All categories

3 years ago

What is the role of equations in this course?

1 answer:

4 0

That will depend on which course you're talking about. It will be a minor role in, say, Maritime Law or Comparitive Religion, but a major one in, say, Particle Physics or Linear Algebra.

You might be interested in

A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull

sdas [7]

In spring mass system we know that angular frequency is given as

$\omega = 2\pi f$

f = 8.38 Hz

$\omega = 2\pi(8.38)$

$\omega = 52.65 rad/s$

now we know that speed of SHM at its extreme position is given by

$v = A\omega$

here we know that

A = 17.5 cm

$v = 0.175 (52.65)$

$v = 9.21 m/s$

so maximum speed is 9.21 m/s

7 0

3 years ago

When an object of weight w is suspended from the center of a massless string?

igomit [66]

I attached a picture of the diagram associated with this question.

Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ

The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ

4 0

3 years ago

Read 2 more answers

A 0.01 kg spring toy is compressed 0.02 m and released vertically. The toy is measured to reach 0.25 m in the air. Determine the

Scorpion4ik [409]

Answer:

122.5 N/m

Explanation:

According to the law of conservation of energy, if there is no air resistance or frictional forces, the initial elastic potential energy of the spring toy is entirely converted into gravitational potential energy when the toy reaches the highest point.

Therefore, we can write:

$\frac{1}{2}kx^2=mgh$

where the term on the left is the initial elastic potential energy while the term on the right is the gravitational potential energy, and where

k is the spring constant

x = 0.02 m is the compression of the spring

m = 0.01 kg is the mass of the toy

h = 0.25 m is the height reached by the toy

$g=9.8 m/s^2$ is the acceleration due to gravity

Solving for k,

$k=\frac{2mgh}{x^2}=\frac{2(0.01)(9.8)(0.25)}{(0.02)^2}=122.5 N/m$

8 0

2 years ago

A contestants spin a wheel when it is their turn in a game show. One contestant gives the wheel an initial angular speed of 3.40

guajiro [1.7K]

Answer:

4.62 s

Explanation:

We are given that

Initial angular speed, $\omega=3.4 rad/s$

$\theta=1\frac{1}{4} rev=\frac{5}{4}\times 2\pi=2.5\pi rad$

$\omega'=0$

$\omega'^2-\omega^2=2\alpha \theta$

Substitute the values

$0-(3.4)^2=2\times 2.5\pi \alpha$

$\alpha=\frac{-(3.4)^2}{2\times 2.5\pi}=-0.736 rad/s^2$

$\omega'=\omega+\alpha t$

$0=3.4-0.736 t$

$-0.736t=-3.4$

$t=\frac{-3.4}{-0.736}=4.62 s$

Hence, the wheel takes 4.62 s to come to rest.

3 0

3 years ago

Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artifici

wariber [46]

Answer:

Hi myself Shrushtee.

Explanation:

Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artificial gravity, human growth is stunted and biological functions break down. An effective way to create artificial gravity is through the use of a rotating enclosed cylinder, as shown in the figure. Humans walk on the inside edge of the cylinder, which is sufficiently large (diameter of 2235 meters) that its curvature is not readably noticeable to the inhabitants. (The space station in the figure is not drawn to the scale of the human.) Once the space station is rotating at the necessary speed, how many minutes would it take the space station to make one revolution?

The distance traveled by the man in one revolution is simply the circumference of the space station, C = 2p R. From this result, you should be able to deduce the time it takes for the space station to sweep out a complete revolution.

<h2>Please mark me as brainleist</h2>

7 0

2 years ago