Select all that apply so it’s not just one answer. please and thanks :))

To do your homework correctly you should (5 points)

pychu [463]

Answer:

C according to me

Explanation:

ITS HARD

4 0

3 years ago

Read 2 more answers

A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. S

Black_prince [1.1K]

Answer:

. The loop is pushed to the right, away from the magnetic ﬁeld

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

8 0

2 years ago

An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s

gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

$Q = ms\Delta T + mL$

here we have

$Q = 40(950)(680 - 32) + 40(450 \times 10^3)$

$Q = 24624 kJ + 18000 kJ$

$Q = 42624 kJ$

Since this is the amount of aluminium per hour

so power required to melt is given by

$P = \frac{Q}{t}$

$P = \frac{42624}{3600} kW$

$P = 11.84 kW$

Since the efficiency is 85% so actual power required will be

$P = \frac{11.84}{0.85} = 13.93 kW$

Part B)

Total energy consumed by the furnace for 30 hours

$Energy = power \times time$

$Energy = 13.93 kW\times 30 h$

$Energy = 417.9 kWh$

now the total cost of energy consumption is given as

$R = P \times 20 \frac{Cents}{kWh}$

$R = 417.9 kWh\times 20 \frac{cents}{kWh}$

$R = 8357.6 Cents$

3 0

2 years ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L) is defined as:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0

2 years ago

Determine the linear velocity of an object with an angular velocity of 5.9 radians per second at a distance of 12 centimeters fr

Andre45 [30]

The linear velocity of a rotating object is the product of the angular velocity and the radius of the circular motion. Angular velocity is the rate of the change of angular displacement of a body that is in a circular motion. It is a vector quantity so it consists of a magnitude and direction. From the problem, the angular velocity is 5.9 rad per second and the radius is given as 12 centimeters. We calculate as follows:

Linear velocity = angular velocity (radius)
Linear velocity = 5.9 (12 ) = 70.8 cm / s

The linear velocity of the body in motion is 70.8 centimeters per second or 0.708 meters per second.

7 0

2 years ago