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3 years ago

Which of the following is the equation for impulse

Physics

1 answer:

anzhelika [568]3 years ago

6 0

Impulse is (force) x (time).

That's the last choice on the list. It could be 'D', or '4', or 'Δ', or 'ד' etc.

If you play with it for a while, you discover that impulse has the same units as momentum, which certainly gives you something to think about.

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"Compared to infrared radiation, does ultraviolet radiation have longer or shorter wavelengths? Does ultraviolet radiation have

balu736 [363]

Answer:Ultraviolet radiation has shorter wavelengths and higher energy than infrared radiation.

Explanation: Electromagnetic radiation radiations which have both electrical and magnetic properties,they can be transmitted through space or through a medium.

It includes Gamma radiation, infra-red, visible light, Ultraviolet radiation etc they occur with different wavelength, the lower the wavelength the higher the Energy dissipated per photon. According to their order of decreasing wavelength and increased energy they are classified as follows.

RADIO WAVE, MICRO WAVE, INFRA-RED, VISIBLE LIGHT, ULTRAVIOLET RAY, X-RAY, GAMMA RAYS.

5 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 $N = 50.9\:\text{N}$