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lions [1.4K]
3 years ago
15

If you travel at 3 m/s for 12 seconds, how far did you travel?

Physics
2 answers:
Rom4ik [11]3 years ago
8 0

Answer:

\boxed {\tt 36 \ meters }

Explanation:

We want to find how far was traveled, or the <u>distance.</u>

The formula for distance is:

d=s*t

where s is the speed and t is the time.

The speed is 3 meters per second and the time is 12 seconds.

s= 3 \ m/s \\t= 12 \ s

Substitute the values into the formula.

d= 3 \ m/s * 12 \ s

Multiply. Note that when multiply, the seconds, or s will cancel.

d= 3 \ m * 12

d= 36 \ m

d= 36 \ meters

The distance traveled was 36 meters.

ella [17]3 years ago
4 0

Answer:

V=3 m/s

t=12 seconds

S=?

S=V×t

S=3×12

S=36meters

So distance you travel is 36meters.

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An athlete kicks a soccer ball that starts at rest so that it leaves their foot with a speed of 10m/s from the top o f a rectang
kirza4 [7]

Answer:

a=500m/s^2

Explanation:

We need only to apply the definition of acceleration, which is:

a=\frac{v_f-v_i}{t_f-t_i}

In our case the final velocity is v_f=10m/s, the initial velocity is v_i=0m/s since it departs from rest, the final time is t_f=0.02s and the initial time we are considering is t_i=0s

So for our values we have:

a=\frac{10m/s-0m/s}{0.02s-0s}=500m/s^2

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A girl is swinging on a rope. Does the rope supporting the swing act on the girl at the very top of her swing?
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A hammer of mass M is moving at speed v0 when it strikes a nail of negligible mass that is stuck in a wooden block. The hammer i
OleMash [197]

Answer:

i think it would be D

8 0
3 years ago
A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top
Advocard [28]

Answer:

a)  Em₀ = 42.96 104 J , b)   W_{fr} = -2.49 105 J , c)  vf = 3.75 m / s

Explanation:

The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

        Em = K + U

a) Let's look for the initial mechanical energy

      Em₀ = K + U

      Em₀ = ½ m v2 + mg and

      Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

      Em₀ = 36 104 + 6.96 104

      Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

    W_{fr} = Em₂ -Em₀

     Em₂ = K + U

     Em₂ = ½ m v₂² + m g y₂

     Em₂ = ½ 50 85 2 + 50 9.8 427

     Em₂ = 180.625 + 2.09 105

     Em₂ = 1,806 105 J

     W_{fr} = Em₂ -Em₀

     W_{fr} = 1,806 105 - 4,296 105

     W_{fr} = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

       W_{fr} = ΔEm

       W_{fr} = Emf - Emo

       -1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

       ½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

       ½ 50.0 vf² = 0.561

       vf = √ 0.561 25

      vf = 3.75 m / s

6 0
4 years ago
suppose you increase your walking speed from 4 m/s to 12 m/s in a period of 1 second what is your acceleration
KiRa [710]

a   = 12 ms⁻¹ - 4ms⁻¹ / 1 s

a  =    8 ms⁻²

7 0
3 years ago
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