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harkovskaia [24]
3 years ago
14

Can someone help me

Physics
1 answer:
lesya692 [45]3 years ago
6 0

Answer:

atleast i tried fella there it is

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You observe a distant galaxy. you find that a spectral line normally found in the visible part of the spectrum is shifted toward
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The galaxy is moving away from you.

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A massless, rigid board is placed across two bathroom scales that are separated by a distance of 1.71 m. A person lies on the bo
Nata [24]

Answer:

Let the weight of the person be W and be located at a distance 'a' from the left scale as shown in the figure

Since the body is in equilibrium we can use equations of statics to analyse the problem.

Taking Sum of Moments about A we have

316\times 1.71-W\times a=0\\\\

Taking Sum of Moments about B we have

475\times 1.71-W\times (1.71-a)=0\\\\

Solving the above 2 equations for W and 'a' we get

316\times 1.71=W\times a\\\\475\times 1.71-W\times 1.71=-Wa\\\\\therefore W=\frac{316+475}{1}=791N\\\\\therefore a=\frac{316\times 1.71}{791}=0.683m

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3 years ago
Weight of an object various replaces on the earth.why?​
german

Answer:

Weight change from place to place because it is a strong function of gravity. Recall that mass is universally constant. Hence weight varies becuas gravity varies. ... Other reason for variation of weight across the earth's surface are, rotation of the earth, altitude and local topography of the area.

5 0
3 years ago
Which of these objects has kinetic energy?
Damm [24]

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A ball moving through the air.

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The ball has momentum which is a form of kinetic energy.

I don't know if that is correct, but I hope it helps!!!!

3 0
3 years ago
Read 2 more answers
A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2
Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S  

2) t = 1 mu S

3) t = 50 mu S

1) 22.4 \mu C

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

C=\frac{Q_0}{V_0}

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

C=4.4\mu F = 4.4\cdot 10^{-6}F

V_0 = 5.1 V

Solving for Q_0,

Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) 20.0\mu C

To find the charge on the capacitor at any other time t, we use the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

where

Q_0 = 22.4 \mu C

t is the time

R=2.0 \Omega is the resistance

C=4.4\mu F is the capacitance

Therefore, at time t=1 \mu s, we have:

Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C

3) 0.08 \mu C

As before, we use again the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

However, here the time to consider is

t=50 \mu C

Substituting into the formula,

Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C

4 0
3 years ago
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