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2 years ago

A ball dropped from a bridge takes 3.0 seconds to reach the water below. How far is the bridge from the water?

1 answer:

8 0

Since it is dropped, initial velocity u = 0

Using s = ut + (1/2)gt^2, putting u = 0, g = 10 m/s^2

s =(1/2)gt^2, t = 3s

s = 0.5 * 10 * 3 *3

s = 45 m.

The bridge is 45m above the water.

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A horizontal force of 350N is exerted on a 2.5 kg ball as it rotates uniformly in a horizontal circle of radius of 0.90m. Calcul

harkovskaia [24]

F=mv^2/R
----> V^2=FR/m=(350x0.9)/2.5=126
----- V=11.22 m/s

5 0

3 years ago

A baseball player throws a baseball with a velocity of 13 m/s north. It is caught by a second player seven seconds later. How fa

guajiro [1.7K]

Answer:

A. 91 meters north

Explanation:

Take +y to be north.

Given:

v₀ = 13 m/s

a = 0 m/s²

t = 7 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (13 m/s) (7 s) + ½ (0 m/s²) (7 s)²

Δy = 91 m

The displacement is 91 m north.

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3 years ago

In 1993 Ileana Salvador of Italy walked 3.0km in under 12min. Suppose that during her walk Salvador is observed to steadily incr

padilas [110]

The distance covered is 115 m

Explanation:

The motion of Ileana is a uniformly accelerated motion (constant acceleration), therefore we can use the following suvat equation:

$s=(\frac{u+v}{2})t$

where

s is the distance covered

u is the initiaal velocity

v is the final velocity

t is the time elapsed

In this problem, we have:

u = 4.20 m/s

v = 5.00 m/s

t = 25.0 s

Therefore, we can re-arrange the equation to find the distance covered:

$s=(\frac{4.20+5.00}{2})(25.0)=115 m$

Learn more about accelerated motion:

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#LearnwithBrainly

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3 years ago

Why are different constellations<br> of stars seen during different<br> seasons?

slamgirl [31]

Actually, they're not. There's a group of stars and constellations arranged
around the pole of the sky that's visible at any time of any dark, clear night,
all year around. And any star or constellation in the rest of the sky is visible
for roughly 11 out of every 12 months ... at SOME time of the night.

Constellations appear to change drastically from one season to the next,
and even from one month to the next, only if you do your stargazing around
the same time every night.

Why does the night sky change at various times of the year ? Here's how to
think about it:

The Earth spins once a day. You spin along with the Earth, and your clock is
built to follow the sun . "Noon" is the time when the sun is directly over your
head, and "Midnight" is the time when the sun is directly beneath your feet.

Let's say that you go out and look at the stars tonight at midnight, when you're
facing directly away from the sun.

In 6 months from now, when you and the Earth are halfway around on the other
side of the sun, where are those same stars ? Now they're straight in the
direction of the sun. So they're directly overhead at Noon, not at Midnight.

THAT's why stars and constellations appear to be in a different part of the sky,
at the same time of night on different dates.

5 0

3 years ago

Read 2 more answers

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

Veronika [31]

Answer:

Explanation:

Given that,

One fragment is 7 times heavier than the other

Let one fragment mass be M

Let this has a velocity v

And the other 7M

And this a velocity V

Initially the fragment is at rest u = 0

Applying conservation of momentum

Momentum is given as p=mv

Initial momentum = final momentum

Po = Pf

(M+7M) × 0 = 7M •V − Mv

0 = 7M•V - Mv

Divide both sides by M

0 = 7V -v

v = 7V

Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block

The workdone by the 7M mass is

Distance moved by 7M mass is 6.8m, Then, d =6.8m

W = fr × d

Where fr = µkN

When N=W =mg, where m=7M

N= 7Mg

fr = −µk × 7mg

Then, W(7m) = −7µk•Mg×d

W(7m) = −7µk•Mg×6.8

W(7m) = −47.6 µk•Mg

Then, same procedure,

Let distance move by the small mass be m

Work done by M mass

W(m) = −µk•Mg×d'

Since it is a wordone by friction, that is why we have a negative sign.

Using conservation of energy

Work done by 7M mass is equal to work done by M mass

W(7m) = W(m)

−47.6 µk•Mg = −µk•Mg×d

Then, M, g and µk cancels out

We are left with

-46.7 = -d

Then, d = 46.7m

7 0

3 years ago

Read 2 more answers