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3 years ago

A satellite of mass 5460 kg orbits the Earth and has a period of 6520 s

Physics

1 answer:

Nikolay [14]3 years ago

8 0

Answer:

what if I do and b then someone else c I don't have enough time pls

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How long does it take lewis hamilton travelling at 156 mph to cover 16 miles?

Paladinen [302]

It takes him
t = 16 miles / 156 mph = 0.1 hours

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3 years ago

You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, it accelerates because t

zepelin [54]

You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, the average force on the jet is due to the catapult is mathematically given as

What is the average force on the jet is due to the catapult?

Generally, the equation for acceleration is mathematically given as

$a=\frac{vf^2-vi^2}{2s}\\\\\Therefore\\\\a=\frac{69.29^2-0^2}{2(90}\\\\a=26.673m/s^2$

In conclusion, The force

F=m*a

F=15100*26.673

F=40272.3N

F 402 KN

Read more about force

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2 years ago

A metaphysical poet is a writer whose

EastWind [94]

Not really sure but...

<span>A metaphysical poet is a writer whose </span>focus is on universal human experiences.
hope this helps!...

8 0

3 years ago

Which part of the electromagnetic spectrum have the highest level of photon energy

Pani-rosa [81]

The energy carried by one photon is directly proportional to its
frequency. So the photon energy is greatest for the electromagnetic
waves with the highest frequency / shortest wavelengths.

That's why when you get past visible light and on up through ultraviolet,
X-rays, and gamma rays, the radiation becomes dangerous ==> each
photon carries enough energy to tear electrons away from their atoms,
ripping molecules apart and damaging cells.

The photon with the highest energy is a gamma-ray photon.

4 0

3 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0

1 year ago