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docker41 [41]
3 years ago
12

two objects are thrown from the top of a tall building and experience no appreciable air resistance. one is thrown up and the ot

her is thrown down, both with the same initial speed. what are their speeds when they hit the street?
Physics
2 answers:
NNADVOKAT [17]3 years ago
5 0

Answer:

Final speeds are equal to each other.

Explanation:

Both objects experiment a change in their velocities because of gravity. Let model each object by following equations:

Object A - Throw Up

v = -\sqrt{v_{o}^{2}+2\cdot g \cdot h}

Object B - Throw Down

v = -\sqrt{v_{o}^{2}+2\cdot g \cdot h}

Where h is the height difference between the top of the building and the street. Let consider upward speed positive.

Final speeds are equal to each other.

Iteru [2.4K]3 years ago
3 0

Answer:

The two objects are traveling at the same speed.

Explanation:

Neglecting air resistance, an object that is thrown up from the top of a tall building has the same speed as the second object thrown down from the top of the same tall building since the initial speed is the same.

The object thrown up is not traveling faster neither is the object thrown down traveling faster.  

Therefore, the two objects will have the same speed when they hit the ground but their time of landing might be different.

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Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

x=3.27\ m

5 0
3 years ago
How much force does it take to accelerate a 2000 kg car at 1m/s^2
Dmitry [639]

Answer:

2000 N

Explanation:

F=ma

m=2000 kg

a=1m/s^2

F=(2000 kg)(1m/s^2)

F=2000 N

4 0
3 years ago
Can you hear it? In the cartoon space rocket, why do you think you would not be able to hear the whoosh of the rocket engine in
Brums [2.3K]

Answer:

Because space is a void with no air flow

Explanation:

7 0
3 years ago
A deuterium atom is a hydrogen atom with a neutron added to its nucleus. Approximate the binding energy of this nucleus, given t
Mademuasel [1]

Answer:

c. 2 MeV.

Explanation:

The computation of the binding energy is shown below

= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV

= 2 MeV

As 1 MeV = (1 u) c^2

hence, the binding energy is 2 MeV

Therefore the correct option is c.

We simply applied the above formula so that the correct binding energy could come

And, the same is to be considered

8 0
3 years ago
If you are traveling around a curve at a constant speed will your velocity change?
marta [7]

Velocity means [ (speed) and (direction) ].

If you're traveling around a curve, then your direction is
always changing.  So your velocity is always changing,
even if your speed isn't.
4 0
3 years ago
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