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Salsk061 [2.6K]
2 years ago
13

A student stays at her initial position for a bit of time, then walks slowly in a straight line for a while, then stops to rest

a while, and finally runs quickly back to her initial position along a straight line. Which of the following statements is true about the average speed and the magnitude of the average velocity of the student during her trip? Her average speed is greater than the magnitude of her average velocity. Her average speed is the same as the magnitude of her average velocity. Her average speed is less than the magnitude of her average velocity. The student's average speed and velocity cannot be compared without knowing the farthest point she reached.
Physics
1 answer:
IRINA_888 [86]2 years ago
5 0

Answer:Average speed is greater than average velocity.

Explanation  :

Given

student walks slowly along a straight line for a while ,then stops to rest a while, and finally runs quickly back to her initial position

Let x be the length of track and the whole process takes t time

For average speed =\frac{distance\ traveled}{time\ taken}

Average speed=\frac{2x}{t}

For average velocity =\frac{Displacement}{time\ taken}

Since displacement is zero as she returns to its initial position.

Average velocity=0

Therefore Average speed is greater than average velocity.

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John runs 1.0 m/s at first, and then accelerates to 1.6 m/s during
erastova [34]

Answer: 0.13m/s^2

Explanation:

Formula: a=\frac{V_2-V_1}{t}

Where;

a = acceleration

V2 = final velocity

V1 = initial velocity

t = time

If John runs 1.0 m/s first, we assume this is V1. He accelerates to 1.6 m/s; this is V2.

a=\frac{1.6m/s-1.0m/s}{4.5s}

a=\frac{0.6m/s}{4.5s}

a=0.13m/s^2

7 0
2 years ago
Read 2 more answers
When the voltage and current have _____ polarities in a pure capacitive circuit, the capacitor is discharging and the energy is
tangare [24]

Answer:opposite

Explanation:for a capacitor to discharge (after charging) the polarities of the current and voltage have to be reversed

6 0
1 year ago
A stone is held at a height h above the ground. A second stone with four times the mass of the first one is held at the same hei
QveST [7]

gravitational potential energy is given by formula

U = mgh

here we need to compare the gravitational potential energy of stone 2 with respect to stone 1

so we will say

\frac{U_2}{U_1} = \frac{m_2gh}{m_1gh}

\frac{U_2}{U_1} =\frac{m_2}{m_1}

given that

m_2 = 4 m_1

now we have

\frac{U_2}{U_1} = 4

5 0
2 years ago
A bucket of water experiencing a gravitational force of 525 N is pulled from a water well. Net force in the Y direction is 45 N
vivado [14]

Answer:

T = 570 N

Explanation:

Given that,

The gravitational force acting on a bucket of water = 525 N

Net force in the Y direction is 45 N

We need to find the magnitude of the force of tension. It can be calculated as :

45 = T - 525

T = 525 + 45

T = 570 N

Hence, the force of tension is 570 N.

7 0
1 year ago
Find the kinetic of a 0.1 kilogram toy truck moving at a speed of 1.1 meters per second
omeli [17]
KE = kinetic energy
PE = potential energy
GPE = gravitational potential energy
energy is always measured in Joules (J)

KE = (0.5) times the mass times the velocity^2
square the velocity first

Mass = (KE x 2) / v^2
square the velocity first, then double the kinetic energy, then divide
mass is measured in kg

velocity = sqrt(KE x 2 / m)
velocity can be called speed, like in the the second problem
remember to find the square root after you double the KE and divide that by the mass.
for example: if after you doubled KE and divided it by the mass you got sqrt(20), the answer would be about 4.47

GPE = mass x gravitational pull (about 9.8 m/s^2 on earth) x height

height = (PE) / (g x m)
do g x m first

So for question 1:
KE = (0.5)0.1 x 1.1^2
always square the velocity first:
KE = (0.5)0.1 x 1.21
KE = 0.0605
so if you rounded it to the nearest hundreths you would get KE = 0.06 J
don't forget the unit of energy is in Joules
5 0
1 year ago
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