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Julli [10]
2 years ago
8

The velocity of a body is increases from 10 m/s ti 15 m/s in 5seconds calculate its acceleration​

Physics
2 answers:
Veseljchak [2.6K]2 years ago
7 0

Answer:

acceleration = v-u/ t

= 15-10/5

= 5/5

= 1 m/s2

Explanation:

hope this helped you.

Romashka-Z-Leto [24]2 years ago
6 0

Answer:

\boxed {\boxed {\sf 1 \ m/s^2}}

Explanation:

Acceleration is the change in velocity over the change in time. Therefore, the formula for calculating acceleration is:

a= \frac{v_f-v_i}{t}

Since the body's velocity increased <em>from</em> 10 meters per second <em>to</em> 15 meters per second 15 m/s is the final velocity and 10 m/s is the initial velocity. The time is 5 seconds.

  • v_f= 15 m/s
  • v_i= 10 m/s
  • t= 5 s

a= \frac{ 15 \ m/s - 10 \ m/s}{5 \ s}

Solve the numerator.

  • 15 m/s - 10 m/s = 5 m/s

a= \frac{ 5 \ m/s }{5 \ s}

Divide.

a= 1 \ m/s/s

a= 1 m/s^2

The acceleration is <u>1 meter per second squared.</u>

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Kobotan [32]
<h3>Question:</h3>

How to find g (acceleration due to gravity)

<h3>Solution:</h3>

We know,

Acceleration due to gravity (g)

=  \frac{GM}{ {R}^{2} }

where, G = Gravitational constant

= 6.67 \times  {10}^{11} N {m}^{2}/k {g}^{2}  \\

M = Mass of the earth

= 6 \times  {10}^{24} \:  kg

R = Radius of the earth

= 6.4 \times  {10}^{6} m

Putting these values of G, M and R in the above formula, we get

g \:  =  \:  \frac{6.67 \times  {10}^{11} N {m}^{2}/k {g}^{2}   \times \: 6 \times  {10}^{24} \:  kg }{(6.4 \times  {10}^{6}m {)}^{2}  }  \\  = 9.8m/ {s}^{2}

So, the value of acceleration due to gravity is

9.8m/s ^{2}

Hope it helps.

Do comment if you have any query.

5 0
3 years ago
A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth
Alex73 [517]

Answer:

F_{Earth}= 15.57 N

F_{Moon}= 2.60 N

F_{Uranus}= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

F_{Earth}=m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

F_{Moon}=m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

F_{Uranus}=m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

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a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight
stiv31 [10]

Answer:

  W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

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Where F is the force of the door and W the weight of water

         W = mg

We use the concept of density

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        m = ρ V

The volume of the water column is

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We replace

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On the other side the cylinder cover has a pressure

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We match the two equations

       P A = ρ A h g

        P = ρ g h

        P = 39200 Pa

The weight of the water column is

       W  = 1000 9.8 4 A

       W / A = 39200 kg / m²

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