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2 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?

1 answer:

Liono4ka [1.6K]2 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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The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from

Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

$L_T=L_S+10log(n)$

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

$L_T=60+10 log(5)$

= 66.98 db

7 0

1 year ago

If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should

liubo4ka [24]

Emily throws the ball at 30 degree below the horizontal

so here the speed is 14 m/s and hence we will find its horizontal and vertical components

$v_x = 14 cos30 = 12.12 m/s$

$v_y = 14 sin30 = 7 m/s$

vertical distance between them

$\delta y = 4 m$

now we will use kinematics in order to find the time taken by the ball to reach at Allison

$\delat y = v_y *t + \frac{1}{2} at^2$

here acceleration is due to gravity

$a = 9.8 m/s^2$

now we will have

$4 = 7 * t + \frac{1}{2}*9.8 * t^2$

now solving above quadratic equation we have

$t = 0.44 s$

now in order to find the horizontal distance where ball will fall is given as

$d = v_x * t$

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

$d = 12.12 * 0.44 = 5.33 m$

so the distance at which Allison is standing to catch the ball will be 5.33 m

8 0

2 years ago

Dylan has two cubes of iron. The larger cube has twice the mass of the smaller cube. He measures the smaller cube. Its mass is 2

liubo4ka [24]

Answer:

The volume of the larger cube is 5.08 g/cm³.

Explanation:

Given that,

Mass of smaller cube = 20 g

Density of smaller cube $\rho= 7.87 g/cm^2$

Dylan has two cubes of iron.

The larger cube has twice the mass of the smaller cube.

$M_{l}=2m_{s}$

Density is same for both cubes because both cubes are same material.

The density is equal to the mass divided by the volume.

$\rho=\dfrac{m}{V}$

$V=\dfrac{m}{\rho}$

Where, V = volume

m = mass

$\rho=density$

We need to calculate the volume of smaller mass

The volume of smaller mass

$V_{s}=\dfrac{m_{s}}{\rho_{s}}$

$V_{s}=\dfrac{20}{7.87}$

$V_{s}=2.54\ cm^3$

Now, We need to calculate the volume of large cube

$V_{l}=\dfrac{m_{l}}{\rho_{l}}$

$V_{l}=\dfrac{2\times20}{7.87}$

$V_{l}=5.08\ g/cm^3$

Hence, The volume of the larger cube is 5.08 g/cm³.

8 0

1 year ago

Aaron Agin nodded off while driving home from play practice this past Sunday evening. His 1500-kg car hit a series of guardrails

Inessa [10]

Answer: 6.48m/s

Explanation:

First, we know that Impulse = change in momentum

Initial velocity, u = 19.8m/s

Let,

Velocity after first collision = x m/s

Velocity after second collision = y m/s

Also, we know that

Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).

5700 = 1500(19.8 - x)

5700 = 29700 - 1500x

1500x = 29700 - 5700

1500x = 24000

x = 24000/1500

x = 16m/s

Also, at the second guard rail. impulse = ft, so that

Impulse = 79000 * 0.12

Impulse = 9480

This makes us have

Impulse = m(x - y)

9480 = 1500(16 -y)

9480 = 24000 - 1500y

1500y = 24000 - 9480

1500y = 14520

y = 14520 / 1500

y = 9.68

Then, the velocity decreases by 3.2, so that the final velocity of the car is

9.68 - 3.2 = 6.48m/s

5 0

1 year ago

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

8 0

1 year ago