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ss7ja [257]
2 years ago
7

(8th grade HELP)

Physics
1 answer:
Shalnov [3]2 years ago
4 0
What is the question your asking
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The car starts from a room with a constant acceleration of 5 ms-2. What path will it pass in 6 seconds and what speed will it re
Kobotan [32]

Answer: The speed will be  30 m/s .

Explanation:

Given: Initial velocity of the car: u = 0 m/s

Constant Acceleration: a = 5 m/s²

Time: t= 6 seconds

To find: Final velocity(v)

Formula:  v = u+at

Substitute values in the formula, we get

v=  0+(5)(6) m/s

⇒ v= 30 m/s

i.e. Final velocity = 30 m/s

Hence, the speed will be 30 m/s .

4 0
2 years ago
A tennis ball, 0.314kg, is accelerated at a rate of 164m/s2 when hit by a professional tennis player. What force does the player
Colt1911 [192]

Newton's 2nd law of motion:

Force = (mass) x (acceleration)

= (0.314 kg) x (164 m/s²)

=  51.5 newtons

(about 11.6 pounds).

Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.

~ I hope this helped, and I would appreciate Brainliest. ♡ ~ ( I request this to all the lengthy answers I give ! )

5 0
3 years ago
Bus starts from rest if the acceleration of the bus is 0.5 MS squared what will be the velocity at the end of two minutes and wh
Nutka1998 [239]

Explanation:

Given that,

Initial speed of the bus, u = 0

Acceleration of the bus, a = 0.5 m/s²

Let v is the velocity at the end of 2 minutes. The change in velocity divided by time equals acceleration.

So,

a=\dfrac{v-u}{t}\\\\v=u+at\\\\v=0+0.5\times 120\\\\v=60\ m/s

Let d is the distance cover during that time. So,

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(60)^2}{2\times 0.5}\\\\d=3600\ m

So, the final speed is 60 m/s and the distance covered during that time is 3600 m.

4 0
3 years ago
What effect did the Cold War have on the American space program?
chubhunter [2.5K]
The answer is D.Competition with the Soviet Union spurred American space missions.

<span>♡♡Hope I helped!!! :)♡♡
</span>
3 0
3 years ago
Read 2 more answers
Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass
kirza4 [7]

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

5 0
3 years ago
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