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eimsori [14]
3 years ago
8

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

a gazelle assumes an acceleration of 4.2 m/s2m/s2 for 6.5 ss , after which the gazelle continues at a steady speed. What is the gazelles top speed?
Physics
1 answer:
Novay_Z [31]3 years ago
5 0

Answer:

v = 27.3 m/s

Explanation:

Given that

Acceleration ,a= 4.2 m/s²

Time ,t= 6.5 s

Lets take the maximum speed gain by Thomson's= v

We know that ,if acceleration is constant then the speed v is given as

v= u +  a t

v=final speed

u=initial speed

a=acceleration

t=time

Here the initial speed of Thomson's ,u = 0 m/s

Now by putting the values in the above equation we get

v= 0 + 4.2 x 6.5 m/s

v = 27.3 m/s

Therefore the maximum speed gain by Thomson will be 27.3 m/s.

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A disk 8.00 cm in radiu: rotates at a constant rate of 1200 revinin about its central axis Determine (a) its angular speed in zadians per second, (b) the tangential speed at a point 3.00 cm from its center, (c) the radial aceleration of a point on the rim, and (d) the total distance a point our tke rim noves ign-2.00 s (E) The moment of inertia if it's mass is 2Kg? is the answer

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The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x
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Answer:

Ff=m\times \dfrac{V_o^2}{2X_1}

Explanation:

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4 0
3 years ago
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

3 0
3 years ago
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