What is the angular momentum of a 0.25 kg mass rotating on the end of a piece of

if a 3-kg object has a momentum of 33 kg��m/s, what's its velocity? a. 99 m/s b. 36 m/s c. 11 m/s d. 9.1 m/s

horsena [70]

M=3kg
p=33kg.m/s
p=m*v
v=p/m
=33/3
=11m/s
thus option (c)

6 0

2 years ago

An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the

Zarrin [17]

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

$x = v_0 t \frac{1}{2} at^2$

Where,

x= Displacement

$v_0$ = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

$L = \frac{1}{2} a t_1 ^2$

For the second cart

$2L \frac{1}{2} at_2^2$

When the tenth car is aligned the length will be 9 times the initial therefore:

$9L = \frac{1}{2} at_3^2$

When the tenth car has passed the length will be 10 times the initial therefore:

$10L = \frac{1}{2}at_4^2$

The difference in time taken from the second car to pass it is 5 seconds, therefore:

$t_2-t_1 = 5s$

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

$\frac{1}{2} = (\frac{t_1}{t_2})^2$

$t_1 = \frac{t_2}{\sqrt{2}}$

From the relationship when the car has passed and the time difference we will have to:

$(t_2-\frac{t_2}{\sqrt{2}}) = 5$

$t_2 (\sqrt{2}-1) = 3\sqrt{2}$

$t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Replacing the value found in the equation given for the second car equation we have to:

$\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Finally we will have the time when the cars are aligned is

$18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2$

$t_3 = 36.213s$

The time when you have passed it would be:

$20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2$

$t_4 = 38.172$

The difference between the two times would be:

$t_4-t_3 = 38.172-36.213 \approx 2s$

Therefore the correct answer is C.

4 0

2 years ago

An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the —z direction. The electric field has an amplitude of

Aloiza [94]

Answer:

a) 6.9*10^14 Hz

b) 9*10^-12 T

Explanation:

From the question, we know that

435 nm is given as the wavelength of the wave, at the same time, we also know that the amplitude of the electric field, E(max) has been given to be 2.7*10^-3 V/m

a)

To find the frequency of the wave, we would be applying this formula

c = fλ, where c = speed of light

f = c/λ

f = 3*10^8 / 435*10^-9

f = 6.90*10^14 Hz

b) again, to find the amplitude of the magnetic field, we would use this relation

E(max) = B(max) * c, magnetic field amplitude, B(max) =

B(max) = E(max)/c

B(max) = 2.7*10^-3 / 3*10^8

B(max) = 9*10^-12 T

c) and lastly,

1T = 1 (V.s/m^2)

6 0

2 years ago

The ph balance is maintained by which organ?

Lena [83]

The Kidneys And Lungs.

7 0

2 years ago

Suppose a truck has a momentum of 40,120 kg • and a mass of 1,180 kg. What is the truck’s velocity?

gladu [14]

Answer:

The velocity of the truck is 34 m/s.

Explanation:

The momentum (p) of an object is the product of its mass (m) and its velocity (v) which can be written as:

p = mv

Here in this problem, we are given a truck which has a momentum of 40120 kg and a mass of 1180 kg and we are to find its velocity.

So substituting the given values in the above formula to find the velocity of the truck.

Truck's velocity = $\frac{40120}{1180}$ = 34 m/s.

3 0

2 years ago