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4 years ago

What is the average velocity of a car starting from 0 and travels 7 meters in 9 seconds?

Physics

1 answer:

Leni [432]4 years ago

5 0

Your question has been heard loud and clear.

Averge velocity formula= Total distance travelled / total time taken.

Total distance=7meters

Total time taken=9 seconds.

Average velocity = 7/9= 0.77 metres/second

Average velocity= 0.77m/s

Thank you.

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PLEASE I NEED HELP I AM REALLY STUCK!!!

AlekseyPX

Answer: Force = Mass X Acceleration

F = 5 x 2

F = 10 N

8 0

3 years ago

An object floats in water with 5 8 of its volume submerged. The ratio of the density of the object to that of water is

ki77a [65]

Answer:

$\dfrac{5}{8}$

Explanation:

m = Mass of object = $\rho v$

m' = Mass of water = $\rho' v'$

$\rho$ = Density of object

$\rho'$ = Density of water

Weight of the water displaced is the force in the case of floating objects

According to the question

$v'=\dfrac{5}{8}v$

In the case of floating objects

$W=W'\\\Rightarrow mg=m'g\\\Rightarrow \rho vg=\rho'v'g\\\Rightarrow \rho v=\rho' \dfrac{5}{8}vg\\\Rightarrow \rho=\rho' \dfrac{5}{8}\\\Rightarrow \dfrac{\rho}{\rho'}=\dfrac{5}{8}$

The ratio of the density of the object to that of water is $\dfrac{5}{8}$

3 0

3 years ago

You pull a wagon with a force of 20 N. The wagon has a mass of 10 kg. What is the wagon's acceleration?

Nana76 [90]

Answer:

<h3>The answer is 2 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

where

f is the force

m is the mass

From the question

f = 20 N

m = 10 kg

We have

$a = \frac{20}{10} \\$

We have the final answer as

Hope this helps you

5 0

3 years ago

A car undergoing uniform acceleration travels 100 meters from a standing start in a given period of time. If the time were incre

iogann1982 [59]

Answer:

d = 1600 m

Explanation:

If car start from rest and accelerates uniformly then the displacement of the car is given as

$d = v_i t + \frac{1}{2}at^2$

now we know

$d = 0 + \frac{1}{2}at^2$

now if the same car will accelerate from rest for four times of the previous time interval

then we will have

$d = 0 + \frac{1}{2}a(4t)^2$

$d = 16(\frac{1}{2}at^2)$

so the distance will be 16 times more

$d = 1600 m$

5 0

4 years ago

three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25

ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0

3 years ago