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Olin [163]
3 years ago
11

A rock is thrown downward from the top of a cliff with an initial speed of 20 m/s. If the rock hits the ground after 5.0 s, what

is the height of the cliff? (Disregard air resistance. a =-g=-9.81 m/s2.)​
(i want step by step)
Physics
1 answer:
Vinil7 [7]3 years ago
7 0

Answer

I hope it's helps you

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An observer measures the length (L), width (w), and height (h) of a box while stationary relative to the box. The observer then
Elenna [48]

Answer:

b. less than w.

Explanation:

In this question, the application of length contraction is what helps us come to our conclusion. When an object moves very fast (relative to the observer), the length of the object seems to be smaller than it actually is (again, for the observer).

This is supported by the length contraction equation below:

L = L_0\sqrt{1-\frac{v^2}{c^2} }

Here, L is the observed length

L_0 is the original length of the object

v is the relative speed between the object and the observer

and c is the speed of light

Using this equation, we can see that as the speed between the object and the observer is increased to be close to that of light, the square root in the equation gives us values less than 1.0

This effectively decreases the length that is observed.

8 0
3 years ago
A football is thrown horizontally with an initial velocity of (16.6 m/s)x^. ignoring air resistance, the average acceleration of
Andrei [34K]

Solution:

According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.

So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.  

v_f=v_o + at ……..(a)

 [where v_f and v_o are final velocity and initial velocity, respectively]

Now ,

Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.  

Applying this value in equation (a)  

v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction  

For calculating the magnitude of the equation we have to square root the given value

         (16.6i - 17.165y)  

\\ \left | V  \right |=sqrt{16.6^{2}+17.165^{2}}\\ = \sqrt{275.56+294.637225}\\= \sqrt{570.197225}\\= 23.87[/tex]

5 0
3 years ago
Read 2 more answers
A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. If the mass is set in motion in a medium that
denpristay [2]

Answer:

d²x/dt² = - 4dx/dt - 4x is the required differential equation.

Explanation:

Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,

F = mg

kx = mg

k = mg/x

= 4 kg × 9.8 m/s²/2.45 m

= 39.2 kgm/s²/2.45 m

= 16 N/m

Now the drag force f = 16v where v is the velocity of the mass.

We now write an equation of motion for the forces on the mass. So,

F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass

-kx - 16v = 4a

-16x - 16v = 4a

16x + 16v = -4a

4x + 4v = -a where v = dx/dt and a = d²x/dt²

4x + 4dx/dt = -d²x/dt²

d²x/dt² = - 4dx/dt - 4x which is the required differential equation

6 0
3 years ago
A 12,000kg. railroad car is traveling at +2m/s when it
ivann1987 [24]

<u>Answer:</u>

The final velocity of the two  railroad cars is 1.09 m/s

<u>Explanation:</u>

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by  

\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}

where

V= Final velocity

M1= mass of the first object in kgs = 12000

M2= mas of the second object in kgs = 10000

V1= initial velocity of the first object in m/s = 2m/s

V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s  

\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}

Which is the final velocity of the two  railroad cars

8 0
3 years ago
Jay bought a toy car. To make the car move, he must turn a key attached to a spring. Turning the key winds the spring tight. The
lapo4ka [179]
The chemical energy in Jay's body, to kinetic energy in the car
5 0
3 years ago
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