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3 years ago

A rock thrown to the right with an initial horizontal velocity of 4 m/s lands after 5

Physics

1 answer:

pickupchik [31]3 years ago

6 0

Answer:

20m

Explanation:

Using the equation s=d/t and re arranging it to solve for distance, d=St, where d=distance, s=speed, ant t=time, we can plug in known variables.

d=4*5

d=20m

there is a distance of 20m traveled

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A football kick returner catches the ball just as a player from the opposing team dives to tackle him. At the time of impact, th

pochemuha

The total momentum of the players after collision is 130 kgm/s.

The given parameters:

Initial momentum of the returner, $P_i_1$ = 0 kgm/s
The initial momentum of the diving player, $P_i_2$ = 130 kgm/s

The total momentum of the players after collision is determined by applying the principle of conservation of linear momentum as follows;

$P_f = P_i_1 + P_i_2\\\\P_f = 0 + 130\\\\P_f = 130 \ kgm/s$

Thus, the total momentum of the players after collision is 130 kgm/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

6 0

2 years ago

A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.

Akimi4 [234]

Answer:

3.59 m/s

Explanation:

The average velocity is defined as:

$\frac{Total displacement}{Total time}$

The turtle first walks 18m south, and then is taken 1,1Km (or 1100m) north. Thus, the total displacement is 1082m north (1100m north - 18m south).

Now we have to calculate the total time, which will be equal to the sum of the time the turtle walked and the time it was taken by truck.

The walking time is 3.5 minutes. Since 1 minute = 60 seconds, then the walking time is 210 seconds.

To calculate the truck time we use the equation:

Time = $\frac{Distance}{Speed}$

Where the distance the truck travelled is 1100m and the speed of the truck is 12m/s.

Thus,

Truck time= $\frac{1100m}{12m/s}$ = 91.67s

The total time is the sum of the walking time and the truck time.

Total time = 210s + 91.67s = 301.67s.

As mencioned previously, the average velocity is equal to total displacement/ total time, thus:

Average velocity = $\frac{1082m}{301.67s}$ = 3.59 m/s North

Since the average velocity is a vector, it has a magnitude and a direction. In this case the magnitude is 3.59 m/s and the direction is north since the turtle's final displacement is north of where it started.

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3 years ago

An electric hoist does 500 joules of work lifting a crate 2 meters. How

LenKa [72]

Hope this answer helps

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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, $(2.0 m)-(1.04 m)=0.96 m$

The catcher needs to hold his glove at a height of 0,96 m from the ground.

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3 years ago

7Which of the following terms describes how glaciers move?

tatuchka [14]

Answer:

D is the answer I think (0 w 0 )

Explanation:

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3 years ago

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