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3 years ago

What provides most of the force to allow the runners to move?

Physics

1 answer:

kobusy [5.1K]3 years ago

4 0

Answer:

When you are running the most important force that you should understand is friction. Friction is a force that opposes movement between two objects, but for runners friction makes you faster. Friction gives you a better and more efficient way to use your energy into speed.

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Which statement describes an example of static electricity?

lana [24]

Answer: Negatively charged particles are repelled by other negatively charged particles

Explanation:

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3 years ago

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A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a

ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω / t

(ω = √g/R)

∝ = $\frac{1}{t } \sqrt{\frac{g}{R} } }$

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = $\frac{2FR}{MR^2 + 2mr^2}$

Now we have ,

∝ = $\frac{1}{t} \sqrt{\frac{g}{R} }$ , ∝ = $\frac{2FR}{MR^2+2mr^2}$

equating both values

We have,

$F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }$

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

$F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N$

Each thruster has to applied a force of 294.5N in tangential direction

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4 years ago

What is the sig fig of 0.00324?

Nonamiya [84]

This number has 3 sig figs.

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3 years ago

Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour

Serhud [2]

For the answer to the question above, first find out the gradient.

m = rise/run 
=(y2-y1)/(x2-x1) 

the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" 
(x1,y1) = (3,53) 
(x2,y2) = (6,62) 

sub values back into the equation 
m = (62-53)/(6-3) 
m = 9/3 
m = 3 

now we use a point-slope form to find the the standard form 
y-y1 = m(x-x1) 
where x1 and y1 are any set of point given 
y-53 = 3(x-3) 
y-53 = 3x - 9 
y = 3x - 9 + 53 
y = 3x + 44 

y is the velocity of the car, x is the time.
I hope this helps.

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3 years ago

What is a good activity for strengthening your stomach muscles?

koban [17]

Answer:

sit-ups

Explanation:

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3 years ago

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