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Nikolay [14]
4 years ago
8

What is the gravitational potential energy with respect to the surface of the water of a 75.0-kilogram diver located 3.00 meters

above the water
Physics
1 answer:
Rama09 [41]4 years ago
5 0

Answer:

2205J

Explanation:

Potential energy = mgh

where g is acceleration due to gravity=9.8m/s^{2}

m is mass of the body

h is height

m=75kg    , h=3m

P.E= 75*9.8*3

 =2205J

You might be interested in
Number of waves that pass a given point in one second
Studentka2010 [4]
<em>number of waves that pass a given point in one second is called <u>frequency..</u></em>
5 0
3 years ago
41. Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 h
romanna [79]

Answer:

138.3 days

Explanation:

Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.

The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.

T^2 is proportional to r^3

That is,

T^2/r^3 = constant

98^2 / 62^3 = T^2 / 78^3

Make T^2 the subject of formula.

T^2 = 98^2 / 62^3 × 78^3

T^2 = 19123.2

T = sqrt ( 19123.2 )

T = 138.2867 days

Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.

4 0
3 years ago
A toy doll and a toy robot are standing on a frictionless surface facing each other. The doll has a mass of 0.2 kg, and the robo
Natali5045456 [20]

Answer:

<h3>1.43m/s²</h3>

Explanation:

According to newtons second law.

F = mass * acceleration

If the doll has a mass of 0.2 kg, and the robot has a mass of 0.5 kg, the resulting mass will be 0.7kg

Force applied = 1N

acceleration = Force/mass

Substitute the values and get acceleration

acceleration = 1/0.7

acceleration = 1.43m/s²

Hence the magnitude of the acceleration of the robot is 1.43m/s²

3 0
3 years ago
1. The gravitational force acting on a falling body and its weight is constant. But the law of universal gravitation tells us th
vagabundo [1.1K]
It's not so much a "contradiction" as an approximation. Newton's law of gravitation is an inverse square law whose range is large. It keeps people on the ground, and it keeps satellites in orbit and that's some thousands of km. The force on someone on the ground - their weight - is probably a lot larger than the centripetal force keeping a satellite in orbit (though I've not actually done a calculation to totally verify this). The distance a falling body - a coin, say - travels is very small, and over such a small distance gravity is assumed/approximated to be constant.
3 0
3 years ago
A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
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