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3 years ago

What is the velocity of a ball dropped from a height of 150 m when it hits the ground? Take the upward direction as positive.

Physics

1 answer:

djyliett [7]3 years ago

7 0

Answer:

The velocity of the ball when its hit the ground will be 54.22 m/sec

Explanation:

We have given height from which ball is dropped h = 150 m

Acceleration due to gravity $g=9.8m/sec^2$

As the ball is dropped so initial velocity will be zero so u = 0 m/sec

According to third equation of motion we know that $v^2=u^2+2gh$

$v^2=0^2+2\times 9.8\times 150$

$v=54.22m/sec$

So the velocity of the ball when its hit the ground will be 54.22 m/sec

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For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci

MAXImum [283]

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

$z = \frac{V_o}{i_o}$

now we will have

$z = \frac{125}{3.20} = 39.06 ohm$

Part b)

Now we also know that

$\frac{R}{z} = cos\theta$

$\theta = 0.982 rad = 56.3 degree$

now we have

$\frac{R}{39.06} = cos56.3$

$R = 21.7 ohm$

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3 years ago

The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

hoa [83]

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It is neccesary calculate the force:

$F=(m+m)*g*sin\theta$

Here, g = gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

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