3 years ago

If you push on a door for 12 seconds with a force of 53 N, how much impulse

Physics

1 answer:

Nataliya [291]3 years ago

5 0

Answer:

C. 636 kg•m/s

Explanation:

53(12) = 636 N•s or kg•m/s

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3. A motorbike travels 45 miles in 15 minutes, what is its speed?

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Hes going 180 mph
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3 years ago

Which statement best describes the energy changes that occur while a child is riding on a sled down a steep, snow-covered hill?

Tema [17]

<span>Kinetic energy increases and potential energy decreases.
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4 years ago

Read 2 more answers

Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher

siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

here we know that

$h\nu$ = energy of the photons incident on the metal

$W$ = minimum energy required to remove photons from metal

$\frac{1}{2}mv^2$ = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

$W = \frac{hc}{351 nm}$

here we know that

$hc = 1242 eV-nm$

now we have

$W = \frac{1242}{351} = 3.54 eV$

now by energy equation above when photon of 303 nm incident on the surface

$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

$4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2$

$(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2$

$8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2$

$v = 4.44 \times 10^5 m/s$

6 0

3 years ago

Imagine that a loudspeaker is producing a quiet tone with a low pitch. How will its vibrations change:

iVinArrow [24]

I believe it is A :) hope this helped

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3 years ago

What is the electric field 4.0 m from the center of the terminal of a Van de Graaff with a 7.10 mC charge, noting that the field

Jobisdone [24]

Answer:

3984875 N/C

Explanation:

Applying,

E = kq/r².................. Equation 1

Where E = Electric field, q = charge, r = distance, k = coulomb's constant.

From the question,

Given: q = 7.10 mC = 0.0071 C, r = 4.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 1

E = 0.0071(8.98×10⁹)/4²

E = 3984875 N/C

Hence the electric field is 3984875 N/C

5 0

3 years ago