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bazaltina [42]
3 years ago
11

what better force is required to accelerate a car at a rate of 2 m/s^2 if the car has a mass of 3000 kg?

Physics
1 answer:
Marizza181 [45]3 years ago
4 0
Acceleration = 2m/s^2
mass = 3000kg

Force = mass × acceleration
= 2 × 3000
= 6000

So, the force required is 6,000N
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A man can lift a mass of 200kg onThe surface of the earth. what is the amount of mass he can lift on the surface of the moon?
Arlecino [84]

Answer:

1,211.1 kg.

Explanation:

the force of gravity is less on the moon than on earth, so if the man can lift 200kg on earth, he could lift a greater amount on the moon because there is less resistance from gravity.

To know the amount of mass he can lift on the moon, we first need to know the amount of weight that is equivalent to those 200kg here on earth. This because the weight of the object is equal to the force that must be applied to lift it, and that force is applied by the man and it will be the same here and on the moon.

We calculate weight using the formula:

w=mg

where w is the weight of the object (the force with which the earth attracts the object) m is the mass and g the acceleration of gravity.

so

w=200g

for earth the acceleration due to gravity is:  g=9.81m/s^2

thus:

w=(200kg)(9.81m/s^2)\\w=1962N

now we use this value to calculate the mass he can lift on the moon, since for the moon g=1.62m/s^2.

we use the same equation, w =mg substituting w = 1962N and g=1.62m/s^2:

w=mg\\\\1962N=m(1.62m/s^2)\\\\m=\frac{1962N}{1.62m/s^2}\\\\ m=1,211.1kg

he can lift 1,211.1 kg.

You can also find the result using the approximate value of the acceleration of gravity on the moon as g/6, where g is the acceleration on earth.

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4 years ago
A venturi meter is a device for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at a speed v2 thro
yarga [219]

Answer:

V_2=6.17\ m/s

Q=0.49\ m^3/s

Explanation:

Given that

A_1=0.03m^2

A_2=0.08m^2

We know that from continuity equation

Q=A_1V_1=A_2V_2

So

A_1V_1=A_2V_2

0.3V_1=0.08V_2

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Now from energy equation

P_1+\dfrac{1}{2}\rho V_1^2=P_2+\dfrac{1}{2}\rho V_2^2

P_2-P_1=\dfrac{1}{2}\rho V_1^2-\dfrac{1}{2}\rho V_2^2

140=\dfrac{1}{2}\times 1.2\times 7.11\times V_2^2-\dfrac{1}{2}\times 1.2\times V_2^2

V_2=6.17\ m/s

Q=A_2V_2

Q=6.17 x 0.08

Q=0.49\ m^3/s

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