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Sergio039 [100]
3 years ago
8

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

ide the building at a constant speed of 1.20m/s. If you wish to drop an egg on your professor's head, how far from the building should the professor be when you release the egg? Assume that the egg is in free fall.
Physics
1 answer:
nekit [7.7K]3 years ago
3 0

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

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A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through
Svet_ta [14]

Answer:

The speed is \sqrt{2}v_{0}.

(a) is correct option.

Explanation:

Given that,

Potential difference V= V_{0}

Speed v = v_{o}

If it were accelerated instead

Potential difference V'=2V_{0}

We need to calculate the speed

Using formula of initial work done on proton

W = q V

We know that,

\Delta W=\Delta K.E

q V=\dfrac{1}{2}mv^2

Put the value into the formula

q V_{0}=\dfrac{1}{2}mv_{0}^2

v_{0}^2=\dfrac{2qV_{0}}{m}....(I)

If it were accelerated instead through a potential difference of 2 V_{0}, then it would gain a speed will be given as :

Using an above formula,

v_{0}'^2=\dfrac{2qV_{0}}{m}

Put the value of V_{0}

v_{0}'^2=\dfrac{2q\times2V_{0}}{m}

v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}

v_{0}'=\sqrt{2}v_{0}

Hence, The speed is \sqrt{2}v_{0}.

6 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
kiruha [24]
I believe the answer is up and down :)
7 0
3 years ago
A block of mass
Gennadij [26K]

(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

<h3>Work done by the applied force</h3>

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

<h3>Work done by the normal force</h3>

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

<h3>Work done force of gravity</h3>

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

<h3> Work done by the net force on the block</h3>

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

6 0
2 years ago
If a force of 12 N is applied to an object
Varvara68 [4.7K]

Answer:

<h3>The answer is 3 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{12}{4}  \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

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Question 8 of 10
Alika [10]
B. Venus hope this helps have a great day
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