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shtirl [24]
2 years ago
7

A car of mass 750kg accelerates away from traffic lights. At the end of the first 100m it has reached a speed of 12 m/s. During

this time, its engine provides an average forward force of 780 N, and the average force of friction on the car is 240 N. Calculate the increase in the car's kinetic energy at the end of the first 100 m
Physics
2 answers:
xenn [34]2 years ago
5 0

Answer:

the answer according to me is 54000j

Explanation:

im not sure why the force and friction are given inthe question because the the formula of ke is 1/mv^2

aliina [53]2 years ago
3 0

Answer:

Hruhriehrvdi8d8edeuebueufbrhddvuszu

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A load of mass 5kg is raised through a height of 2m. calculate the work done against (g=10mls)​
Illusion [34]

The work done against gravity is 100 J

Explanation:

The work done against gravity in order to lift an object is equal to the change in gravitational potential energy of the object:

W=mg\Delta h

where

m is the mass of the object

g is the acceleration of gravity

\Delta h is the change in height of the object

For the object in this problem, we have:

m = 5 kg

g=10 m/s^2

\Delta h = 2 m

Substituting into the equation,

W=(5)(10)(2)=100 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0
3 years ago
The Escape speed at the surface of a certain planet is twice that of the earth. what is its mass in unit of earth's mass?
VLD [36.1K]

Answer:

22Km/sec

Explanation:

6 0
2 years ago
Which term refers to energy derived from heat inside the earth
tino4ka555 [31]
I think the answer is Geothermal energy. 
3 0
3 years ago
A refrigeration cycle has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle.
DENIUS [597]

Answer:

The coefficient of performance for the cycle is 2.33.

Explanation:

Given that,

Output energy Q_{out}=1000\ Btu

Work done W_{cycle}=300\ Btu

We need to calculate the coefficient of performance

Using formula of  the coefficient of performance

COP=\dftrac{Q_{in}}{W_{cycle}}

We need to calculate the Q_{in}

W_{cycle}=Q_{out}-Q_{in}

Put the value into the formula

300=1000-Q_{in}

Q_{in}=300-1000

Q_{in}=700\ Btu

Now put the value of Q_{in} into the formula of COP

COP=\dfrac{700}{300}

COP=\dfrac{7}{3}=2.33

Hence, The coefficient of performance for the cycle is 2.33.

5 0
3 years ago
An 80-kg hiker climbs to the top of a tall hill and builds up 470,000 J of gravitational potential energy. How high did the hike
nekit [7.7K]

Answer:

599 meters is the answer rounded to the nearest whole number and 599.489795918 meters is the complete answer

Explanation:

to find gravitational potential energy you multiply mass x acceleration due to gravity (always 9.8 on earth) x hight

since we know the gravitational potential energy and want to find out the hight, we take the gravitational potential energy (470,000) and divide it by the product of acceleration due to gravity x mass (9.8 x 80)

so how high the hiker climbed is equal to 470,000 divided by (9.8 x 80)

hight = 470,000 / (9.8 x 80)

hight = 470,000 / 784

hight = 599.489795918 meters

as for rounding, if the decimal is less than 5 you round "down" and keep the current whole number, if the decimal is 5 or greater you round "up" and add 1 to get your new number

5 0
3 years ago
Read 2 more answers
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