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3 years ago

Please please please please help

Physics

2 answers:

aivan3 [116]3 years ago

4 0

Answer:

I am pretty sure its the second one but I could be wrong sorry if I am.

Explanation:

Nikitich [7]3 years ago

3 0

Second one i think yeah

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A 65 kg box is lifted by a person pulling a rope a distance of 15 meters straight up at a constant speed. How much Power is requ

Y_Kistochka [10]

Answer:

Power is 1061.67W

Explanation:

Power=force×distance/time

Power=65×9.8×15/9 assuming gravity=9.8m/s²

Power=3185/3=1061.67W

8 0

2 years ago

Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that

Leno4ka [110]

Answer:

8.829 m/s²

Explanation:

M = Mass of Earth

m = Mass of Exoplanet

$g_e$ = Acceleration due to gravity on Earth = 9.81 m/s²

g = Acceleration due to gravity on Exoplanet

$m=M-0.1M\\\Rightarrow m=0.9M$

$g_e=G\frac{M}{r^2}$

$g=G\frac{0.9M}{r^2}$

Dividing the equations we get

$\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2$

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

3 0

3 years ago

On a 100km track , a train travels the first 30km with a speed of 30km/h . How fast the train travel the next 70 km if the avera

nirvana33 [79]

Solution :-

Given :

Distance 1 = 30 km

Distance 2 = 70 km

We know that speed = distance/time

and, Average speed = total distance/total time taken

When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour

Average speed = 9distance 1 + distance 2)/(time 1 + time 2)

AS time 2 or t2 is time taken for the second part of the journey of 70 km

⇒ 40 = 100/(1 + t2)

⇒ 40 + 40t2 = 100

⇒ 40t2 = 100 - 40

⇒ 40t2 = 60

⇒ t2 = 60/40

⇒ t2 = 1.5

So, t2 or time taken to travel the second part of the journey is 1.5 hours.

Speed of the second part of the journey = distance 2/time 2

⇒ 70/1.5

⇒ 46.666 km/hr or 46.7 km/hr.

Hence the answer is = 46.666 km/hr or 46.7 km/hr.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

3 0

3 years ago

g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

r = $\frac{mv}{qB}$

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × $10^{-5}$ T, v = 121 m/s, Θ = $90^{0}$ (since it enters perpendicularly to the field), q = e = 1.6 × $10^{-19}$ C and m = 9.11 × $10^{-31}$ Kg.