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jeka94
2 years ago
5

Two metal balls are the same size, but one weighs twice as much as the other. The balls are dropped from the top of a two story

building at the same instant of time. The time it takes the balls to reach the ground below will be:
Physics
1 answer:
seropon [69]2 years ago
4 0

The time taken by the metal balls of the same size but different weight, to reach the ground will be the same.

Reason behind:

Two metal balls are the same size, but one weighs twice as much as the other. The balls are dropped from the top of a two-story building at the same instant of time. It is required to find the time taken by the balls to reach the ground.

In free fall, all objects experience the same acceleration owing to gravity when they are close to the earth.

g=9.8 \text{ m/s}^2 is the measure of gravitational acceleration.

Because of this, the two metal balls are the same size but have different masses. The air resistance for both balls will be the same due to their similar sizes. Let, g' be the acceleration in the presence of air resistance. The balls are both discharged at once.

From a height of h, both balls descend due to gravity G'.

Therefore, the time taken by both balls is:

\begin{aligned}&s=u t+\frac{1}{2} g t^{2} \\&u=0, s=h \\&t=\sqrt{\frac{2 h}{g}}\end{aligned}

Therefore, the time is independent of mass. Thus the time taken by both the balls, will be about the same.

Learn more about time taken by metal ball to reach the ground here,

brainly.com/question/22719691

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Explanation:

  • Mass(m)= 20kg
  • Acceleration (a)= 5m/s²
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We know that,

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6 0
1 year ago
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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

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A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
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Answer:

a) 1.2*10^{-3}cos(1.25t)

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where we have used the values given by the information of the problem for N B and A.

b)

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emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

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3 years ago
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mario62 [17]

Answer:

V=\dfrac{3}{2}\ m/s

Explanation:

Two identical bodies are sliding toward each other on a frictionless surface.

Initial speed of body 1, m₁ = 1 m/s

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They collide and stick.

We need to find the speed of the combined mass. Let V is the speed of the combined mass.

Using the conservation of momentum.

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We have, m₁ = m₂ = m

m\times 1+m\times 2=(m+m)V\\\\m+2m=2m\times V\\\\3m=2mV\\\\V=\dfrac{3}{2}\ m/s

So, the speed of the combined mass is \dfrac{3}{2}\ m/s.

6 0
2 years ago
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