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Alex787 [66]
2 years ago
15

a car of mass 800 kg is travelling at 12m/s the car accelerates over a distance of 240 m there force causing this acceleration i

s 200 newton determine the work done on the car and its final velocity car
Physics
1 answer:
alisha [4.7K]2 years ago
5 0

Answer:

See below

Explanation:

Force X distance = work

 200 N * 240 m = 48 000 N-m = 48 000 j

F = ma

200 = 800 a

a = 1/4 m/s^2

d = v t  + 1/2 a t^2

12 m/s t  + 1/2  1/4 m/s^2  t^2 = 240

                shows t = 17 sec  (via Quadratic Formula)

final velocity = 12 + at = 12 + 1/4(17) = 16.26 m/s

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At a certain altitude above the Earth's surface, the electric field has a magnitude of 135 V/m. How much energy is stored in 1.0
Paul [167]

Answer:

Explanation:

Given that,

Electric field E=135V/m

Energy stored in 1m³of air=?

The energy stored in an electric field is given as

u = ½ εo E²

Where

U is the energy stored

εo is permissivity and it value is 8.85×10^-12C²/N..m²

And E is the electric field

Then,

U=½×8.85×10^-12×135²

U=8.06×10^-8J/m³

Then, the energy stored in 1m³ of air is 8.06×10^-8 J/m³

4 0
3 years ago
One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet
Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

               \frac{G M m}{r^{2}}=m \omega^{2} r

The orbit’s period is given by,

               T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}

Where,

T_{e} = Earth’s period

T_{p} = planet’s period

M_{s} = sun’s mass

r_{e} = earth’s radius

Now,

             T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}

As, planet mass is equal to 0.7 times the sun mass, so

            T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}

Taking the ratios of both equation, we get,

             \frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}

            \frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}

           \frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}

Given T_{p}=9.5 \text { days } and T_{e}=365 \text { days }

          \frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}

         r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}

7 0
3 years ago
A ball is thrown at a launching angle of 52 o above the horizontal from one meter above the ground, with a velocity of 18 m/s.
Vladimir79 [104]

i dont really know, im sorry

5 0
3 years ago
The amount of work done by a heat engine equals the amount of thermal energy _____. added to the engine plus the waste heat adde
RideAnS [48]

added to the engine minus the waste heat

Explanation:

The amount of work done by a heat engine equals the amount of thermal energy added to the engine minus the waste heat.

Thermodynamics is used to measure the energy changes that accompanies physical and chemical process.

 In a system;

  Internal energy = Quantity of heat added - Work done on it

 internal energy is designated as ΔU

  quantity of heat as Q

   work done as W

                ΔU = Q - W

  so       W = Q  -   ΔU

some of the heat added added to the system goes as waste heat as no system is 100% efficient.

   the work done on a system goes in as thermal energy, internal energy and waste heat

learn more:

Thermodynamics brainly.com/question/3564634

#learnwithBrainly

3 0
3 years ago
Why does the satellite not fall while revolving the earth​
hoa [83]

Answer:

Satellites don't fall from the sky because they are orbiting Earth. Even when satellites are thousands of miles away, Earth's gravity still tugs on them. Gravity--combined with the satellite's momentum from its launch into space--cause the satellite go into orbit above Earth, instead of falling back down to the ground.

8 0
3 years ago
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