculate the distance traveled by the car in 60 s (1/60 h). Let´s place the origin of the frame of reference at the front of the train when it starts to pass the car so that the initial position of the car is 0 (x0 = 0 m):

x = 0 m + 72 km/h · (1/60) h

x = 1.2 km.

Then, if the whole train passes the car at that time, the position of the front of the train at that time will be 1.2 km + 0.18 km = 1.38 km.

Then using the equation of position we can obtain the velocity:

x = x0 + v · t

1.38 km = 0 m + v · (1/60) h

1.38 km / (1/60) h = v

v = 82.8 km/h

The velocity of the train is 82,8 km/h

The same result could be obtained using the rear of the train. You only have to identify where the rear is at t = 0 and where it is at t = 60 s.

Try it!

6 0
3 years ago
Determine the force of gravitational attraction between the Earth and the moon. Their masses are 5.98 x 1024 kg and 7.26 x 1022
monitta

Answer:

F=1.95\times 10^{20}\ N

Explanation:

Mass of Earth, m_e=5.98 \times 10^{24}\ kg

Mass of Moon, m_m=7.26\times 10^{22}\ kg

The distance between Earth and the Moon is, d=384,400\ km

We need to find the force of gravitational attraction between the Earth and the moon. The force of gravity is given by :

F=G\dfrac{m_em_m}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{5.98 \times 10^{24}\times 7.26\times 10^{22}}{(384400 \times 10^3)^2}\\\\F=1.95\times 10^{20}\ N

So, the required force is 1.95\times 10^{20}\ N.

4 0
3 years ago
How long did it take Fernando to drive to
barxatty [35]
How do you calculate distance over speed?
Image result for if your going 30 m/s to go somewhere thats 1680 miles away
The formula can be rearranged in three ways:
speed = distance ÷ time.
distance = speed × time.
time = distance ÷ speed.

1680÷30 = 56

So it would take around 56 minutes to get to Kroger.


I hope this helps !! :)
3 0
1 year ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m
r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

Now we know

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2

KE = 4.55 \times 10^{-27} kg

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0
3 years ago
I need help with this work
san4es73 [151]
What work??? I don’t see anything
7 0
3 years ago
Read 2 more answers
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