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3 years ago

14

How much time does it take for tweety’s bird cage to hit the ground after it was dropped if it reached a velocity of 22 meters p

er second right before impact?

1 answer:

guajiro [1.7K]3 years ago

7 0

It matters on the weight

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On his way off to college, Russell drags his suitcase 19 m from the door of his house to the car at a constant speed with a hori

Mashcka [7]

Answer:

The work done on the suitcase is, W = 1691 J

Explanation:

Given data,

The force on the suitcase is, F = 89 N

The distance Russell dragged the suitcase, S = 19 m

The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction

The work done on the suitcase by Russell is given by the formula

W = F · S

Substituting the given values,

W = 89 N x 19 m

W = 1691 J

Hence, the work done on the suitcase is, W = 1691 J

8 0

3 years ago

In designing circular rides for amusement parks, mechanical engineers must consider how small variations in certain parameters c

Bess [88]

Answer:

Part a)

$dF = -\frac{mv^2}{r^2} dr$

Part b)

$dF = \frac{2mvdv}{r}$

Part c)

$dT = - \frac{2\pi r}{v^2} dv$

Explanation:

Part a)

As we know that force on the passenger while moving in circle is given as

$F = \frac{mv^2}{r}$

now variation in force is given as

$dF = -\frac{mv^2}{r^2} dr$

here speed is constant

Part b)

Now if the variation in force is required such that r is constant then we will have

$F = \frac{mv^2}{r}$

so we have

$dF = \frac{2mvdv}{r}$

Part c)

As we know that time period of the circular motion is given as

$T = \frac{2\pi r}{v}$

so here if radius is constant then variation in time period is given as

$dT = - \frac{2\pi r}{v^2} dv$

8 0

3 years ago

What would be some benefits of living on a planet with less surface gravity

zepelin [54]

You could jump high!

3 0

3 years ago

Read 2 more answers

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

How do hydrogen bonds affect boiling points?

wel

Answer:

I would say A but am not sure

7 0

3 years ago