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attashe74 [19]
3 years ago
7

Suppose a planet has twice the mass of the earth and twice the earth's radius. What is the acceleration of an object in free fal

l on this planet?
2.5 m/s2

4.9 m/s2

9.8 m/s2

4.9 × 106 m/s2

...?
Physics
1 answer:
ira [324]3 years ago
6 0
We use the equation:

g = G m / r2 
where
<span>G=constant universal gravitational = 6.67 x 10^-11 n m^2/kg^2 </span>
<span>m=planet mass=5.9736 x 10^24 kg (twice=11.9472) </span>
<span>r radium planet=6.372 km (twice=12.744) 
</span>
<span>g= 6.67 x 10^-11 n m^2/kg^2 x 11.9472 10^24 kg/(12.744.000m)^2 </span>
<span>g=4.90 m/s^2 (1/2 of Earth gravity)  <------- second option</span>
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An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.
SCORPION-xisa [38]

Answer: 6.175 km

Explanation:

from the question, we have the following

velocity of the automobile = 95 km/g

velocity of the train = 75 km/h

length of the train = 1.30 km

since the automobile and the train are moving in the same direction, we need to find the velocity of the car relative to the train which will be their difference in speed = 95 - 75 = 20 km/h

we need to find the time it takes the automobile to overtake the train using the formula time = distance / speed , with the distance being the length of the train.

time (t) = 1.3 / 20

= 0.065 hour

now we can find the distance traveled by the automobile using the the time taken for it to overtake the train and the speed of the automobile.

therefore, distance = speed x time

     distance = 95 x 0.065 =6.175 km

5 0
4 years ago
Ok sooo I need help I don’t understand this lol
Burka [1]

Answer:

I dont either but I'm sorry and hopefully you figured it out

3 0
3 years ago
A uniform metre rule of mass 100g balance the 40cm mark when a mass x is placed at the 20cm mark
vivado [14]

Answer:

X = 50 g

Explanation:

Please see attached photo for explanation.

From the attached photo,

Anticlock–wise moment = X × 20

Clockwise moment = 100 × 10

Anticlock–wise moment = clockwise moment

X × 20 = 100 × 10

X × 20 = 1000

Divide both side by 20

X = 1000 / 20

X = 50 g

Therefore, the value of X is 50 g

6 0
3 years ago
In what direction does an object move when affected by an unbalanced force
gayaneshka [121]

Answer:

An unbalanced force means that there are more than one forces with different directions or magnitudes affecting the object. In this case, the vector addition of the forces should be done. The vector addition will result a net force. The object will move in the direction of the net force.

8 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
4 years ago
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