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Jet001 [13]
3 years ago
6

You hold block A with a mass of 1 kg in one hand and block B with a mass of 2 kg in the other. You release them both from the sa

me height above the ground at the same time. (Air resistance can safely be ignored as they fall.) Which of the following describes and explains the motion of the two objects after you release them?
A) Before A hits the ground before the block B. The same force of gravity acts on both objects, but because block A has a lower mass, it will have a larger acceleration
B) Block B hits the ground before the block A. The larger force on block B causes it to have a greater acceleration than block A
C) The two blocks hit the ground at the same time, because they both ackelerate at the same rate while falling. The greater gravitational force on block B is compersated giving it the same acceleration
D) The two blocks hit the ground at the same time because they move at the same constant speed while falling.The force on the blocks are irrelevent once they start falling
Physics
1 answer:
Dominik [7]3 years ago
5 0

ok so im not that good when it comes to physics but i think its C)

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In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
3 years ago
During the first 50 s a truck traveled at constant speed of 25 m/s. Find the distance that it is traveled.
Allushta [10]
Time=50s
speed=25m/s

Distance = speed×time
=25×50
=1250m

DISTANCE TRAVELLED IS =1250m

6 0
3 years ago
A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0
oksian1 [2.3K]

Answer:

Explanation:

Given

mass of box m=2\ kg

speed of box v=1.9\ m/s

distance moved by the box x=10\ cm

coefficient of kinetic friction \mu _k=0.66

Friction  force f_r=\mu_kN

f_r=0.66\times mg

f_r=0.66\times 2\times 9.8=12.936 \N

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2

\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2

3.61-1.2936=0.005\times k

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3 0
3 years ago
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How does mining affect the environment
weqwewe [10]
The environmental impact of mining affects, nesting habitats for wildlife. Wildlife either moves on, or dies.
hope this helps:)


5 0
3 years ago
Given uranium 235 go through an alpha, beta, beta, alpha, gamma, neutron and alpha. What do you have now?
Licemer1 [7]

Answer:

b. 88, 222

Explanation:

235U₉₂ ----→ Alpha --------→  231P₉₀  ----→- beta -----→ 231Q₉₁ ------→-beta -------→231R₉₂--------→-alpha ------→-227S₉₀ ------→ gamma -----→-227S₉₀ ----------→ neutron ------→-226T₉₀-----------→ alpha --------→222 X ₈₈

Atomic No is 88 , atomic weight = 222 .

3 0
3 years ago
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