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Aloiza [94]
3 years ago
13

interval (from lowering the body to his feet taking off) lasts for 0.3 s and the mass of the player is 90 kg. Ignore air resista

nce. Part A Determine the force that the the player exerts on the surface when he jumps. Express your answer to one significant figure and include the appropriate units. Enter positive value if the force is upward and negative value if the force is downward.
Physics
1 answer:
steposvetlana [31]3 years ago
8 0

Answer:

The force applied to the surface is 9 kilo Newton.

Explanation:

While jumping on the surface the player applies the force that is equal to its weight on the surface.

The mass of the player is given as 90 kg.

Force applied by the player = weight of the player

Force applied by the player = m × g

Where m is the mass of the player and g is acceleration due to gravity

Force applied by the player = 90 × 9.8

Force applied by the player = 882 Newton

Expressing your answer to one significant figure, we get

Force applied by the player =0. 9 kilo Newton

The force applied to the surface is 0.9 kilo Newton.

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The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in
chubhunter [2.5K]

Answer:

The angular separation equals 0.35^{o}

Explanation:

We have according to Snell's law

n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})

\therefore \theta _{2}=sin^{-1}(\frac{n_{1}sin(\theta _{1})}{n_{2}})

Using this equation for both the colors separately we have

\theta _{red}=sin^{-1}(\frac{sin(23.90^{o})}{1.62})\\\\\theta _{red}=14.48^{o}

Similarly for violet light we have

\theta _{violet}=sin^{-1}(\frac{sin(23.90^{o})}{1.660})\\\\\theta _{violet}=14.13^{o}

Thus the angular separation becomes

\Delta \theta =14.48-14.13=0.35^{o}

6 0
3 years ago
the frequency of a beam of uv light is 1.0 ×10 ^15hz what is the energy in one quantum of this light express it in ev? ​
kap26 [50]

Answer:

4.14 eV

Explanation:

f = 1.0 ×10^15 Hz

h= 6.63×10^-34 J s (  this is called PLANCK 'S CONSTANT)

ENEGY = E = ?

E = hf  ( THIS IS FORMULA FOR ENERGY OF ONE QUANTA OR ONE PHOTON )

E= 6.63×10^-34×1.0 ×10^15

E = 6.63×10^-19 J

As 1eV = 1.6×10^-19 J so changing energy in eV from joules we will divide energy by 1.6×10^-19

hence E in eV = 6.63×10^-19/(1.6×10^-19)

          E = 4.14 eV

7 0
3 years ago
the maximum displacement of a particle, within a wave, above or below its equilibrium position, is called__________
Gre4nikov [31]

Answer:

the amplitude.

Explanation:

3 0
3 years ago
Read 2 more answers
PLEASE ASAP ILL GIVE BRAINLIEST.
taurus [48]

Answer:

applied force

Explanation:

any force where you push or pull is always applied force.

4 0
2 years ago
An inventor claims to have invented a heat engine that receives 750kJ of heat from a source at 400K and produces 250kJ of net wo
IRISSAK [1]

Answer:

the claim is not valid or reasonable.

Explanation:

In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:

η(max) = 1 - T₁/T₂

where,

η(max) = maximum efficiency = ?

T₁ = Sink Temperature = 300 K

T₂ = Source Temperature = 400 K

Therefore,

η(max) = 1 - 300 K/400 K

η(max) = 0.25 = 25%

Now, we calculate the actual frequency of the engine:

η = W/Q

where,

W = Net Work = 250 KJ

Q = Heat Received = 750 KJ

Therefore,

η = 250 KJ/750 KJ

η = 0.333 = 33.3 %

η > η(max)

The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.

<u>Therefore, the claim is not valid or reasonable.</u>

3 0
3 years ago
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