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sergejj [24]
2 years ago
7

Question 8 of 25

Physics
1 answer:
Dima020 [189]2 years ago
8 0

Answer:

B.

Explanation:

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A 2.0 kg block, initially moving at 10.0 m/s, slides 50.0 m across a sheet of ice beforecoming to rest. What is the magnitude of
inna [77]

Answer:

The magnitude of the average frictional force on the block is 2 N.

Explanation:

Given that.

Mass of the block, m = 2 kg

Initial velocity of the block, u = 10 m/s

Distance, d = 50 m

Finally, it stops, v = 0

Let a is the acceleration of the block. It can be calculated using third equation of motion. It can be given by :

v^2-u^2=2ad

-u^2=2ad

a=\dfrac{-u^2}{2d}\\\\a=\dfrac{-(10\ m/s)^2}{2\times 50\ m}\\\\a=-1\ m/s^2

The frictional force on the block is given by the formula as :

F = ma

F=2\ kg\times (-1)\ m/s^2\\\\F=-2\ N

|F| = 2 N

So, the magnitude of the average frictional force on the block is 2 N. Hence, this is the required solution.

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What is 40% of 230 i need to know soon plz im only 7
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Answer:it's 92 my friend

Explanation:

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2 years ago
Read 2 more answers
(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1510 m/s? (c) With wh
makvit [3.9K]

Answers:

(a) 2509.98 m/s

(b) 397042.215 m

(c) 1917.76 m/s

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is 700 km  and whose gravitational acceleration at the surface is a_{g}=4.5 m/s^{2}

Knowing this, let's begin:

a) In this part we need to find the escape speed V_{e} on the asteroid:

V_{e}=\sqrt{\frac{2GM}{R}} (1)

Where:

G is the universal gravitational constant

M is the mass of the asteroid

R=700 km=700(10)^{3} m is the radius of the asteroid

On the other hand we know the gravitational acceleration is a_{g}=4.5 m/s^{2}, which is given by:

a_{g}=\frac{GM}{R^{2}} (2)

Isolating GM:

GM=a_{g}R^{2} (3)

Substituting (3) in (1):

V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R} (4)

V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)} (5)

V_{e}=2509.98 m/s (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

E_{o}=E_{f} (7)

Being:

E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R} (8)

E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h} (9)

Where:

E_{o} is the initial mechanical energy

E_{f} is the final mechanical energy

K_{o} is the initial kinetic energy

K_{f}=0 is the final kinetic energy

U_{o} is the initial gravitational potential energy

U_{f} is the final gravitational potential energy

m is the mass of the object

V=1510 m/s is the radial speed of the object

h is the distance above the surface of the object

Then:

\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h} (10)

Isolating h:

h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R (11)

h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m (11)

h=397042.215 m (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance h=981.8 km=981.8(10)^{3} m. This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2} (13)

Isolating V:

V=\sqrt{2a_{g} R(1-\frac{R}{R+h})} (14)

V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})} (15)

Finally:

V=1917.76 m/s

7 0
2 years ago
Apply: The earth's gravity is pulling on you. Are you pulling on the earth? Explain your
coldgirl [10]

Answer:

Yes, you are pulling on Earth. Reasoning. Third Newton's law of motion, action and reaction law, sates that for every action force, there is an equal (in magnitude) and opposite reaction force.

Explanation:

goo gle.

5 0
2 years ago
The most common purpose for fracking in the u.s is to get
sveta [45]
Natural gas is the answer.
5 0
3 years ago
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