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Setler [38]
3 years ago
12

A ball is dropped from a height of 20m and renounce with the velocity which is 3/4 of d velocity with which it hits the ground.

What is d tym interval btw the first and second bounces​
Physics
1 answer:
Mashutka [201]3 years ago
3 0

Answer:

1.73 seconds

Explanation:

The velocity the ball first hits the ground with is:

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (-10 m/s²) (-20 m)

v = -20 m/s

The velocity it rebounds with is 3/4 of that in the opposite direction, or 15 m/s.

The time it takes to return to the ground is:

Δx = v₀ t + ½ at²

0 = (15 m/s) t + ½ (-10 m/s²) t²

0 = t (15 − 5t²)

t = √3

t ≈ 1.73 seconds

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A roller coaster car is going over the top of a 18-mm-radius circular rise. At the top of the hill, the passengers "feel light,"
Andre45 [30]

Answer:

0.29713 m/s

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity

r = Radius = 18 mm

By balancing the forces in the system we have

mg-N=\dfrac{mv^2}{r}\\\Rightarrow mg-\dfrac{mg}{2}=\dfrac{mv^2}{r}\\\Rightarrow v=\sqrt{r(g-\dfrac{g}{2})}\\\Rightarrow v=\sqrt{0.018\times (9.81-\dfrac{9.81}{2})}\\\Rightarrow v=0.29713\ m/s

The velocity of the coaster is 0.29713 m/s

7 0
3 years ago
how does spatial pattern of heights illustrate the relationship between temperature density and the rate of vertical pressure ch
Anika [276]

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

Generally, the relationship between temperature, density and rate of vertical pressure is given as;

\rho = \frac{PM}{RT}

\frac{dP}{dz} = -\rho g\\\\

where;

  • <em>ρ is density</em>
  • <em>T is temperature</em>
  • <em>dP is rate of change of vertical  pressure</em>

Thus, from the formula above, we can conclude the following relationship between temperature, density and the rate of vertical pressure change in spatial pattern of heights.

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

Learn  more here:brainly.com/question/25395377

5 0
2 years ago
What happens with alpha, beta, and gamma radiation?
tresset_1 [31]
I'm sorry but I don't really understand the question. What is the quest actually asking???
4 0
3 years ago
What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne
irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance h= 1790 km = 1.790\times10^{6}\ m

Magnetic field B=4\times10^{-8}\ T

Mass of proton m=1.673\times10^{-21}\ Kg

Radius of earth R =6.38\times10^{6}\ m

Radius of orbit r=R+h

r=6.38\times10^{6}+1.790\times10^{6}

r=8170000\ m

We need to calculate the speed

Using formula of magnetic field

Bvq=\dfrac{mv^2}{r}

v=\dfrac{Bqr}{m}

Put the value into the formula

v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}

v=31.25\ m/s

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0
3 years ago
A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i
Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

v^2=u^2+2as

0^2=14.5^2+2\times a\times 18.25

a=-5.76m/sec^2

We know that acceleration is given by

a=\mu g

5.76=\mu\times  9.81

\mu =0.587

So coefficient of friction will be 0.587

6 0
3 years ago
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