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zzz [600]
3 years ago
8

A student rapidly rubs the palms of both hands

Physics
1 answer:
Lapatulllka [165]3 years ago
3 0

Answer:

Option a

Explanation:

The action of the rapid rubbing of the hands are a result of chemical reaction and responses that takes place inside our body.

When we rub our hands together, that shows the mechanical motion of the body or we can say that while rubbing hands chemical energy gets converted to mechanical energy (here, Kinetic energy).

While rubbing, due to friction, heat is produced and hence we can say that this mechanical energy has transformed to heat energy.

Thus the conversion here is from chemical to mechanical to heat energy.

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A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?
garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

U_i +K_i = U_f + K_f

where:

U_i is the initial potential energy, at the bottom

K_i is the initial kinetic energy, at the bottom

U_f is the final potential energy, at the top

K_f is the final kinetic energy, at the top

We can rewrite the equation as:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m is the mass of the hopper

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0 is the initial height

u is the launch speed of the hopper

h_f = 1.20 m is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s

Learn more about kinetic energy and potential energy:

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4 0
3 years ago
Which object has the most kinetic energy? a bus a car a plane a bicycle
loris [4]

Answer:I would guess a plane

Assuming they all Thad the same velocity....

4 0
3 years ago
1)a persons ________ will change if they move from the earth to the moon
fgiga [73]

A person's weight will change if they move from the earth to the moon. This does not however, change the person's mass. Mass is the amount of matter that makes up an object, and volume is how much space it takes up. On the moon, there is a lighter gravitational pull on said person, so they will not weigh as much if they stepped on a scale.

3 0
3 years ago
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What is a net force on an object that has a mass of 20.0 kg, an applied force of 100 n moving on a surface with a friction coeff
sergiy2304 [10]

The net force on the object as described is; 58.84N

Two forces acting on the object are;

  • The <em>applied force and the frictional force.</em>

In essence; the frictional force can be evaluated as;

  • Frictional force; = coefficient × Weight of object.

  • Frictional force = 0.21 × 20 × 9.8.

  • Frictional force = 41.16N

  • The Net force = Applied force - frictional force

  • Net force = 100 - 41.16N

Net Force = 58.84 N.

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5 0
2 years ago
What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio
pantera1 [17]
If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m*(v²/R),
 
where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite. 

Masses cancel out and we have:

G*(M/r²)=v²/R, R=r so:

G*(M/r)=v²

r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G*(M/ω²r²),

r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G*(M/(4π²/T²)), and finally we take the third root to get r:

r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite. 
3 0
3 years ago
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