Ask question

Ask question

All categories

3 years ago

7

A man can use a snowblower to move snow in 5 minutes. If he moves it with a shovel, will take 20 minutes. He will use more power

with which process?

1 answer:

Neko [114]3 years ago

6 0

Answer:

the second process of using the the shovel to move the snow as it takes a longer time meaning more power is expended

You might be interested in

Explain why a person who has runny nose is unable to detect smells clearly.

Orlov [11]

If you have a runny nose then the mucus will block the path that smell goes through. So if you are trying to smell a flower then you will have a hard time because the mucus is in the way, or blocking the entrance path. Hope this helps:)

May I please have brainliest?

3 0

3 years ago

Read 2 more answers

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

3 years ago

A solenoid with an inductance of 8 mH is connected in series with a resistance of 5 Ω and an EMF forming a series RL circuit. A

monitta

Answer:

induced EMF = 240 V

and by the lenz's law direction of induced EMF is opposite to the applied EMF

Explanation:

given data

inductance = 8 mH

resistance = 5 Ω

current = 4.0 A

time t = 0

current grow = 4.0 A to 10.0 A

to find out

value and the direction of the induced EMF

solution

we get here induced EMF of induction is express as

E = - L $\frac{dI}{dt}$ ...................1

so E = - L $\frac{I2 - I1}{dt}$

put here value we get

E = - 8 × $10^{-3}$ $\frac{10 - 4}{0.2*10^{-3}}$

E = -40 × 6

E = -240

take magnitude

induced EMF = 240 V

and by the lenz's law we get direction of induced EMF is opposite to the applied EMF

5 0

3 years ago

Introduction to Symbolic Answers, Part B, Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.

lesantik [10]

2cos2(o)-1 is the answer

4 0

3 years ago

A beam of light strikes one face of a window pane with an angle of incidence of 30.0°. The index of refraction of the glass is 1

natka813 [3]

Answer:

(a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

Explanation:

Given that,

Angle of incidence = 30.0°

Index of reflection of glass = 1.52

(a). We need to calculate the angle of refraction for the ray inside the glass

Using snell's law

$\dfrac{\sin i}{\sin r}=\mu$

$\sin r=\dfrac{\sin i}{\mu}$

Put the value into the formula

$\sin r=\dfrac{\sin 30}{1.52}$

$r=\sin^{-1}(0.329)$

$r=19.26^{\circ}$

(b). We know that,

The incident ray and emerging ray is equal then the ray will be parallel.

We need to prove that the rays in air on either side of the glass are parallel to each other

Using formula for emerging ray

$\dfrac{\sin e}{\sin r}=\mu$

$\sin e=\sin r\times \mu$

Put the value into the formula

$\sin e=0.3289\times 1.52$

$e=\sin^{-1}(0.499)$

$e=29.9\approx 30^{\circ}$

So, $\sin i=\sin e$

This is proved.

Hence, (a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

6 0

3 years ago