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Zolol [24]
3 years ago
10

What's the number of protons

Physics
1 answer:
ad-work [718]3 years ago
5 0

It would be equal to atomic number

OR mass number - No. of neutrons

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What is the kinetic energy of a 200 kg satellite as it follows a circular orbit of radius 8x106m around the earth?
motikmotik
Given the equation for the Speed of a Satellite

v = SqRt{Gravitational Constant}{Mass of Earth} divided by the radius given in your problem

we have:


(square root whole term on right side)

v = G Me
———
r


so. (6.67x10^-11)(5.97x10^24)
___________________
(8.0x10^6)


v = 7055 m/s (which is reasonable)


so utilize the Kinetic Energy Formula


KE = 1/2mv^2


KE = 1/2(200)(7055)^2


KE = 4.977x10^9 J


4 0
3 years ago
Read 2 more answers
A student who takes a multiple-choice test by reading the stem of each item, generating the correct response before looking at t
Ludmilka [50]

Answer:

(A) a heuristic

Explanation:

A heuristic:It is a reasoning strategy to find answers, make judgement about any something."

It is possible to choose between the options given, we have the following interpretation as; Interpretation: "A heuristic: finding answers is a thinking technique, evaluating something." Mechanism: In the heuristic approach the method is to find solutions or answers to a question by choosing the right and optimal compositions.

5 0
3 years ago
A jet aircraft is traveling at 262 m/s in hor-
NeTakaya

Solution :

Speed of the air craft, $S_a$ = 262 m/s

Fuel burns at the rate of, $S_b$ = 3.92 kg/s

Rate at which the engine takes in air, $S_{air}$ = 85.9 kg/s

Speed of the exhaust gas that are ejected relative to the aircraft, $S_{exh}$ =921 m/s

Therefore, the upward thrust of the jet engine is given by

$F=S_{air}(S_{exh}-S_a)+(S_b \times S_{exh})$

F = 85.9(921 - 262) + (3.92 x 921)

   = 4862635.79 + 3610.32

   = $4.8 \times 10^6 \ N$

Therefore thrust of the jet engine is $4.8 \times 10^6 \ N$.

3 0
3 years ago
A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi
grin007 [14]

Explanation:

It is given that,

Magnitude of charge, q=15\ \mu C=15\times 10^{-6}\ C

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, B=0.08\ j

Velocity, v=(5\ cos25)i+(5\ sin25)j

v=[(4.53)i+(2.11)j]\ m/s

We need to find the magnitude of force on the charge. Magnetic force is given by :

F=q(v\times B)

F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]

<em>Since</em>, i\times j=k\ and\ j\times j=0

F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]

F=0.00000543\ kN

F=5.43\times 10^{-6}\ kN

So, the force acting on the charge is 5.43\times 10^{-6}\ kN and is moving in positive z axis. Hence, this is the required solution.

6 0
3 years ago
What is the potential energy of a 2500 g object suspended 5 m above the earth's surface?
alexandr402 [8]

Answer:

Explanation:

Potential energy, which is the energy a body assumes at a position, can be calculated using the formula:

P.E = m × g × h

Where;

m = mass (kg)

g = acceleration due to gravity (10m/s²)

h = height (m)

7 0
3 years ago
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