Momentum quiz for physics

How can the frequency with which the direction of a current changes be regulated?

sesenic [268]

Answer:

In an inductive circuit, when frequency increases, the circuit current decreases and vice versa.

Explanation:

5 0

2 years ago

Intermolecular forces hold

blagie [28]

I believe Intermolecular forces hold, <span>molecules, ions, and atoms? But I would see if that doesn't sound familiar check it with a site or something?</span>

3 0

2 years ago

A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is

Rudik [331]

Answer:

$a=16.2m/s^{2}$

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

$F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}\\$

When fixed,F=o

Hence

$masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}$

The value of the acceleration is 16.2m/s^2

7 0

2 years ago

43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a

olga2289 [7]

Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

$P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J$

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J$

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

$\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N$

The frictional force that acts on the sliding bear is 378.47166 N

5 0

2 years ago

Read 2 more answers

Let A be the second to last digit and let B be the last two digits of your 8-digit student ID. Example: for 20245347, A = 4 and

Andrej [43]

Answer:

7.35 m/s

Explanation:

Using y - y' = ut - 1/2gt², we find the time it takes the ball to fall from the 1.2 m table top and hit the floor.

y' = initial position of ball = 1.2 m, y = final position of ball = 0 m, u = initial vertical velocity of ball = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground.

So, substituting the values of the variables into the equation, we have

y - y' = ut - 1/2gt²

0 - 1.2 m = (0 m/s)t - 1/2(9.8 m/s²)t²

- 1.2 m = 0 - (4.9 m/s²)t²

- 1.2 m = - (4.9 m/s²)t²

t² = - 1.2 m/- (4.9 m/s²)

t² = 0.245 s²

t = √(0.245 s²)

t = 0.49 s

Since d = vt where d = horizontal distance ball moves = 3.6 m, v = horizontal velocity of ball = unknown and t = time it takes ball to land = 0.49 s.

So, d = vt

v = d/t

= 3.6 m/0.49 s

= 7.35 m/s

Since the initial velocity of the ball is 7.35 m/s since the initial vertical velocity is 0 m/s.

It is shown thus V = √(u² + v²)

= √(0² + v²)

= √(0 + v²)

= √v²

= v

= 7.35 m/s

7 0

2 years ago