4 years ago

How much force is required to accelerate a car with a mass 2000 kg to 3m/s^2?

Physics

2 answers:

asambeis [7]4 years ago

5 0

The answer is C. 6000N

Cloud [144]4 years ago

5 0

Answer:

Option a) 6000 kg.m/s

Explanation:

mass = 2000 kg

acceleration = 3m/s^2

force = mass × acceleration

= 2000 × 3

= 6000 kg.m/s

Hope this helps

plz mark as brainliest!!!!!

You might be interested in

The primary source of evidence proposed by many scientist to support the theory of an ancient earth is ____ dating.

vova2212 [387]

It depends on your definition of “ancient.” Radiometric dating using Carbon-14 can reliably date back to about 50,000 years, uranium-lead or lead-lead dating can date back multiple millions, potassium-argon dating can reach 1.5 billion, and rubidium-strontium can reach 50 billion (nearly 4x the age of the universe). It depends on the context in which this question is being asked.

7 0

3 years ago

Where is all of none law found

alexgriva [62]

Website on Google,use well!

7 0

3 years ago

A graduated cylinder has 50ml of water in it before A marble is dropped in after the marble is dropped into the graduated cylind

Vikentia [17]

Answer:

0.05L

Explanation:

if i understqnd this correctly the water line should increase with the mass of the marble

100-50=50ml

=0.05L

5 0

2 years ago

Why the friction between an object and a surface is less than when the surface is titled than when the same surface is horizonta

torisob [31]

Answer:

nvjch

ffufuxuhxjcjfbfdhdh

7 0

3 years ago

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) a = 2i − 5j + k

EastWind [94]

Answer:

Angle=70°

Explanation:

Given data

a=2i-5j+k

b=9i-k

To find

Angle Θ

Solution

As we know from dot product rule that

$a.b=|a||b|cos\alpha \\cos\alpha=\frac{a.b}{|a||b|}\\\alpha=cos^{-1} (\frac{a.b}{|a||b|})$

First we need to find a.b

i.i=j.j=k.k=1

i.j=i.k=j.k=0

$a.b=(18)+(-1)\\a.b=17$

Now for |a| and |b|

$|a|=\sqrt{x^{2} +y^{2}+z^{2} }\\|a|=\sqrt{(2)^{2} +(-5)^{2}+(1)^{2} }\\ |a|=5.477\\And\\|b|=\sqrt{x^{2} +y^{2}+z^{2} }\\|b|=\sqrt{(9)^{2} +(0)^{2}+(-1)^{2} }\\|b|=9.055$

So the angle is given as

$\alpha=cos^{-1} (\frac{a.b}{|a||b|})\\\alpha=cos^{-1} (\frac{17}{(5.477)*(9.055)})\\\alpha =70^{o}$

4 0

3 years ago