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Phantasy [73]
1 year ago
10

A ball on the end of a rope is moving in a vertical circle near the surface of the earth. Point A is at the top of the circle; C

is at the bottom. Points B and D are exactly halfway between A and C. Which one of the following statements concerning the A boy is whirling a stone at the end of a string around his head. The string makes one complete revolution every second, and the tension in the string is FT. The boy increases the speed of the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions per second. What happens to the tension in the sting?
Physics
1 answer:
mrs_skeptik [129]1 year ago
3 0

Answer:

Tension in the string will increase

Explanation:

As we know that tension in the string at any angle with the vertical is given as

T - mgcos\theta = m\omega^2 R

now we have

T = mgcos\theta + m\omega^2 R

also we know that

angular speed of the stone is directly depending on the time period of the motion

so it is given as

\omega = \frac{2\pi}{T}

since the frequency of the revolution is increased from n = 1 rev/s to 2 rev/s

so the angular speed would be doubled

So here we can say that

tension in the string will increase when we will increase the frequency of revolution.

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Gnesinka [82]

Answer:

It increased by a factor of 3.

Explanation:

The gravitational potential energy of an object is given by

U=mgh

where

m is the mass

g is the gravitational acceleration

h is the heigth of the object relative to some reference point (for instance, the ground)

As we see from the formula, the gravitational potential energy is directly proportional to the mass, m: therefore, if the mass of the cylinder is increased by a factor 3, then the gravitational potential energy will also increase by a factor 3.

6 0
1 year ago
A ball is thrown through the air.What condition(s) would enable the ball to continue in its state of motion?
Aleksandr-060686 [28]
I think that the answer is c but I’m not sure
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Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward
KATRIN_1 [288]

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal  =30 degree

Density \rho \ of\  water= 1000 kg/m3

 Drag force FD is given as

F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A

        = 0.5\times 1000\times 4.32\times  1.2\times 1.3

Drag force = 14422.2 N acting opposite to the motion

As cable made angle  of 30 degree with horizontal  thus horizontal component is take into action to calculate drag force

TCos30 = F_D

T = \frac{F_D}{cos30}

T =\frac{ 14422.2}{cos 30}

T = 16653.32 N

7 0
1 year ago
A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-di
Nataly [62]

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

F . t = m(v_f - v_i)

so here we know that object is at rest initially so we have

80 (12) = 80( v_f - 0)

v_f = 12 m/s

Now for next 6 s the force decreases to ZERO linearly

so we can write the force equation as

F = 80 - \frac{40}{3} t

now again by same equation we have

\int F .dt = m(v_f - v_i)

\int (80 - (40/3)t) dt = 80(v_f - 12)

80 t - \frac{40t^2}{6} = 80(v_f - 12)

put t = 6 s

480 - 240 = 80(v_f - 12)

v_f = 12 + 3

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6 0
1 year ago
A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (t
Mekhanik [1.2K]

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature T_1=323K

Volume V_1=V\ and\ V_2=\frac{V}{2}

As there is no heat transfer so this is an adiabatic process

For and adiabatic process TV^{\gamma -1}=constant

Here \gamma =1.4

So T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}

T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1

T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K

4 0
2 years ago
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