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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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Hummingbirds may seem fragile, but their wings are capable of sustaining very large forces and accelerations. (Figure 1) shows d
gulaghasi [49]

Acceleration is mathematically defined as; ΔV/t.

<h3>What is acceleration?</h3>

Acceleration is defined as the ratio of the change in velocity to time. It is defined as the extent to which the velocity changes within a given time interval.

The question is incomplete so we can not arrive at the final answer. However, acceleration is mathematically defined as; ΔV/t.

Learn more about acceleration: brainly.com/question/2437624

5 0
2 years ago
An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 ​units/sec. What is the force on the
Dmitriy789 [7]

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2).  Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature.  This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

7 0
3 years ago
Calculate the work done on a Pressure / Volume (PV) isothermal expansion where 400 Pa and 0.08 volume in m3?
Ainat [17]
PV = 400 x 0.08 = 32 J

Hope this helps
7 0
3 years ago
A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he
zmey [24]

Answer:

851.33 N

Explanation:

Using newton;s equation of motion,

v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

Note: As the Diver touches the water,  u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

8 0
3 years ago
Joanne drives her car at a speed of 20 m/s. when she applied her breaks, a frictional force of 2000 N brought her car to a compl
Papessa [141]

Answer:

A) 1000 kg

Explanation:

vf = vi + at

0 = 20 + (a)(10)

a = -2.0 m/s^2

F = ma

2000 = (m)(2)

m = 1000 kg

8 0
1 year ago
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