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3 years ago

PLEASE HELP I WILL GIVE BRALINEST

Physics

2 answers:

Tcecarenko [31]3 years ago

7 0

1. Position over time graph. Time vs position graph.

2. Velocity

3. It’s accelerating from the start, it remains at a constant velocity, and it decelerates until it stops.

maria [59]3 years ago

3 0

1. acceleration i believe
2.more time the father they get (if that makes sense)

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What part of the atom takes up most of the atoms space?

Ahat [919]

Answer:

The electrons

Explanation:

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3 years ago

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All BUT one statement applies to electromagnetic radiation. That is

koban [17]

A) Radio waves are a form of electromagnetic radiation.

TRUE

It is the type of electromagnetic wave with least energy range and longest wavelength range among all

B) Electromagnetic radiation does not require a medium to travel.

TRUE

This radiation can travel through vacuum also and its speed is maximum in vacuum

C) In a vacuum, electromagnetic radiation travels at the speed of light.

TRUE

speed of all electromagnetic radiation is same as speed of light in vacuum

D) Electromagnetic radiation travels in the form of longitudinal waves.

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All electromagnetic radiation is always in form of transverse waves

So correct options are

4 0

3 years ago

In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75kN. Knowing that v = 0.30 and E =

VladimirAG [237]

Answer:

(a) Elongation of rod = 0.19732 mm

(b) Change in diameter = 0.00651 mm

Explanation:

Circular area at end of steel rod = pi * diameter^2 / 4

$Area = pi * (22 * 10^-^3)/4$

Area = 3.801 * 10^(-4) meter squared

Stress = force / area

Stress = 75000 / (3.801 * 10^-4)

Stress = 197316495 Pa OR 0.197 GPa

Modulus of elasticity = stress / strain

200 = 0.19732 / Strain

Strain = 0.0009866 (longitudinal)

(a) Strain = change in length / initial length

0.0009866 = Elongation of rod / 200

Elongation of rod = 0.19732 mm

(b) Poisson ratio = lateral strain / longitudinal strain

0.3 = lateral strain / 0.0009866

Lateral strain = 0.000296

Lateral strain = Change in diameter / original diameter

0.000296 = Change in diameter / 22

Change in diameter = 0.00651 mm

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3 years ago

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A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0

n200080 [17]

Answer:

This a closed system because the mass of the system is conserved

The energy system that undergoes change is the Potential energy system

The energy system diagram is shown on the first uploaded image

Work done = Change in gravitational potential energy

So solving algebraically for work done would be

Work done = $m*g*h$

where m is mass

g is acceleration due to gravity

and h is the height

Work done in terms of force and distance is = mg

where m is mass of bucket and

g is acceleration due to gravity

Explanation:

a) At the start, potential and kinetic energy were zero. so, energy is zero.

As the person pulls the bucket up, the potential energy becomes mgh.

so,final energy will be consisting of only potential energy.

B) Here work done is equal to change in gravitational potential energy.

W = $\Delta P.E$

W = m*g*h

where g = 9.9 m/ $s^2$

C) Work = force * distance

mgh = force * h

force = mg

force = weight of bucket

6 0

3 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

3 years ago