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3 years ago

What would always be true of an object possessing a potential energy of 0 joules

1 answer:

6 0

<em>if the object has 0 potential energy it means it is on ground ..
<u>explanation :
</u>to measure potential energy we take ground as a reference point but when object is placed on reference point i.e ground then height is 0..
as potential energy = mgh .. here h becomes 0 so potential energy is 0 joules..</em>

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Which of the following is a benefit of physical activity?

inessss [21]

Hello There!

Physical activity definitely controls weight.
Also, it does promote better sleep.
Studies have also shown that it improves energy.
Based on these facts, i'm going with D.

Hope This Helps You!
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8 0

3 years ago

Read 2 more answers

20 POINTS!!!

bogdanovich [222]

a) Average velocity for her run north: 2.68 m/s north

b) The average velocity for the whole trip is zero

c) The average speed for the whole trip is 1.95 m/s

Explanation:

In this part of the problem we only want to consider the part when Micha is running north.

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the displacement (a vector connecting the initial position to the final position of motion)

t is the time

For the part running north, we have

$d=2 mil \cdot 1609 = 3218 m$ north is the displacement

$t = 20 min \cdot 60 = 1200 s$ is the time interval

Substituting,

$v=\frac{3218}{1200}=2.68 m/s$ north (we have to specify also the direction, since velocity is a vector)

As we said, average velocity is given by

$v=\frac{d}{t}$

where

d is the displacement

t is the time

If we consider the whole trip, however, the displacement is zero:

d = 0

Because Micha returns home, so the initial position corresponds to the final position of motion. Therefore, the average velocity is also zero:

v = 0

The average speed is given by

$s=\frac{d}{t}$

where

d is the distance covered (the total length of the path, regardless of the direction)

t is the time interval

Since MIcha ran 2 miles north and then 2 miles back, the total distance is

$d=2 mi + 2mi = 4 mi \cdot 1609 = 6436 m$

And the time taken is

$t=20 min + 15 min + 20 min = 55 min \cdot 60 = 3300 s$

So, the average speed is

$s=\frac{6436}{3300}=1.95 m/s$

And since speed is a scalar, there is no need to specify a direction.

Learn more about speed and velocity:

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brainly.com/question/5063905

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5 0

3 years ago

A cylindrical bucket, open at the top, is 28.0 cm high and 11.0 cm in diameter. A circular hole with a cross-sectional area 1.55

svlad2 [7]

Answer:

so height is 0.1283 m

Explanation:

given data

height = 28 cm

diameter = 11 cm

cross-sectional area = 1.55 cm2

water flow rate = 2.46×10^−4 m3/s

to find out

How high will the water in the bucket rise

solution

we know that here

potential energy = kinetic energy

mgh = 1/2 mv²

multiply both sides by the 2 and we get

2mgh=mv²

solve it we get

√(2gh) = v ....................1

h = v²/2g ...............2

and

flow rate = A V

2.46×10^−4 = V 1.55×10^−4

V = 1.5870 m/s

so from 2

h = v²/2g

h = 1.5870²/ 2(9.81)

h = 0.1283 m

so height is 0.1283 m

6 0

4 years ago

A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi

EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is, $\rm d_1 = 2.2 \ cm$

initial radius,

$r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm$

The initial crossection area;

$\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2$

The final crossection area;

$\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2$

The initial flow rate is;

R = density ×velocity ×area

$\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1 = 3.947 \ m/sec$

The speed of the water in the wider part will be;

From the continuity equation;

$\rm A_1 V_1 = A_2V_2 \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec$

Hence, the speed of the water in the wider part will be 1.194 m/sec.

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7 0

2 years ago

We can compare these two interactions on the basis of impulse (see above), but sometimes, we are more interested in the forces (

kiruha [24]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$I = 476 \ N \cdot s$

$I_1 = 14.21 \ N\cdot s$

$F = 20300 \ N$

Explanation:

Considering the first question

From the question we are told that

The force produced is $F = 3400 \ N$

The duration of the punch is $t = 0.14 \ s$

Generally the impulse delivered is mathematically represented as

$I = F * t$

=> $I = 3400 * 0.14$

=> $I = 476 \ N \cdot s$

Considering the second question

The approaching velocity of the ball is $v_b = 45 \ m/s$

The leaving velocity of the ball is $v_l = -53 \ m/s$

The mass of the ball is $m_b = 0.145 \ kg$

Generally the magnitude of the impulse delivered is mathematically represented as

$I_1 = m* v_b - m * v_l$

=> $I_1 = [0.145 * 45] - [0.145 * -53]$

=> $I_1 = 14.21 \ N\cdot s$

Considering the third question

The duration of the impact of the bat is $t _1 = 0.7 \ ms = 0.7 *10^{-3} \ s$

Generally the average force exerted by the bat is mathematically represented as

$F = \frac{I_1}{t_1}$

=> $F = \frac{14.21 }{0.7 *10^{-3}}$

=> $F = 20300 \ N$

7 0

4 years ago