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3 years ago

9

What is the force required accelerate a 5.0 kg object at 3.0m/s^ 2?

1 answer:

NeX [460]3 years ago

6 0

The equation to find force is f=ma. So, if you plug in the information that you have you'll get F=5x3 and that'll equal F=15N

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What is the net work done on the 20kg block while it moves the 4 meters?

Vinil7 [7]

Answer:

The answer to your question is 784.8 J. None of your answer, did you forget some information?

Explanation:

Data

mass = 20 kg

distance = 4 m

work = ?

Formula

Work = force x distance

Force = mass x gravity

Process

1.- Calculate the weight of the block

Weight = 20 x 9.81

Weight = 196.2 N

2.- Calculate the work done

Work = 196.2 x 4

Work = 784.8 J

5 0

3 years ago

Consider an electron that is 10-10 m from an alpha particle (9 = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electri

Studentka2010 [4]

Answer:

a) $E=2.88*10^{11}N/C$

b) $E=1.44*10^{11}N/C$

c) $F=4.61*10^{-8}N$

Explanation:

We use the definition of a electric field produced by a point charge:

$E=k*q/r^2$

<u>a)Electric Field due to the alpha particle:</u>

$E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(10^{-10})^2=2.88*10^{11}N/C$

<u>b)Electric Field due to electron:</u>

$E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(10^{-10})^2=1.44*10^{11}N/C$

<u>c)Electric Force on the alpha particle, on the electron:</u>

The alpha particle and electron feel the same force but with opposite direction:

$F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(10^{-10})^2=4.61*10^{-8}N$

4 0

2 years ago

A police car is at rest parallel to the highway and measures the speed of cars. It sends the signal with a frequency of 1200 Hz,

masha68 [24]

Answer:

a) The car was moving at a speed of $29.167\ m.s^{-1}$

b) The negative sign of $v_o$ denotes that the observer is coming towards the police car which is the source of the sound.

c) $f_o=1283.33\ Hz$

Explanation:

Given:

original frequency of the source, $f=1200\ Hz$

speed of the source, $v_s=0\ m.s^{-1}$

velocity of the obstacle car be, $v_o$

speed of sound, $s=350\ m.s^{-1}$

observed frequency, $f_o=1100\ Hz$

<u>Using the equation from the Doppler's effect:</u>

$\frac{f_o}{f} =\frac{(s+v_o)}{(s-v_s)}$

$\frac{1100}{1200} =\frac{(350+v_o)}{350-0}$

$v_o=-29.167\ m.s^{-1}$

a)

The car was moving at a speed of $29.167\ m.s^{-1}$

b)

The negative sign of $v_o$ denotes that the observer is coming towards the police car which is the source of the sound.

c)

Now when, $v_s=50\ m.s^{-1}$

Then, $f_o=?$

Using the Doppler's eq.:

$\frac{f_o}{1200} =\frac{(350+(-29.167))}{(350-50)}$

$f_o=1283.33\ Hz$

3 0

2 years ago

Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in

iVinArrow [24]

Answer:

The value is $P_2 = 40.54 \ psla$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 14\ psla$

The initial temperature is $T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K$

The final temperature is $T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K$

Generally the equation for adiabatic process is mathematically represented as

$PT^{\frac{\gamma}{1- \gamma} } = Constant$

=> $P_1T_1^{\frac{\gamma}{1- \gamma} } = P_2T_2^{\frac{\gamma}{1- \gamma} }$

Generally for a monoatomic gas $\gamma = \frac{5}{3}$

So

$14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }$

=> $14 * 283^{-2.5} =P_2 * 433^{-2.5}$

=> $P_2 = 40.54 \ psla$

8 0

2 years ago

A hockey puck has a coefficient of kinetic friction of μk = .35. If the puck feels a normal force (FN) of 5 N, what is the frict

alina1380 [7]

Answer:

The frictional force is $F_f = 1.75 \ N$

Explanation:

From the question we are told that

The coefficient of kinetic force is μk = 0.35

The normal force felt by the puck is $F_N = 5 \ N$

Generally the frictional force that acts on the puck is mathematically represented as

$F_f = \mu_k * F_N$

=> $F_f = 0.35 * 5$

=> $F_f = 1.75 \ N$

3 0

2 years ago