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3 years ago

Brainliest!!! We observe that it takes 5,3 seconds to cross the distance of 35

Physics

1 answer:

Umnica [9.8K]3 years ago

7 0

Answer:

A) speed = distance / time

= 35 / 5.3

= 6.6 = 7 m/s

b) 7 m/s = 25.2 km/h = 25 km/h

speed = distance / time

time = distance / speed

= 1000 / 25

= 4 hrs = 240 min

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How do base-isolators, the rubber and steel pads under some buildings, reduce earthquake damage?

nika2105 [10]

They absorb the energy of seismic waves.

Explanation:

Base - isolators, rubber and steel pads under buildings reduces earthquake damage by absorbing the energy of seismic waves.

Earthquakes are produced by the sudden release of energy within the earth which causes displacement and sharp movement on earth surface.
There are four main types of waves that are propagated by seismic waves which wrecks havoc.
Understanding the waves and their properties has been useful in designing earthquake resistant structures.
Base-isolators, rubber and steel pads helps to absorb seismic waves.
When seismic waves are absorbed, they are cut off.
This way they cannot cause damage.

learn more:

Earthquake brainly.com/question/6520403

#learnwithBrainly

8 0

3 years ago

Read 2 more answers

This should be pretty easy i think.

timama [110]

Answer:

its a sure for not but i think its a 192583

3 0

2 years ago

Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Dimas [21]

Answer:

Abby is standing (4.5^2 + 2.3^2)^1/2 from the far speaker

D2 = 5.05 m from the far speaker

The difference in distances from the speakers is

5.05 - 4.5 = .55 m (Let y be wavelength, lambda)

n y = 4.5

(n + 1) y = 5.05 for the speakers to be in phase at smallest wavelength

y = .55 m subtracting equations

f = v / y = 340 / .55 = 618 / sec should be the smallest frequency

8 0

3 years ago

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

$x=ut$

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

$t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s$

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

$y=\frac{1}{2} gt^2$

Substitute 9.8 m/s² for g and 0.461 s for t.

$y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m$

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by, $(2.0 m)-(1.04 m)=0.96 m$