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skad [1K]
2 years ago
15

All objects, regardless of their mass, fall with the same rate of acceleration on

Physics
2 answers:
tamaranim1 [39]2 years ago
4 0

Answer:

A -TRUE

Explanation:

The mass, size, and shape of the object are not a factor in describing the motion of the object. So all objects, regardless of size or shape or weight, free fall with the same acceleration.

crimeas [40]2 years ago
3 0
True this is true because the acceleration is equal
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A motorist traveling at 12 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are ca
Vera_Pavlovna [14]

Answer:

Explanation:

Given

Motorcyclist speed=12 m/s  

maximum acceleration=-6 m/s^2

distance=39 m

Let x be the distance traveled by motorist in his reaction time

therefore remaining 39-x will be traveled with -6m/s^2 acceleration

v^2-u^2=2as

s=39-x

v=0

u=12 m/s

0-12^2=2\left ( -6\right )\left ( 39-x\right )

x=27 m

Therefore he traveled 27 m in his reaction time

27=12\times t

t=2.25 s

(b)If his reaction time is 2.56 sec

then distance traveled in his reaction time

x_0=12\times 2.56=30.72 m

Remaining distance 39-30.72=8.28 m

therefore its velocity when it reaches the deer

v^2-u^2=2as

v^2=12^2+2\times \left ( -6\right )\times 8.28=44.64

v=6.681 m/s

8 0
3 years ago
What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You c
Maksim231197 [3]

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

\displaystyle{t=D/v}

and from Hubble's Law:

v =H_0D

Therefore:

\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}

With the given value for the Hubble's constant we have:

H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s

and thus,

t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years

6 0
3 years ago
‏Explain how an electron emitted by the photoelectric effect can have kinetic energy less than threshold energy ?
xenn [34]

Answer:

the photons (quanta of light) collide with the electrons, these electrons have to overcome the threshold energy that is the energy of union with the metal, and the energy that remains is converted to kinetic energy.

          K = E - Ф

Explanation:

The photoelectric effect is the emission of electrons from the surface of a metal.

This was correctly explained by Einstein, in his explanation the energy of the photons (quanta of light) collide with the electrons, these electrons have to overcome the threshold energy that is the energy of union with the metal, and the energy that remains is converted to kinetic energy.

          E = hf

          E = K + Ф

          K = E - Ф

The energy of the photons is given by the Planck relation E = hf and according to Einstein the number of joints must be added

            E = n hf

Therefore, depending on the value of this energy, the emitted electrons can have energy from zero onwards.

4 0
2 years ago
What factors contribute to global winds
serg [7]
Temperature and elevation, if it is cold in Idaho and warm on the eastern end of a mountain side in california (or if warm air is going in that direction) then the cold air, being more dense, will go towards california while the cold air in Idaho will become warm. Same goes for the rest of the world 
3 0
3 years ago
How high up is a 5kg object that has 300j of energy
Eddi Din [679]

-- If the object is moving with speed of 10.954 meters per second, then
it has 300J of kinetic energy no matter where it may be located.

-- If the object is 6.118 meters above somewhere, then it has 300J of
gravitational potential energy relative to that place. 


7 0
3 years ago
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