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andreyandreev [35.5K]

3 years ago

13

2 kg steel ball is dropped straight down ont a hard horixontal florr and bounces straight up . the balls speed just before and j

ust after impact with the floor is 10/s determine the magnitude of the impluese delivered to the floor by the steel ball?

1 answer:

Rainbow [258]3 years ago

5 0

Answer:

$40 \frac{kg*m}{s}$

Explanation:

Impulse-momentum theorem states that impulse is equal to the change of momentum:

$\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ (1)

with pf the final momentum and pi the initial momentum. Knowing that momentum is mass (m) times velocity (1) is:

$\overrightarrow{J}=m \overrightarrow{v}_{f}- m \overrightarrow{v}_{i}$

It's important to note that we're dealing with vector quantities so direction matters. If we choose towards the floor positive direction then the initial velocity is positive and the final velocity is negative, so:

$\overrightarrow{J}=-(2kg)(10\frac{m}{s}) - (2.0kg)(10\frac{m}{s})$

$\overrightarrow{J}=-40 \frac{kg*m}{s}$

So, the impulse delivered to the floor is $40 \frac{kg*m}{s}$

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3 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

b)

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

d)

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

e)

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

f)

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

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3 years ago

To wait until the oncoming vehicle passes before completing a left turn is known as:

Reika [66]

Answer:

Risk rejection

Explanation:

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5 0

3 years ago

otez555 [7]

A. Impulse is simply the product of Force and time. Therefore,

I = F * t ---> 1

where I is impulse, F is force, t is time

However another formula for solving impulse is:

I = m vf – m vi ---> 2

where m is mass, vf is final velocity and vi is initial velocity

Therefore using equation 2 to solve for impulse I:

I = 2000kg (0) – 2000kg (77 m/s)
I = -154,000 kg m/s

B. By conservation of momentum, we also know that Impulse is conserved. That means that increasing the time by a factor of 3 would still result in an impuse of -154,000 kg m/s. So,

I = F’ * (3 t) = -154,000 kg m/s

Since t is multiplied by 3, therefore this only means that Force is decreased by a factor of 3 to keep the impulse constant, therefore:

(F/3) (3t) = -154,000 kg m/s

Summary of Answers:

A. I = -154,000 kg m/s

B. Force is decreased by factor of 3

8 0

3 years ago

You are given a parallel plate capacitor that has plates of area 27 cm^2 which are separated by 0.0100 mm of nylon (dielectric c

deff fn [24]

Answer:

Applied voltage should be 13.5396 kV

Explanation:

The charge stored by a capacitor when subjected to a potential difference 'v' across it's plates is given by

$Q=\frac{K\epsilon _{o} A}{d}V$

Solving for V we get

$V=\frac{Qd}{K\epsilon _{o}A}$

thus to store a charge of 0.11mC we solve for V as follows

Applying values we get

$V=\frac{0.11\times 10^{-3}\times 0.01\times 10^{-3}}{27\times 10^{-4}\times 3.4\times 8.85\times 10^{-12}}$

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7 0

3 years ago