All categories

1 year ago

a ball on a string makes 30.0 revolutions in 14.4s, in a circle of radius 0.340m. what is its period.(unit=s)

Physics

1 answer:

STALIN [3.7K]1 year ago

7 0

Answer:

0.480 seconds

Explanation:

The period is the time for 1 revolution. Writing a proportion:

14.4 s / 30.0 rev = t / 1 rev

t = 0.480 s

The period is 0.480 seconds.

You might be interested in

Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for

g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0

2 years ago

A skateboarder rides swiftly up the edge of a bowl-shaped surface and leaps into the air. While in the air, the skateboarder fli

Katen [24]

Answer:

Option(a) is the correct answer to the given question .

Explanation:

The main objective of the angular momentum is evaluating however much the rotational movement as well as the angular velocity in the entity does have.The angular momentum is measured in terms of $kgm^{2 }\ / s$ .

In the given question the skateboarder rides quickly up the bottom of a bowl-shaped surface and climb into the air.it means it is rotational movement also it is not touching anything so it is angular momentum.
All the other option is incorrect because it is not follows the given scenario

3 0

1 year ago

robby skateboards 0.50 blocks to his friend's house in 1.2 minutes. what is his speed? 6.0 blocks/min 6.0 blocks/min, in the dir

Annette [7]

That would be 0.42 blocks per minute

4 0

1 year ago

Read 2 more answers

A particle of charge 2.3 ✕ 10−8 C experiences an upward force of magnitude 4.6 ✕ 10−6 N when it is placed in a particular point

Marysya12 [62]

<h2>Answer:</h2>

(a) +2 x 10² N/C (upwards)

(b) -2.2μN or -2.2 x 10⁻⁶N (downwards)

<h2>Explanation:</h2>

The force (F) acting on a particle of charge (Q) at a particular point is related to its electric field (E) by the following;

F = Q x E ----------------------(i)

This means that the force acting on the charged particle is the product of its charge and the electric field at that point.

<em>(a) Given</em>;

Q = charge of the particle = 2.3 x 10⁻⁸ C

F = force acting on the particle = 4.6 x 10⁻⁶N

<em>Substitute these values into equation (i) as follows;</em>

=> 4.6 x 10⁻⁶ = 2.3 x 10⁻⁸ x E

=> E = 4.6 x 10⁻⁶ ÷ 2.3 x 10⁻⁸

=> E = 2 x 10² N/C

Since the value is positive, the electric field is directed upwards.

Therefore, the electric field at that point is +2 x 10² N/C

<em>(b) If a charge of q = -1.1 x 10⁻⁸ is placed there, the force (F) acting is calculated as follows;</em>

<em>Substitute Q = q into equation (i) as follows;</em>

F = q x E

<em>Substitute the value of q and E = 2 x 10² N/C into the equation above as follows;</em>

F = -1.1 x 10⁻⁸ x 2 x 10²

F = -2.2 x 10⁻⁶ N

F = -2.2μN

Since the value of the force, F, is negative, it means it is directed downwards.

Therefore the force on the charge is -2.2μN

3 0

2 years ago

Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a

Alchen [17]

The height of the roof is <u>3.57m</u>

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

$s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2$

The length of the window s is given by,

$s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2$

The first drop is at the bottom and it takes 5t seconds to reach down.

The height of the roof h is the distance traveled by the first drop and is given by,

$h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m$

the height of the roof is 3.57 m

8 0

2 years ago

Read 2 more answers