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3 years ago

Take another look at line 1. Suppose that you use distance and time between any pair of neighboring dots to calculate speed:

Physics

1 answer:

Nat2105 [25]3 years ago

6 0

Answer:

The time of travel and the distance between neighboring dots is always the same. So, the speed between neighboring dots is constant. This speed will be the same as the average speed in part F.

Explanation:

This is the sample answer provided, do not copy word-for-word, or you will most likely get in trouble for cheating.

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Here is a graph of speed vs time. If the object is moving to the east, which BEST describes the speed and velocity of the graph?

Artemon [7]

Answer:

Both speed and velocity are changing.

Explanation:

They are both going up so both are changing

5 0

2 years ago

Micah knows that a car had a change in velocity of 15 m/s.To determine acceleration ,Micah also needs the blank of the total tri

Fittoniya [83]

Answer: To determine acceleration ,Micah also needs the Time of the total trip in seconds.

Explanation:

Acceleration can be defined as rate of change of velocity.

$a = \frac{dv}{dt}$

for calculating acceleration, initial and final velocity are required in meter per second and the total time of the trip in seconds. Then acceleration is measured in meter per second square.

Thus, Micah knows that a car had a change in velocity of 15 m/s.To determine acceleration ,Micah also needs the <u>Time</u> of the total trip in seconds.

7 0

3 years ago

Read 2 more answers

At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Fi

Nataliya [291]

Answer:

The distance from the charge is 3.35 m.

Explanation:

Given that,

Electric potential, V = 635 V

Magnitude of electric field, E = 189 N/C

We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{V}{d}$

d is the distance from charge

$d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m$

So, the distance from the charge is 3.35 m. Hence, this is the required solution.

8 0

2 years ago

Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec

SOVA2 [1]

Answer:

$F_{net} = 6.879\times 10^{- 7}\ N$

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

$F_{net} = F + F'$

$F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N$

6 0

3 years ago

Help please! Thanks so much!

qaws [65]

You pretty much just have to convert and be mindful of significant figures.

5 km = 3 miles
0.3 cm = 0.12 in

4 0

3 years ago