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fiasKO [112]
3 years ago
11

A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed

is doubled, what happens to the tension in the string?
Physics
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

Explanation:

The tension in the string is the centripetal force. This force is given by

F = \dfrac{mv^2}{r}

m is the mass, v is the velocity and r is the radius.

It follows that F \propto v^2, provided m and r are constant.

When v is doubled, the new force, F_1, is

F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F

Hence, the tension in the string is quadrupled.

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A short current element dl⃗ = (0.500 mm)j^ carries a current of 4.80 A in the same direction as dl⃗ . Point P is located at r⃗ =
sammy [17]

Answer:

Magnetic field in Tesla: 1.65*10^−10 T,   0,   3.09*10^−10 T

Explanation:

Given data:

i = 4.8 A ,

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dBz = 3.09 * 10^-10 T

Attached below is the detailed solution

7 0
3 years ago
A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte
vaieri [72.5K]

Answer:

Part a)

H = 44.1 m

Part b)

y = 13.48 m

Part c)

d = 8.86 m

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

t = \frac{v sin\theta}{g}

now we have

3 = \frac{vsin\theta}{9.8}

v sin\theta = 29.4 m/s

Now maximum height above ground is given as

H = \frac{v^2sin^2\theta}{2g}

H = \frac{29.4^2}{2(9.8)}

H = 44.1 m

Part b)

Height of the fence is given as

y = (vsin\theta) t - \frac{1}{2}gt^2

y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)

y = 13.48 m

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

x = v_x t

97.5 = v_x (5.5)

v_x = 17.72 m/s

now total time of flight is given as

T = 3 + 3 = 6 s

so range is given as

R = v_x T

R = (17.72)(6)

R = 106.4 m

so the distance from the fence is given as

d = 106.4 - 97.5

d = 8.86 m

7 0
3 years ago
A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto
sesenic [268]
<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float=l\times b\times h=5.2\times 2.4\times h=12.48h

Density of water=10^3kg/m^3

Weight of water displaced by barge=Buoyant force=-Weight of horse

Volume\;of\;water\times density\;of\;water\times g=410\times g

12.48h\times 1000=410

h=\frac{410}{12.48\times 1000}=0.03 m

1 m=100 cm

0.03 m=0.03\times 100=3cm

Hence, the depth of barge float=3 cm

<h2 />
4 0
3 years ago
Read 2 more answers
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