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REY [17]
2 years ago
6

Calculate how fast the ball would be moving at the instant it leaves the projectile launcher of the spring is compressed by 3.75

cm. Use a value of k = 500 N/m for the spring constant, 10 g for the mass of the ball, and 75 g for the effective mass of the ball holder. Show your work.
Physics
1 answer:
Mnenie [13.5K]2 years ago
6 0

Answer:

V = 8.34m/s

Explanation:

Given that

1/2ke^2 = 1/2mv^2 ......1

Where e = 3.75cm = (3.75/100)m

e = 0.0375m

K = 500 N/m

m = 10g = 10/1000

= 0.01kg

Substitute the values into equation 1

0.5×500×(0.0375)^2 = 0.5×0.01×v^2

250×0.001395 = 0.005v^2

0.348 = 0.005v^2

v^2 = 0.348/0.005

v^2 = 69.6

V = √69.6

V = 8.34m/s

The ball launches at the speed of V = 8.34m/s

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Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a
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Solution :

a). B at the center :

     $=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

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Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

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3 0
2 years ago
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A lion chasing a zebra sprints 200 meters in ten seconds what is the lions average speed
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3 years ago
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zimovet [89]
At surface,
v = kq/r

And potential energy of an electron is given by,
PE = -ev = -ekq/r

At escape velocity,
PE + KE = 0.
Therefore,
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1/2mv^2 = ekq/r
v = Sqrt [2ekq/mr], where v = escape velocity, e = 1.6*10^-19 C, k = 8.99*10^9 Nm^2/C^2, m = 9.11*10^-31 kg, r = 1.1*10^-2 m, q = 8*10^-9 C

Substituting;
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