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scoray [572]
3 years ago
15

A car traveled at an average speed of 60 MPhil for two hours how far did it travel

Physics
2 answers:
masha68 [24]3 years ago
8 0
<span>120 miles is the answer

</span>
astra-53 [7]3 years ago
5 0
Here, Speed = 60 Miles/hour
Time = 2 hour

Substitute it in the expression: Distance = Speed * Time
Distance = 60*2 = 120 Km

So, your final answer is 120 Km

Hope this helps!
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What is the horse power of an electric motor which can do by 1250 joule of work in 5 seconds​
Ksju [112]
1250 J in 5 sec= 250 Joule(s) per second (1250/5 0

250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)

250 Watts output = 250/0.65 efficiency = 384 Watts input

1 Horsepower = 732 Watts

Motors 1 Horsepower and under are made in certain step sizes like

3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.

3/4 Horsepower is 549 Watts

1/2 Horsepower is 366 Watts

so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
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2 years ago
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babymother [125]

your answer is make up artist

4 0
3 years ago
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MariettaO [177]
1. Encoding Information
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6 0
3 years ago
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A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a
Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

    (v^2)/(R) = g×μ

               μ = (v^2)/(R×g)

                  =  ((25)^2)/((100)×(9.8))

                  = 0.64

Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

4 0
3 years ago
A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the
andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

F = mg

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

\sum \tau = 0

Regarding the forces we have,

3Mg-1mg=0

Re-arrange to find M,

M = \frac{m}{3}

M = \frac{50}{3}

M = 16.67Kg

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0
3 years ago
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