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Katena32 [7]
3 years ago
12

You are running on a perfectly circular track that has a radius of 102 meters. What is the largest displacement you will ever at

tain?
Physics
1 answer:
tiny-mole [99]3 years ago
8 0

The largest possible displacement on a circular track is the straight-line distance between the starting point and the point directly opposite it, half-way around the circle. That's the diameter of the track ... 204 meters.

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In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a
katovenus [111]

The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

Vx = 10.3 cos 22.8 = 9.5 m/s

Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

R = u²sin2θ /g

where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.

Substituting the values, we get

R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81

R = 7.75m

At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

Vxo = 2 x 5.5 /78 x 9.5

Vxo = 1.34 s

Maximum height gained when final velocity is zero

Vy = 0 = Vyo -gt

time t = Vyo/g = 4/9.8 = 0.41s'

Increase in range by using of halters is

ΔR = ΔVx' x t

ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

Learn more about projectile.

brainly.com/question/11422992

#SPJ1

3 0
2 years ago
Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete
Sonbull [250]

Answer:

<em>work done in pumping the entire fuel is 466587 J</em>

<em></em>

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical,<em> we assume that the fuel within also takes the cylindrical shape.</em>

<em>Also, we assume that the fuel completely fills the tank.</em>

volume of a cylinder = \pi r^{2}l

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x 1.5^{2} x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 14.139 = 93317.4 N

work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = <em>466587 J</em>

8 0
3 years ago
What is the answer to these questions?
Oksana_A [137]

Answer:

THES IS NOT

Explanation:

THIS PAPAER IS A FAKE PAPAER BEACISE POGI TALAGA AKO

7 0
3 years ago
You are pushing on a heavy desk with a force of 65 N the desk does not slide the force of friction between the desk and the floo
astra-53 [7]

When an object does not move even on pushing , static frictional force acts on in opposite direction of the applied force to stop the object from moving. static frictional force is a self adjusting force and it adjust its value according to the applied force if the applied force is smaller than the maximum value of static frictional force. The object starts moving once the applied force on it becomes greater than the maximum static frictional force.  hence the statement is true.

7 0
3 years ago
I need the answer to this question what has the student plotted on the vertical axis?
expeople1 [14]

Answer:

The correct option is D

Explanation:

In trying to achieve what the student wanted to see, which is to see the relationship between the weight the cord can hold and how long the cord will stretch. Since the origin of the graph is from zero, the value plotted on the vertical axis would be just the length caused by each weights. Thus, <u>the original length would have to be subtracted from the measured length to determine the actual length caused by the weight added to the cord</u>.

7 0
3 years ago
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