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3 years ago

Calculate the density of a solid cube that

1 answer:

4 0

Ok so if each side is 4.53 cm, we can multiply 4.53 x 4.53 x 4.53 to get the volume (since v= l x w x h). Density equals mass/volume, so

519 g/4.53 cm
114.57 g/cm^3 (since none of the units cancel)

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The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the w

Citrus2011 [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The magnetic field is $B_{net} = \frac{1}{4} * mT$

And the direction is $-\r k$

Explanation:

From the question we are told that

The magnetic field at the center is $B = 1mT$

Generally magnetic field is mathematically represented as

$B = \frac{\mu_o I}{2R}$

We are told that it is equal to 1mT

$B = \frac{\mu_o I}{2R} = 1mT$

From the first diagram we see that the effect of the current flowing in the circular loop is (i.e the magnetic field generated)

$\frac{\mu_o I}{2R} = 1mT$

This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

$B_1 = \frac{1}{2} \frac{\mu_o I}{2R}$

and for the larger semi-circular loop is

$B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

So the net magnetic field would be

$B_{net} = B_1 - B_2$

$= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

$=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}$

$=\frac{\mu_o I}{8R}$

$= \frac{1}{4} \frac{\mu_o I}{2R}$

Recall $\frac{\mu_o I}{2R} = 1mT$

$B_{net} = \frac{1}{4} * mT$

Using the Right-hand rule we see that the direction is into the page which is $-k$

3 0

3 years ago

Your own car has a mass of 2000 kg. If your car produces a force of 5000 N, how fast will it accelerate?

Sidana [21]

Answer:

2.5m/s²

Explanation:

Given parameters:

Mass of car = 2000kg

Force produced by the car = 5000N

Unknown:

Acceleration of the car = ?

Solution:

According to Newton's second law of motion, Force is a product of mass and acceleration.

Force = mass x acceleration

Now, insert the parameters and find the unknown;

5000 = 2000 x acceleration

Acceleration = $\frac{5000}{2000}$ = 2.5m/s²

7 0

3 years ago

How is lightning related to static electricity?

Naily [24]

Lighting is the static electricity stored in the clouds that is disposed to the earth.

8 0

3 years ago

Two objects are moving in the xy-plane. Object A has a mass of 3.2 kg and has a velocity = (2.3 m/s)i+ (4.2 m/s)j and object B h

prisoha [69]

The total momentum of the system is 2.14i + 21.27j.

A vector quantity with both direction and magnitude is momentum. Kg m/s (kilogram meter per second) or N s serve as its units (newton second).

The total starting momentum of a system must match the entire final momentum of the system since momentum is a conserved quantity. The overall momentum does not change.

The total momentum of the system is defined as follows:

As momentum is vector quantity and vectors can be added, so, the momentum of a system is given by

P = Pₓ + P'

where Pₓ is the x-component of momentum

P' is the y-component of the momentum

Also, we know that

P=mv

where m is mass

v is velocity

Thus,

P = Pₓ + P'

P = m₁vₓ + m₂v'

vₓ is the x-component of the velocity

v' is the y- component of the velocity

Given, m₁= 3.2kg

m₂ = 2.9kg

Now,

P = 3.2 (2.3i + 4.2j) + 2.9 (-1.8i +2.7j)

P = (7.36i + 13.44j) + (-5.22i + 7.83j)

P = 2.14i + 21.27j

Thus, the total momentum of the given system is 2.14i + 21.27j.

Learn more about the momentum here:

brainly.com/question/4956182

#SPJ1

8 0

1 year ago

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago