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Elodia [21]
3 years ago
15

Why it is difficult to run fast in sand​

Physics
2 answers:
balu736 [363]3 years ago
8 0
Running on sand requires 1.6 times more energy spent than running on hard surface, so the force applied by our foot on sand is less.
prohojiy [21]3 years ago
5 0

Explanation:

The force exerted by our foot on the sand is less because time taken is large so it is difficult to run fast on sand.

hope it will help you

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Match each prefix to the correct multiple of 10. 1. centi- 10 2. kilo- 100 3. deka- 1000 4. milli- 1/100 5. hecto- 1/10 6. deci-
Mice21 [21]

Explanation :

There are some basic metric conversions.

Prefix                   Multiple of 10

centi                            \frac{1}{100}

Kilo                              1000

deka                              10

milli                              \frac{1}{1000}

hecto                            100

deci                              \frac{1}{10}

For example :

1cm=\frac{1}{100}m\\\\1Km=1000m\\\\1mm=\frac{1}{1000}m\\\\1Km=10\text{ hecto meter}\\\\1Km=100\text{ deka meter}


7 0
3 years ago
Read 2 more answers
A constant horizontal F force began to act on the initially immovable body placed on a horizontal surface. After t time the forc
mel-nik [20]

Answer:

The coefficient of friction is (F/(19.6·m)

Explanation:

The given parameters are;

The force applied to the immovable body = F

The time duration the force acts = t

The time the body spends in motion = 3·t

The acceleration due to gravity, g = 9.8 m/s²

From Newton's second law of motion, we have;

The impulse of the force = F × t = m × Δv₁

Where;

Δv₁ = v₁ - 0 = v₁

The impulse applied by the force of friction, F_f is F_f × (3·t - t) =  F_f × (2·t)

Given that the motion of the object is stopped by the frictional force, we have;

The impulse due to the frictional force = Momentum change = m × Δv₂ = F_f × (2·t)

Where;

Δv₂ = v₂ - 0 = v₂

Given that the velocity, v₂, at the start of the deceleration = The velocity at the point the force ceased to  act, v₁, we have;

m × Δv₂ = F_f × (2·t) = m × Δv₁ = F × t

∴ F_f × (2·t) = F × t

F_f = F × t/(2·t) = F/2

The coefficient of dynamic friction, \mu _k = Frictional force/(The weight of the body) = (F/2)/(9.8 × m)

\mu _k = (F/(19.6·m)

The coefficient of friction, \mu _k = (F/(19.6·m)

5 0
3 years ago
A velocity selector has a magnetic field of magnitude 0.22 T perpendicular to an electric field of magnitude 0.51 MV/m.
ohaa [14]

Answer:

speed of a particle  = 2.31 × 10^{6} m/s

energy of proton required = 27.77 KeV

energy of electron required =  15.171 eV

Explanation:

given data

magnetic field of magnitude = 0.22 T

electric field of magnitude = 0.51 MV/m

to find out

speed of a particle and energy must protons have to pass through undeflected and  energy must electrons have to pass through undeflected

solution

we know that force due to magnetic and electric field is express as

force due to magnetic field B = qvB    ..............1

and force due to electric field E = qE   .....................2

so without deflection force due to magnetic field  = force due to electric field  

so here qvB = qE

and V = \frac{E}{B}    ...................3

put here value

V =  \frac{0.51*10^6}{0.22}

speed of a particle  = 2.31 × 10^{6} m/s

and

now energy of proton required will be here as

energy of proton required = mass of proton × \frac{V^2}{2}

put here value

energy of proton required = 1.67 × 10^{-27} × \frac{(2.31*10^6)^2}{2}

energy of proton required = 4.45 × 10^{-15} J

energy of proton required = 4.45 × 10^{-15} J ÷ (1.602 × 10^{-19}

energy of proton required = 27777.777 eV

energy of proton required = 27.77 KeV

and

now we get here energy of electron required that is

energy of electron required = mass of electron × \frac{V^2}{2}

put here value

energy of electron required = 9.11 × 10^{-31} × \frac{(2.31*10^6)^2}{2}

energy of electron required =  24.305× 10^{-19} J

energy of electron required =  24.305 × 10^{-19} J ÷ (1.602 × 10^{-19}  

energy of electron required =  15.171 eV

5 0
4 years ago
Read 2 more answers
It takes Serina 1.72 hours to drive to school. Her route is 47 km long. What is Serina's average speed on her drive to school? Y
denis-greek [22]

Explanation:

Average speed = distance / time

v_avg = 47 km / 1.72 hr

v_avg = 27.3 km/hr

7 0
3 years ago
Which of the following is an element
Ilia_Sergeevich [38]
D oxygen i believe is your answer 
5 0
4 years ago
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