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joja [24]
3 years ago
14

Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered

by it in this time
Physics
1 answer:
Soloha48 [4]3 years ago
6 0

Answer:

\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}

\boxed{\sf Distance \ covered \ (s) = 100 \ m}

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

\sf (i) \ From \ 1^{st} \ equation \ of \ motion:

\sf \implies v = u + at

\sf \implies 20 = 0 + a(10)

\sf \implies 10a = 20

\sf \implies \frac{10a}{10}  =  \frac{20}{10}

\sf \implies a = 2 \: m/ {s}^{2}

\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:

\sf \implies s = ut +  \frac{1}{2} a {t}^{2}

\sf \implies s = (0)(10) +  \frac{1}{2}  \times 2 \times  {(10)}^{2}

\sf \implies s =  \frac{1}{ \cancel{2}}  \times  \cancel{2} \times  {(10)}^{2}

\sf \implies s =  {10}^{2}

\sf \implies s = 100 \: m

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What is the magnitude of the momentum of a 3400 kg airplane traveling at a speed of 450 miles per hour?
Vitek1552 [10]

Answer:

The magnitude of momentum of the airplane is 6.83\times 10^5\ kg-m/s.

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An electron is released a short distance above earth's surface. a second electron directly below it exerts an electrostatic forc
slamgirl [31]

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Clean up the units first:

Cancel the C² on the right side, then divide each side by Newton:

8.926 x 10⁻³⁰ = (8.98755 x 10⁹ m²) x (1.602 x 10⁻¹⁹)² / D²

Now, let's multiply both sides by (D² x 10²⁹) :

D² x 8.926 x 10⁻¹ = (8.98755 m²) x (1.602)²

Divide each side by (0.8926):

D² = (8.98755 x 1.602²) / (0.8926)  meter²

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I am shocked, impressed, and amazed !

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3 0
3 years ago
The 1350-kg car has a velocity of 24 km/h up the 8-percent grade when the driver applies more power for 18 s to bring the car up
Brrunno [24]

Answer:

Explanation:

We shall apply concept of impulse to solve the problem  .

Impulse = force x time

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4 0
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