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3 years ago

A bird on a long migration flies 63 kilometers per hour for 2900

Physics

1 answer:

Fed [463]3 years ago

5 0

Answer:

Explanation:

2900 divided by 63 (2900/63)

- 46.03174603

Rounded to the nearest whole number

is 46.

THE ANSWER IS 46

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Which group of words do you think is an idiom?

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Explanation:

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If the average time it takes for the cart from point 1 to point 2 is 0.2 s, calculate the angle θ from the horizontal of the tra

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Answer:

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Melissa's favorite exercise equipment at the gym consists of various springs. in one exercise, she pulls a handle grip attached

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The formula for force exerted on/by a spring is

F = k*e where k is the spring constant and x is the distance stretched from unstrained position. This should allow you to find what you need.

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3 years ago

An archer defending a castle is on an 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take

kodGreya [7K]

Answer:

about 602 milliseconds

Explanation:

The motion can be approximated by the equation ...

y = -4.9t^2 -22.8t +15.5

where t is the time since the arrow was released, and y is the distance above the ground.

When y=0, the arrow has hit the ground.

Using the quadratic formula, we find ...

t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))

= (22.8 ± √823.64)/(-9.8)

The positive solution is ...

t ≈ 0.60195193

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8 0

3 years ago

A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.260

ruslelena [56]

Answer:

The tension in the rope is 41.38 N.

Explanation:

Given that,

Mass of bucket of water = 14.0 kg

Diameter of cylinder = 0.260 m

Mass of cylinder = 12.1 kg

Distance = 10.7 m

Suppose we need to find that,

What is the tension in the rope while the bucket is falling

We need to calculate the acceleration

Using relation of torque

$\tau=F\times r$

$I\times\alpha=F\times r$

Where, I = moment of inertia

$\alpha$ = angular acceleration

$\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r$

$F=\dfrac{M}{2}a$ ...(I)

Here, F = tension

The force is

$F=m(g-a)$ ...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

$\dfrac{M}{2}a=m(g-a)$

$a=\dfrac{g}{1+\dfrac{M}{2m}}$

Put the value into the formula

$a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}$

$a=6.84\ m/s^2$

We need to calculate the tension in the rope

Using equation (I)

$F=\dfrac{M}{2}a$

Put the value into the formula

$F=\dfrac{12.1}{2}\times6.84$

$F=41.38\ N$

Hence, The tension in the rope is 41.38 N.

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3 years ago