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2 years ago

I WILLL MARK YOUUUUUUU HELPPPP

Physics

2 answers:

Ksivusya [100]2 years ago

3 0

Answer:

10.75 kg

Explanation:

moon = (weight on earth/9.81m/s^2)*1.622m/s^2

Firlakuza [10]2 years ago

3 0

Answer:

weight of the moon = mass × gravity

= 65kg × 1•6

=104N/kg

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What is the net force acting on the buggy. ?The net force is pointing to the ?

Papessa [141]

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The A note has a frequency of 440 Hz. What is the wavelength if the speed of sound is 343 m/s?

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The wavelength of a wave (λ) is given by λ $= \frac{c}{f}$
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2 years ago

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A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter

kifflom [539]

Answer:

a). 87.5 mA or $87.5 x10^{-3}$ A

b). 1.78 $\frac{m}{s}$

Explanation:

$d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}$

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

$I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\ I=87.5 x10^{-3}$ A

b).

$I=n*abs (q)*V_{d}*A$

$A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2} \\A=5.309x10^{-6}$

$V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}$

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2 years ago

An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V0, and the

tamaranim1 [39]

Answer:

V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)

Explanation:

Attached is the full solution

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2 years ago

Newton's third law can be summarized as "every action has an equal and opposite reaction". In this problem, consider the action

scoundrel [369]

Answer:

Action force: Would be the force of your feet against the Earth given by the weight defined as:

$W = mg$

Where g is a constat who represent the gravity $g =9.8 m/s^2$ in Earth

Reaction force: Would be the force of the Earth pushing against your feet. And on this case is represented by the normal force defined as:

$N = \mu f_f$

Where $\mu$ represent the friction coefficient between the ground and the object.

And $f_f$ the friction force.

If we don't have any other forces involved in the y axis we can conclude that:

$W=N= mg$

And as we can see we have that Action force = Reaction force

So then the third Law of Newton is satisfied.

Explanation:

For this case we have this:

Action force: Would be the force of your feet against the Earth given by the weight defined as:

$W = mg$

Where g is a constat who represent the gravity $g =9.8 m/s^2$ in Earth

Reaction force: Would be the force of the Earth pushing against your feet. And on this case is represented by the normal force defined as:

$N = \mu f_f$

Where $\mu$ represent the friction coefficient between the ground and the object.

And $f_f$ the friction force.

If we don't have any other forces involved in the y axis we can conclude that:

$W=N= mg$

And as we can see we have that Action force = Reaction force

So then the third Law of Newton is satisfied.

7 0

3 years ago