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Dvinal [7]
3 years ago
14

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car

Physics
1 answer:
Novay_Z [31]3 years ago
0 0

Answer:

4.5 *10^{3}\ kgm/s^{2}

Explanation:

Given:

Mass= 1.00 * 10^{3} \ kilograms,\\Acceleration =4.5\  meters/second^{2}

As we know that

Force=Mass * Acceleration

Putting the value of mass and Acceleration  we get ,

Force\  = \ 1.00 * 10^{3} * 4.5\ m/s^{2}

Force\  =4.5 * \ 10^{3} \  kgm/s^{2}

Therefore Net force is :4.5 *10^{3}\ kgm/s^{2}

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4 0
3 years ago
The position of a dragonfly that is flying parallel to the ground is given as a function of time by r⃗ =[2.90m+(0.0900m/s2)t2]i^
Arlecino [84]

The solution for this problem is:

 
r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s² 
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7 0
3 years ago
Read 2 more answers
Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh
mel-nik [20]

Answer:u=\frac{v}{2}\sqrt{5-4sin\phi }

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

m(2v)+m(-vsin\phi )=2m(ucos\theta )

2ucos\theta =v(2-sin\theta )------1

Conserving momentum in x direction

mvcos\phi =2musin\theta-----2

squaring and adding 1 &2

(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2

4u^2=4v^2+v^2-4v^2sin\phi

4u^2=5v^2-4v^2sin\phi

u=\frac{v}{2}\sqrt{5-4sin\phi }

7 0
3 years ago
If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?
leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

<span>[density * volume]_salt water = [density * volume]_fresh water                 ---> 1</span>

 

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

 

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

<span>h = 0.72 m        (ANSWER)</span>

8 0
3 years ago
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VikaD [51]

Answer:

vrzjxuxuxjcckhcouvou

Explanation:

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5 0
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