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3 years ago

10

If WEIGHT depends on gravity and the pull of gravity

1 answer:

deff fn [24]3 years ago

4 0

Answer:

The weight will be more on planet B because greater is the gravity more is the weight of a body

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You are on the roof of the physics building, 46.0 above the ground . Your physics professor, who is 1.80 tall, is walking alongs

nekit [7.7K]

There is one mistake in the question as unit of height of building is not given.So I assume it as meter.The complete question is here

You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 m tall, is walking alongside the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor’s head, where should the professor be when you release the egg? Assume that the egg is in free fall.

Answer:

d=3.67 m

Explanation:

Height of building=46.0 m

First we need to find time taken by egg to reach 1.80 m above the surface

So to find time use below equation

$S=vt+\frac{1}{2} gt^{2}\\ (46.0-1.80)m=(om/s)t+\frac{1}{2}(9.8m/s^{2} )t^{2}\\t=\sqrt{\frac{(46.0-1.80)m}{4.9} }\\ t=3.06s$

As velocity 1.20m/s is given and we have find time.So we can easily find the distance

So

$distance=velocity*time\\d=v*t\\d=(1.20m/s)*(3.06s)\\d=3.67m$

3 0

3 years ago

A) A wire made from iron with a cross-section of diameter 0.800 mm carries a current of 14.0 A. Calculate the "areal current den

Veseljchak [2.6K]

Answer:

A) ρ= $1.74x10^{26}$

B) μ= $1.68x10^{29}\frac{electron}{m^3}$

C) v= $1.03x10^{-3}\frac{m}{s}$

D)e= $8.99x10^-9$

Explanation:

A)

The magnetic field can be find knowing the current is the charge per second

β= $\frac{14eC*s}{1.6x10^{-19}C}\\$

β= 8.75x10^{19}e*s

Electron density

ρ= $\frac{8.75x10^{19}}{\pi*0.400x10^{-3}m} = 1.74x10^{26}$

B)

μ= $\frac{7.86}{56.2}\frac{g}{cm^3} \frac{mol}{g}*6.022x10^{23}\frac{molecules}{mol} *2 \frac{electron}{molecule}$

μ $=1.68x10^{23} \frac{electron}{cm^3}= 1.68x10^{29} \frac{electron}{m^3}$

C)

The drift speed using last information found

$v= \frac{J}{n*q} \\v= \frac{14A}{\pi*((0.40x10^-3)^2)*1.68x10^29*1.60x10^-19)} = 1.03x10^-3(\frac{m}{s} )$

D)

To compared the random thermal motion and the current's drift speed

$e=\frac{1.03x10^-3}{1.15x10^5} = 8.99x10^-9$

8 0

3 years ago

A driver of a car enters a new 110 km/h speed zone on the highway. The driver begins to accelerate immediately and reaches 110 k

levacccp [35]

Answer:

30Km/h

Explanation:

acceleration is the change of speed in a given time so when we substract the accelerations we can know how much the car goes per an hour

3 0

3 years ago

When mantle rocks near the radioactive core are heated, they become less dense than the cooler, upper mantle rocks. These warmer

avanturin [10]

C.convection cells is the right answer

4 0

3 years ago

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A sound recording is made of a train as it blows its whistle for 10 seconds. The pitch of the sound appears to be decreasing dur

____ [38]

Answer:

the actual answer is C i did it on usatestprep

Explanation:

7 0

3 years ago

Read 2 more answers