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3 years ago

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A skillet is removed from an oven whose temperature is 450°F and placed in a room whose temperature is 70°F. After 5 minutes, th

e temperature of the skillet is 400°F. How long will it be until its temperature is 150°F?

1 answer:

scZoUnD [109]3 years ago

3 0

Answer:

attached below

Explanation:

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You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l

sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

$h= v_it + \frac{1}{2}gt^2\\$

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}$

<u>vi = 9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

$h= v_it + \frac{1}{2}gt^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

$17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0$

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

4 0

2 years ago

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

3 years ago

One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00∘C by ice–water mixture. The ro

Hitman42 [59]

Answer:

153.0815W/m.K

Explanation:

Heat transferred in phase is changed is expressed as:

$Q=\pm mL$ (m-mass, Q-heat, L-Latent heat of phase change)

Latent heat is the heat required to change the phase of 1kg of the material.

#The rate of heat flow(by conduction) per unit time:

$H=\frac{\bigtriangleup Q}{\bigtriangleup t}=kA\frac{T_H-T_C}{L}\\\\L_f=334\times10^3J/kj\\\\H_{ice}=\frac{Q_{ice}}{t}=\frac{m_{ice}L_f}{t}$ #Heat flowing through melting ice.

$H_{ice}=\frac{5.0\times10^-^3kg)334\times10^3J/kg)}{10\times60s}\\\\=2.7833J/s$

To solve for k:

$H=\frac{\bigtriangleup Q }{\bigtriangleup t}\\\\=kA\frac{T_H-T_C}{L}$ $H=\frac{\bigtriangleup Q }{\bigtriangleup t}\\\\=kA\frac{T_H-T_C}{L}$ $H=\frac{\bigtriangleup Q }{\bigtriangleup t}\\\\=kA\frac{T_H-T_C}{L}\\\\k=\frac{H}{A\frac{T_H-T_C}{L}}=\frac{HL}{A(T_H-T_C)}$

$=\frac{(2.7833J/s\times 0.55)}{1.00\times10^-^4m^2\times 100K}\\\\=153.0815W/m.K$

The thermal conductivity k of the metal is 153.0815W/m.K

8 0

3 years ago

Which best describes how geothermal energy is used to make electricity?. . A.. Rocks are split to release energy that drives a s

In-s [12.5K]

Answer is b that is Heat energy from below the ground converts water to steam to drive a steam turbine attached to an electrical generator.. .

6 0

3 years ago

a) If a proton moved from a location with a 5.0 V potential to a location with 7.5 V potential, would its potential energy incre

tekilochka [14]

Answer:

A. The potential energy of the positively charge proton from 5v to 7.5v will increase because in an electric field work is done on the proton to move it to a point of higher potential

B. when an electron is moved from 7.5v to 5v it loses potential energy due to the charges lost so as to bring it to a point of lower potential

Explanation:

3 0

3 years ago