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4 years ago

Describe briefly difference between conduction, convection, and radiation.

Physics

1 answer:

Fed [463]4 years ago

3 0

The three types of heat transfer are conduction convection and radiation. Explanation:

Heat transfer happens only when there is said to be a temperature difference. (i.e) heat transfer occurs with a higher temperature region to a lower temperature region.
Heat is transferred usually via some medium in the case of convection and conduction whereas in the radiation no medium is required ( i.e it take place in a vacuum ).
Conduction is a type of heat transfer that is mainly a character for the conducting and semiconducting solid material.
Convection is the heat transfer method that usually takes place in the liquids and gases.
Radiation is the heat transfer process in which heat is transferred via infrared or electromagnetic waves.

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A slab of glass 8.0 cm thick is placed upon a printed page. If the refractive index of the glass is 1.5, how far from the surfac

7nadin3 [17]

Answer:

5.3 cm

Explanation:

This question is an illustration of real and apparent distance.

From the question, we have the following given parameters

Real Distance, R = 8.0cm

Refractive Index, μ = 1.5

Required

Determine the apparent distance (A)

The relationship between R, A and μ is:

μ = R/A

i.e.

Refractive Index = Real Distance ÷ Apparent Distance

Substitute values in the above formula

1.5 = 8/A

Multiply both sides by A

1.5 * A = A * 8/A

1.5A = 8

Divide both side by 1.5

1.5A/1.5 = 8/1.5

A = 8/1.5

A = 5.3cm

Hence, the letters would appear at a distance of 5.3cm

8 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

Calculate the change in velocity of a 0.070kg tennis ball hit by Serena with a force of 140 N over 0.020 s

Brilliant_brown [7]

V=at and a=F/m

140/.070 = 2000m/s^2

2000*.020 = 40m/s

The ball’s velocity increased by 40m/s.

6 0

3 years ago

The GPE of a 70kg person standing on a chair 1m off the ground is how many joules?

Artemon [7]

GPE I am assuming is gravitational potential energy. I'll denote it as U for simplicity.

U = mgy
U = (70kg)(9.81m/s^2)(1m) = 686.7J

U = 686.7J

Hope this helps!

8 0

4 years ago

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendu

erica [24]

Answer:

a) v = 16.57 m / s, b) a = 19.6 m / s², d) N = 1.76 10³ N, N / W = 3

Explanation:

This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

Em_f = K = ½ m v²

as they indicate that there is no friction

Em₀ = em_f

mgh = ½ m v²

v = $\sqrt{2gh}$

let's calculate

v = $\sqrt{2 \ 9.8 \ 14.0}$

v = 16.57 m / s

b) the centripetal acceleration has the formula

a = v² / r

a = 16.57² / 14.0

a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

N-W = m a

N - mg = m ar

N = m (g + a)

let's calculate

N = 60.0 (9.8 + 19.6)

N = 1.76 10³ N

the relationship with weight is

N / W = 1.76 10³/( 60 9.8)

N / W = 3

normal is three times greater than body weight

e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it

6 0

3 years ago