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3 years ago

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A mechanic uses a jack to lift up a car. He exerts a force of 11,000 N at a distance of 3m from the axis of rotation. How much t

orque was put on the car?

1 answer:

pshichka [43]3 years ago

6 0

Answer:

<h2>The amount of torque put on the car is 33,000Nm</h2>

Explanation:

Formula for calculating torque is expressed as T = rFsin $\theta\\$ where;

r is the radius of the of the arm of the jack = 3m

F is the force exerted = 11000

$\theta\\$ is the angle of rotation = 90°

On substituting;

$T = 3*11000sin90^{o} \\T = 3*11000 (sin90^{o} =1)\\T = 33000Nm$

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John has a boat that will travel at the rate of 15 kph in still water. he can go upstream for 35 km in the same time it takes to

MA_775_DIABLO [31]

Let F = the downstream speed of the water.

<span>Then the boat's upstream speed is: 15 - F </span>
<span>The boat's downstream speed is: 15 + F </span>

<span>Assume both the journeys mentioned take T hours, then using "speed x time = distance" we get: </span>

<span>Downstream journey: (15 + F)T = 140 </span>
<span>Upstream journey: (15 - F)T = 35 </span>

<span>Add the two formulae together: </span>

<span>(15 + F)T + (15 - F)T = 140 + 35 </span>

<span>15T + FT + 15T - FT = 175 </span>

<span>30T = 175 </span>

<span>T = 35/6 </span>

<span>Use one of the equations to find F: </span>

<span>(15 + F)T = 140 </span>

<span>15 + F = 140/T </span>
<span>F = 140/T - 15 </span>
<span>F = 140/(35/6) - 15 </span>
<span>F = 24 - 15 </span>
<span>F = 9 </span>

<span>i.e. the downstream speed of the water is 9 kph </span>

<span>Therefore, the boat's speed downstream is 15 + F = 15 + 9 = 24 kph.
the answer is: *24kph*</span>

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4 years ago

When 224-nm light falls on a metal, the current through a photoelectric circuit is brought to zero at a stopping voltage of 1.84

OLga [1]

Answer:

3.71 eV

Explanation:

λ = Wavelength of light = 224 nm = 224 x 10⁻⁹ m

c = speed of electromagnetic wave = 3 x 10⁸ m/s

V₀ = stopping potential = 1.84 volts

W₀ = Work function of the metal = ?

Using the equation

$\frac{hc}{\lambda } = eV_{o} + W_{o}$

$\frac{(6.63\times 10^{-34})(3\times 10^{8})}{224\times 10^{-9} } = (1.6\times 10^{-19})(1.84) + W_{o}$

$W_{o}$ = 5.94 x 10⁻¹⁹

$W_{o}$ = 3.71 eV

7 0

3 years ago

Clasifica los siguientes enunciados de acuerdo a si se describen movimientos

Alenkasestr [34]

Answer:

- translation

- rotation, traslation

- traslation, rotation

- vibrating

Explanation:

El movimiento de un cuerpo cae por su propio peso <u>traslación</u>.

El movimiento de las ruedas de una bicicleta al ser pedaleada <u>rotación, traslación</u>.

El movimiento de la Tierra alrededor de sol <u>traslación, rotación</u>.

El movimiento de la cuerda de una guitarra cuando se está tocando música <u>vibración</u>.

- - - - - - - - - - - - - - - - - - - - - - - - - - - -

The movement of a body falls under its own weight <u>translation</u>.

The movement of the wheels of a bicycle when being pedaled <u>rotation, translation.</u>

The movement of the Earth around the sun, <u>translation, rotation</u>.

The movement of a guitar string when playing music <u>vibrating</u>.

7 0

3 years ago

Why is static electricity worse in winter?

Ratling [72]

Because static electricity build up in our homes because we have the heaters on which suck moisture out of the air.

8 0

3 years ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago