The current produced by electromagnetic induction is greater when...

Two blocks of the same mass but made of different material slide across a horizontal, rough surface and eventually come to rest.

Akimi4 [234]

Answer:

a. $\mathbf{F_{f_2} = 2 F_{f1}}$ , $\mathbf { since \ the \ force \ o f \ friction \ is \ represented \ as \ the \ slope \ for \ each \ of \ the \ two \ curves.}$

Explanation:

From the information given;

Using the work-energy theorem

ΔKE = W = $\mathbf{ F_f \times r}$

K = $\mathbf{ F_f \times r}$

∴

$\dfrac{K_1}{K_2} = \dfrac{F_{f1}}{F_{f2}} (\dfrac{r_1}{r_2})$

Since $K_1 = K_2$ and r_1 = 4, and r_2 = 2 (from the missing diagram which is attached below)

Then;

$1 = \dfrac{F_{f1}}{F_{f2}} (\dfrac{4 \ m}{2 \ m})$

$\mathbf{F_{f_2} = 2 F_{f1}}$

5 0

3 years ago

Power can be defined as?

Tasya [4]

B. How much work can be done in a given time. That’s why it’s measured occasionally I. “Horsepower.” It’s your ability to work fast and far.

7 0

3 years ago

A man standing in front of a mountain beats a drum at regular intervals. The rate of drumming is generally increased and he find

Likurg_2 [28]

I think it is d I hope this help you if not let me know if it is not right

3 0

3 years ago

1. What is the relationship between wavelength and the penetrating power of an electromagnetic wave?

Andreyy89

#1

As we know that energy of electromagnetic wave is given by

$E = \frac{hc}{\lambda}$

so here we know that penetrating power will directly depends on its energy and energy inversely depends on wavelength

So here we can say correct answer will be

C) The penetrating power decreases as the wavelength increases.

#2

Speed of sound is maximum in solids and minimum in gas

so here as ice melts into water the speed of sound must have to decrease

so correct answer will be

D) The speed of sound would decrease because sound travels faster through solids than liquids.

#3

mechanical waves required medium to travel while non mechanical waves do not require any medium to travel

so here correct answer will be

A) sound

6 0

3 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s