Ask question

Ask question

All categories

3 years ago

7

To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant a

ngular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.6 mm.
If the thrower takes 1.0 s to complete one revolution, starting from rest, what will be the speed of the discus at release?

1 answer:

yawa3891 [41]3 years ago

6 0

Answer:

Explanation:

Time taken to complete one revolution is called time period.

So, Time period, T = 1 s

Diameter = 1.6 mm

radius, r = 0.8 mm

Let the angular speed is ω.

The relation between angular velocity and the time period is

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

ω = 2 x 3.14 = 6.28 rad/s

The relation between the linear velocity and the angular velocity is

v = r x ω

v = 0.8 x 10^-3 x 6.28

v = 0.005 m/s

You might be interested in

14. Which one of the following pictures shows the object that is the most dense? *

rewona [7]

Answer:

B

Explanation:

Density is about how closely compact molecules are. (^-^)

5 0

2 years ago

A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

3 0

3 years ago

Question 1 of 10

Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field $E$ at a distance $R$ from a charge $Q$ is given by

$E = k\dfrac{Q}{R^2}$

where $k = 9*10^9Nm/C$ is the coulomb's constant.

Now, in our case

$R = 0.0075m$

$Q = 0.0035C$ ;

therefore,

$E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}$

$\boxed{E = 5.6*10^{11}N/C.}$

which is choice C from the options given<em> (at least it resembles it).</em>

6 0

3 years ago

What is the kinetic theory of matter? How does it relate to the motion of molecules?

Charra [1.4K]

Answer:

Explanation:

Kinetic energy is energy that an object has because of its motion. The Kinetic Molecular Theory explains the forces between molecules and the energy that they possess. This theory is based on three theories about matter. Matter is composed of small particles (atoms, molecules, and ions).

7 0

2 years ago

Sladkaya [172]

Answer:
I think the answer is
C) iron nails are attracted towards all materials

8 0

2 years ago