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2 years ago

A car is driving on the road. The car starts to decelerate at -3 m/s2. It stops moving after 125

Physics

1 answer:

galina1969 [7]2 years ago

8 0

Answer:

27.39 m/s

Explanation:

From the question given above, the following data were obtained:

Deceleration (a) = –3 m/s²

Distance (s) = 125 m

Final velocity (v) = 0 m/s

Initial velocity (u) =?

The initial velocity (u) of the car can be obtained as follow:

v² = u² + 2as

0² = u² + (2 × –3 × 125)

0 = u² + (– 750)

0 = u² – 750

Collect like terms

0 + 750 = u²

750 = u²

Take the square root of both side

u = √750

u = 27.39 m/s

Thus, the initial velocity of the car was 27.39 m/s

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A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict

madreJ [45]

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

8 0

3 years ago

Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their

Nataliya [291]

Answer:

Though the question is not specified here, but this information can determine the following quantity: period T= 6 secs, Frequency F=1/6 Hz, speed of rotation V= 2 pi ft/sec and wave length =pi/3 ft

Explanation:

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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

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3 years ago

What happens to a sound wave when it interferes with another sound wave having the same frequency but traveling in the opposite

Zolol [24]

its called destructive interference.

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2 years ago

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noname [10]

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Thus,
star 1 will rise 1 hour before star 2

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2 years ago

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