3 years ago

Someone help me with these questions

Physics

1 answer:

Andreas93 [3]3 years ago

6 0

Answer:

Im not 100% sure so don't completly rely on these answer s

#1 is B

#2 ia B

#3 is A

#4 is A

#5 is A

and i think #6 is A

im 100% sure that 3 and 4 are A tho

Explanation:

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Velocity has two pieces of information.what are they

seraphim [82]

They are speed and direction.

3 0

4 years ago

a ball is dropped from rest at a height of 80 m above the ground. what is the speed just as it hits the ground?

Ghella [55]

Δx=Vi(t)+.5at²
Δx=80
a=9.8
Vi=0 since its from rest, the initial velocity is 0.
80=0+.5(9.8)t²
t=16.33
You can use one of the big five equations: Vf=Vi+at
Vf=0+9.8(16.33)
Vf=160.034 m/s

4 0

4 years ago

A dragster in a race accelerated from rest to 60 m/s by the time it reached the finish line. The dragster moved the distance fro

Andreyy89

Answer:

l don't now but l think the is 160

Explanation:

160 or 810

6 0

4 years ago

An automobile accelerates uniformly from a speed of 60 km/hr (16.67 m/s) to a speed of 80 km/hr (22.2 m/s) in 5 s. Determine the

monitta

Answer:

The acceleration is $1.106\ m/s^2$ and the distance covered is 97.17 m.

Explanation:

Given that,

Initial speed of an automobile, u = 60 km/hr = 16.67 m/s

Final speed of an automobile, v = 80 km/hr = 22.2 m/s

Time, t = 5 s

We need to find the acceleration of the car and the distance traveled in this 5 sec interval. Let a is the acceleration. Using the definition of acceleration as :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{22.2-16.67}{5}\\\\a=1.106\ m/s^2$

Let d is the distance covered. Using the third equation of motion to find it as follows :

$v^2-u^2=2as\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{(22.2)^2-(16.67)^2}{2\times 1.106}\\\\s=97.17\ m$

So, the acceleration is $1.106\ m/s^2$ and the distance covered is 97.17 m.

7 0

3 years ago

Plzzzzzz plzzzzz help will give brainlist

ad-work [718]

Either the first one or the second one but i believe it’s the stricter on

4 0

3 years ago