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2 years ago

8

A car speeds up as it rolls down a hill. Which is this an example of? A. positive acceleration B. negative acceleration C. relat

ive velocity

1 answer:

alina1380 [7]2 years ago

4 0

I believe it is A. positive acceleration since it is speeding up

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An upward force of 32.6 N is applied via a string to lift a ball with a mass of 2.8 kg. (a) What is the gravitational force acti

Igoryamba

Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

There are two forces acting on the ball, one directed upward (assuming this direction as positive, along the y-axis) which is the tension on the string (lifting force), and another aimed downward, which is the attractive force due to gravity.
Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
So, we can write the following expression for Fg:

$F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)$

b)

The net force on the ball, will be just the difference between the lifting force (32.6 N) and the force due to gravity, Fg:

$F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N (2)$

c)

According Newton's 2nd Law, the acceleration caused by a net force on a point mass (we can take the ball as one) is given by the following expression:

$a = \frac{F_{net} }{m} = \frac{5.2N}{2.8kg} = 1.9 m/s2 (3)$

3 0

2 years ago

During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying

adelina 88 [10]

The maximum height attained is 460 m.

<h3>What is the maximum height?</h3>

We know that the final velocity of a body is 0 m/s at the maximum height which is the greatest height that is attained by the body. We now use the formula;

v^2 = u^2 -2gh

Given that v = 0 m/s

u^2 = 2gh

h = u^2/2g

v = final velocity

u = initial velocity

h = maximin height

g = acceleration due to gravity

h = (95)^2/2 * 9.8

h = 460 m

Learn more about maximum height:brainly.com/question/6261898

#SPJ1

6 0

1 year ago

How is it possible for a single sedimentary rock to contain rock particles animal shells and minerals that crystallized from wat

shutvik [7]

"Sedimentary rocks are types of rock that are formed by the deposition of material at the Earth's surface and within bodies of water. Sedimentation is the collective name for processes that cause mineral and/or organic particles (detritus) to settle and accumulate or minerals to precipitate from a solution."

7 0

2 years ago

Read 2 more answers

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback ca

love history [14]

Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. $s_{o}$ = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

Truck:

$v_{i}$ = 33.2 m/s, a = 0 (since the velocity is constant), $s_{o}$ = 0

Using $s =s_{o}+v_{i}t+1/2at^{2}$

s = 33.2t .......... eq (1)

Hatchback:

$a=5m/s^{2}$ , $v_{i}$ = 0 m/s (since initial velocity is zero), $s_{o}$ = 0

Using $s =s_{o}+v_{i}t+1/2at^{2}$

putting in the data we will get

$s=(1/2)(5)t^{2}$

now putting 's' value from eq (1)

$2.5t^{2}-33.2t=0$

which will give,

t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

7 0

3 years ago

Read 2 more answers

You normally drive a 12-h trip at an average speed of 100 km/h . Today you are in a hurry. During the first two-thirds of the di

kherson [118]

Answer:

78 km/h

Explanation:

If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:

100 km/h · 12 h = 1,200 km

Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.

1,200 * 2/3 = 800 km

I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.

To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.

116 km/1 h = 800 km/? h
800 = 116 · ?
? = 800/116
? = 6.89655172

I spent 6.89655172 hours driving during the first 2/3 of the distance.

Now, I need to subtract this value from 12 hours to find the remaining time I have left.

12 h - 6.89655172 h = 5.10344828 h

Using this remaining time and my remaining distance, I can calculate my average speed.

? km/1 hr = 400 km/5.10344828 h
5.10344828 · ? = 400
? = 400/5.10344828
? = 78.3783783148

My average speed during the last third of the distance is around 78 km/h.

8 0

2 years ago