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2 years ago

If you double the velocity of an object, you increase its kinetic energy by a factor of what ?

1 answer:

Ivan2 years ago

3 0

Energy is proportional to the square of the velocity. So if you double the velocity you increase the Kinetic Energy by a factor of four

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A police officer uses a radar gun to determine the speed of a car. A specialized radar gun uses ultraviolet light to determine t

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Answer:

A transverse and D electromagnetic

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Explanation:

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2 years ago

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Jesse wants to know how well a particular brand of car wax protects his car from dirt. What is the independent variable ?

miv72 [106K]

The brand of car wax

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2 years ago

What is the polarization ice cap how long is yes

lubasha [3.4K]

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3 years ago

94. The diagram shows the orbit of a satellite

Valentin [98]

Answer:

7.65x10^3 m/s

Explanation:

The computation of the satellite's orbital speed is shown below:

Given that

Earth mass, M_e = 5.97 × 10^24 kg

Gravitational constant, G = 6.67 × 10^-11 N·m^2/kg

Orbital radius, r = 6.80 × 10^6m

Based on the above information

the satellite's orbital speed is

V_o = √GM_e ÷ √r

= √6.67 × 10^-11 × 5.97 × 10^24 ÷ √6.80 × 10^6

= 7.65x10^3 m/s

4 0

3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

2 years ago