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Natali5045456 [20]
3 years ago
5

A 2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per seco

nd east. What will be the cart's velocity after a 6.0-newton westward force acts on it for 2.0 seconds?
(1) 2.0 m/s east
(2) 10.0 m/s east
(3) 2.0 m/s west
(4) 10.0 m/s west
Physics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

(2) 10.0 m/s east

Explanation:

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Two satellites A and B orbit the Earth in the same plane. Their masses are 5 m and 7 m, respectively, and their radii 4 r and 7
Dmitry [639]

Answer:

The ratio of their orbital speeds are 5:4.

Explanation:

Given that,

Mass of A = 5 m

Mass of B = 7 m

Radius of A = 4 r

Radius of B = 7 r

The orbital speed of satellite A,

v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}......(I)

The orbital speed of satellite B,

v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}......(I)

We need to calculate the ratio of their orbital speeds

Using equation (I) and (II)

\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}

Put the value into the formula

\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}

\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}

Hence, The ratio of their orbital speeds are 5:4.

8 0
3 years ago
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B​
andreyandreev [35.5K]

Explanation:

Let \textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}} and \textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}

The sum of the two vectors is

\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}

= 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}

The difference between the two vectors can be written as

\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}

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8 0
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1 poir
djverab [1.8K]

Answer:

D) 735 J(oules)

Explanation:

Work is defined as force * distance

Force is defined as mass * acceleration

Given a mass of 15 kg and a gravitational acceleration of 9.8 m/s² since the box is being lifted up, the force being applied to the box is 15 kg * 9.8 m/s² = 147 N

Since the distance is 5 meters, the work done is 147 N * 5 m = 735 N/m = 735 J, making D the correct answer.

4 0
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Which formula is used to find fluctuation of the shape of body
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Answer:

varn=n1+1ehvkT–1

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This is Einstein's equation.

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So we can just substitute our data.
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