PHYSICS HIGH SCHOOL

A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity is 9.8 m/s 2 . What would be his maximum range on the Moon, where the free-fall acceleration is g 6 ? Answer in units of m

Answers

Answer 1

Answer:

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle $=45^(\circ)$

Acceleration due to gravity on earth is $9.8 m/s^2$

Acceleration due to gravity on moon is $(9.8)/(6)=1.63 m/s^2$

Range of projectile is given by

$R=(u^2\sin 2\theta )/(g)$

$R_(earth)=(u^2\sin 2\theta )/(g)=5$ ----1

$R_(moon)=(u^2\sin 2\theta )/((g)/(6))$ -----2

Divide 1 & 2

$(5)/(R_(moon))=(1)/(6)$

$R_(moon)=30 m$

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Answers

The answer is longitudinal. There are two types of waves: transverse and longitudinal. Transverse waves are those whose vibrations are perpendicular to the direction of the wave. Longitudinal or compression waves are those whose vibrations are in the same direction as the wave. Sound is caused by the compression of vibrations of molecules.

Answer:

longitudinal

Explanation:

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Answers

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Which of the following types of electromagnetic radiation is most dangerous? a. x rays
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Answers

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Answers

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Therefore, the correct option is D.

Learn more about Ultrasound here:

brainly.com/question/14902329

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Answer:

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Explanation:

on edg

A train starting from a distance of velocity of 90 km.h-1 in 10 minutes assuming that the acceleration is uniform find the acceleration and the distance travelled by the train for attending the velocity

Answers

To make this problem solvable and you can get the help you need, I'll complete and arrange some data.

Answer:

Acceleration: $0.0417\ m/s^2$ , Distance=7,500 m

Explanation:

Uniform Acceleration Motion

It's a type of motion in which the velocity of an object changes uniformly over time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

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Using the equation [1] we can solve for a:

$\displaystyle a=(v_f-v_o)/(t)\qquad\qquad [3]$

The problem will be rewritten as follows:

A train starting from rest reaches a velocity of 90 km/h in 10 minutes. Assuming that the acceleration is uniform, find the acceleration and the distance traveled by the train for attending the velocity.

Let's take the relevant data:

vo=0

vf=90 Km/h*1000/3600 = 25 m/s

t = 10 minutes = 10*60 = 600 seconds

Now compute the acceleration by using [3]:

$\displaystyle a=(25-0)/(600)=0.0417$