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3 years ago

The heat from the bonfire is transferred to the students hands through the process of???

Physics

2 answers:

soldi70 [24.7K]3 years ago

5 0

I believe it would be radiation.
Hope it helps!

Allisa [31]3 years ago

3 0

I believe it is Convection, then again I'm not in college.

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A circular loop of wire 75 mm in radius carries a current of 113 A. Find the (a) magnetic field strength and (b) energy density

Roman55 [17]

The magnetic field strength is 9.47 ×10⁻⁴ T

The energy density at the center of the loop is 0.36 J/m³

<h3>Calculating Magnetic field strength & Energy density </h3>

From the question, we are to find the magnetic field strength

The magnetic field strength of a loop can be calculated by using the formula,

$B = \frac{\mu_{0} I}{2R}$

Where B is the magnetic field strength

$\mu_{0}$ is the permeability of free space $(\mu_{0}=4\pi \times 10^{-7} \ N/A^{2})$

$I$ is the current

and R is the radius

From the give information,

$R = 75 \ mm= 75 \times 10^{-3} \ m$

and $I = 113 \ A$

Putting the parameters into the formula, we get

$B = \frac{4\pi \times 10^{-7} \times 113}{2 \times 75 \times 10^{-3} }$

$B = 9.47 \times 10^{-4} \ T$

Hence, the magnetic field strength is 9.47 ×10⁻⁴ T

Now, for the energy density

Energy density can be calculated by using the formula,

$u_{B} = \frac{B^{2} }{2\mu_{0} }$

Where $u_{B}$ is the energy density

Then,

$u_{B}= \frac{(9.47\times 10^{-4} )^{2} }{2 \times 4\pi \times 10^{-7} }$

$u_{B} = 0.36 \ J/m^{3}$

Hence, the energy density at the center of the loop is 0.36 J/m³

Learn more on Magnetic field stregth & Energy density here: brainly.com/question/13035557

7 0

2 years ago

Assuming the earth is a uniform sphere of mass M and radius R, show that the acceleration of fall at the earth's surface is give

Keith_Richards [23]

Explanation:

The weight of an object on the surface of the earth is equal to the gravitational force exerted by the earth on the object.

$W=F_G$

$mg = G \dfrac{mM}{R^2}$

which gives us an expression for the acceleration due to gravity <em>g</em> as

$g = G\dfrac{M}{R^2}$

At a height h = R, the radius of a satellite's orbit is 2R. Then the acceleration due to gravity $g_h$ at this height is

$mg_h = G \dfrac{mM}{(2R)^2}= G \dfrac{mM}{4R^2}$

Simplifying this, we get

$g_h= G \dfrac{M}{4R^2} = \dfrac{1}{4} \left(G \dfrac{M}{R^2} \right) = \dfrac{1}{4}g$

3 0

3 years ago

A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.

N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

$\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2$

(c) Moment of inertia :

The net torque acting on it is, $\tau=I\alpha$ , I is the moment of inertia

Also, $\tau=Fr$

So,

$I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2$

Hence, this is the required solution.

3 0

3 years ago

b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force

Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

$= force \times radius$

$= f \times r$

And,

Torque is also

$= moment\ of\ inertia \times angular\ acceleration$

$= I \times \alpha$

So,

We can say that

$f \times r = I \times \alpha$

$f \times 0.05 = 0.2 \times 2$

0.05f = 0.4

f = 8 N

We simply applied the above formulas

8 0

4 years ago

There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. A pilot fl

sattari [20]

Answer:

2.0 s, 200 m

Explanation:

Time to hit the ground depends only on height. Since the plane is at the same height, the weight lands at the same time as before, 2.0 s.

Since the plane is going twice as fast, the weight will travel twice as far (ignoring air resistance). So it will travel a horizontal distance of 200 m.

5 0

4 years ago

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