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Elina [12.6K]
3 years ago
14

A car has a mass of 1000 kg and accelerates at 2 meters per second per second. what is the magnitude of the net force exerted on

the car
Physics
2 answers:
9966 [12]3 years ago
4 0
F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N
olga_2 [115]3 years ago
3 0

Answer:

The magnitude is 2000N

Explanation:

To solve this question we are going to use the second Newton law.

The second Newton law states that the acceleration of an object is directly proportional to the net force which acts on the object and inverse proportional to the mass of the object ⇒

a=\frac{F}{m}

Which can also be written as :

F=m.a (I)

Where F is the force

Where m is the mass of the object and where ''a'' is the acceleration.

In the problem, we know that

m=1000kg

a=2\frac{m}{s^{2}}

If we replace in the equation (I) :

F=(1000kg).(2\frac{m}{s^{2}})=2000(kg.\frac{m}{s^{2}})

The unit kg.\frac{m}{s^{2}} is defined as N which is called Newton. N is an unit of force.

Finally, the net force exerted on the car is 2000N

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Distance travelled = 200 metre

Time taken = 24 second

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6 0
3 years ago
Two astronauts of mass 100 kg are 2 m apart in outer space. What is the
fredd [130]

The force of gravity between the astronauts is 1.67\cdot 10^{-7}N

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between them

In this problem, we have two astronauts, whose masses are:

m_1 = 100 kg\\m_2 = 100 kg

While the separation between the astronauts is

r = 2 m

Substituting into the equation, we can find the gravitational force between the two astronauts:

F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N

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4 0
3 years ago
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You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body
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A) The change in internal chemical energy is 1.15\cdot 10^7 J

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

P=Fv

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

P=(80)(8.0)=640 W

The energy output is related to the power by the equation

P=\frac{E}{t}

where:

P = 640 W is the power output

E is the energy output

t = 30 min \cdot 60 = 1800 s is the time elapsed

Solving for E,

E=Pt=(640)(1800)=1.15\cdot 10^6 J

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

E=1.15\cdot 10^7 J

Here we want to find the time t' needed to convert an amount of chemical energy of

E'=3.8\cdot 10^5 J

So we can setup the following proportion:

\frac{t}{E}=\frac{t'}{E'}

And solving for t',

t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min

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pantera1 [17]

Answer:

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Answer:

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