Suppose you increase your walking speed from 7 m/s to 15 m/s in a period of 3 s. What is your acceleration?

A 1.5kg cart moves along a track at 0.20m/s until it hits a fixed bumper at the end of the track. Find the average force exerted

DochEvi [55]

Answer:

Net force will be 4.875 N

Explanation:

We have given mass of the cart m = 1.5 kg

Initial velocity of the cart = 0.2 m/sec

And final velocity of the car = - 0.125 m/sec ( Negative direction is due to opposite direction )

Instant of time $\Delta t=0.1sec$

Change in momentum is given by

$\Delta P=1.5\times (0.2-(-0.125))=1.5\times 0.325=0.4875kgm/sec$

Now force is given by

$F=\frac{\Delta P}{\Delta t}=\frac{0.4875}{0.1}=4.875N$

Net force will be 4.875 N

6 0

3 years ago

Whats the answer<br> ----------------------------

GenaCL600 [577]

Answer:

repel each other

Explanation:

The magnitude of the charge of an electron is called... ... If a positively-charged glass rod is suspended so that it turns easily and another positively-charged glass rod is brought close to it, the two rods will... Repel each other.

8 0

2 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

3 years ago

Why is an element considered a pure substance?

anastassius [24]

Answer:

An element is a pure substance and is made of only one type of atom.

Explanation:

It cannot be broken down into a simpler substance.

3 0

2 years ago

Find the cost of excavating a space 84 ft long, 42 ft wide, and 9 ft deep at a cost of $39/yd3. (simplify your answer completely

m_a_m_a [10]

The cost of excavating a space of 84 ft long, 42 ft wide, and 9 ft deep is $45864

Information about the problem:

Space long= 84 ft
Space wide= 42 ft
Space deep= 9 ft
Cost by yard3 = $39/yd3
Total cost= ?

To solve this problem, we have to state the equation using the information of the problem:

Calculating the volume of the total space:

space volume = space long * space wide * space deep

space volume = 84 ft * 42 ft * 9 ft

space volume = 31752 ft3

By converting the volume from ft3 to yd3, we have:

31752 ft3 * (0,037037 yd3 / 1 ft3) = 1176 yd3

Calculating the cost of excavating the volume space:

Total cost = space volume * cost by yard3

Total cost = 1176 yd3 * $39/yd3

Total cost = $45864

<h3>What is volume?</h3>

It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.

Learn more about volume at: brainly.com/question/12628341

#SPJ4

4 0

1 year ago