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ollegr [7]
3 years ago
8

A 1000kg truck traveling at 108km/h skids 10m before it stops. What is the magnitude of the frictional force acting on the car?

Physics
1 answer:
Leokris [45]3 years ago
3 0

Given that,

Mass of truck = 1000 kg

Speed = 108 km/h = 30 m/s

Distance = 10 m

We need to calculate the acceleration

Using equation of motion

v^2=u^2+2as

Where, v = final speed

u = initial speed

s = distance

Put the value into the formula

(0)^2=30^2+2\times a\times 10

a=-\dfrac{30^2}{2\times10}

a= -45 m/s^2

We need to calculate the magnitude of the frictional force acting on the truck

Using formula of force

F=ma

Where, m = mass of truck

a = acceleration

Put the value into the formula

F=1000\times(-45)

F= -45000\ N

Negative sign shows the opposite direction of motion.

Hence, The magnitude of the frictional force acting on the truck is 45000 N.

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Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner(v_1)=3.1 m/s

speed of second runner(v_2)=4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner t=\frac{295.6}{4.65}=63.56 s

time required by first runner to reach finish line=\frac{250}{3.1}=80.64 s

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0
4 years ago
A large spool in an electrician's workshop has 65 m of insulation-coated wire coiled around it. When the electrician connects a
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Answer:

40.34\ \text{m}

Explanation:

L_1 = Length of wire = 65 m

I_1 = Initial current = 1.8 A

I_2 = Final current = 2.9 A

We know

R\propto \dfrac{1}{I}

and

R\propto L

\dfrac{V}{I}\propto L\\\Rightarrow L\propto \dfrac{1}{I}

so

\dfrac{L_2}{L_1}=\dfrac{I_1}{I_2}\\\Rightarrow L_2=\dfrac{I_1}{I_2}L_1\\\Rightarrow L_2=\dfrac{1.8}{2.9}\times 65\\\Rightarrow L_2=40.34\ \text{m}

The length of the wire remaining on the spool is 40.34\ \text{m}.

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If the object on the right gained mass so that it had as much mass as the object on the left, how would the gravitational force
just olya [345]
The object would stay constant
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Waves in a fish bowl jostled by the Thingamajigger move to the sides at an average velocity of 0.50 m/s. If they occur once ever
Colt1911 [192]

Answer:

0.125 m

Explanation:

In this problem, we have:

v = 0.50 m/s is the average velocity of the wave

T = 0.25 s is the period of the wave

We can find the frequency of the wave, which is equal to the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{0.25 s}=4 Hz

The problem is asking us to find the distance between two crests of the wave: this is equivalent to the wavelength. The wavelength is related to the average velocity and the frequency by the formula:

\lambda=\frac{v}{f}

Substituting the numerical values, we find

\lambda=\frac{0.5 m/s}{4 Hz}=0.125 m

4 0
3 years ago
Please do all of i will give you brainlest and thanks to best answer plz do it right
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Answer:

winter solstice i think

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