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Mrac [35]
2 years ago
14

Calculate the mass of a planet that exerts a 1.5 X 10 13 N force on a 5.4 x 10

Physics
2 answers:
astraxan [27]2 years ago
8 0

Answer:

umm I also need help sorry

Explanation:

help..

kodGreya [7K]2 years ago
8 0

Answer:

1.07 X 10^15 kg

Explanation:

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Conductors allow electric charges to move freely

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3 years ago
A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

7 0
2 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
2 years ago
A EXAMPLE OF WHEN BOTH PHYSICAL CHANGE AND CHEMICAL CHANGE OCCUR 3 EXAMPLES
kvasek [131]

Answer:

I know 1, that is in the case of a burning of a candle.

Explanation:

5 0
3 years ago
Please help me! Uniform acceleration problem sheet:
Brrunno [24]

From the calculation, the value of the acceleration is 5.8 m/s^2.

<h3>What is uniform acceleration?</h3>

The term uniform acceleration refers to a situation in which the velocity increases by equal amounts in equal time intervals.

Given the fact that the car started from rest and reached a velocity of  780.34 mph or 348.84 m/s in 1 minute of 60 seconds;

v = u + at

a = v/t

a = 348.84 m/s/ 60 seconds

a = 5.8 m/s^2

Learn more about acceleration:brainly.com/question/12550364?

#SPJ1

7 0
2 years ago
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