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ella [17]
2 years ago
11

POTENTIAL I KINETIC ENERGY

Physics
1 answer:
TEA [102]2 years ago
4 0

7kinetic energy is decreasing in B

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A wave created by shaking a rope up and down is called a.
Sidana [21]
A wave created by shaking a rope up and down is called a:
(According to me and certified experts) Transverse wave.

If this helps, then please give Brainliest!
5 0
2 years ago
Mark and David are loading identical cement blocks onto David’s pickup truck. Mark lifts his block straight up from the ground t
Pepsi [2]

Answer:

b) true. The jobs are equal

Explanation:

The work on a body is the scalar product of the force applied by the distance traveled.

    W = F. d

Work is a scalar, the work equation can be developed

    W = F d cos θ

Where θ is the angle between force and displacement

Let's apply these conditions to the exercise

a) False, if we see the expression d cosT is the projection of the displacement in the direction of the force, so there may be several displacement, but its projection is always the same

b) true. The jobs are equal dx = d cosθ

c) False, because the force is equal and the projection of displacement is the same

d) False, knowledge of T is not necessary because the projection of displacement is always the same

e) False mass is not in the definition of work

5 0
3 years ago
Both independent and dependent clauses
Assoli18 [71]

Answer: have a subject and a verb

Explanation:

5 0
3 years ago
Read 2 more answers
A truck on the freeway originally moving at 6.6 meters/second accelerates uniformly with acceleration a = 2.8 meters/second2 for
Sonja [21]

Answer:

139.514 metres

Explanation:

Initial velocity of the truck = 6.6 m/s

Acceleration of the truck = 2.8 m/s^2

Time interval = 7.9 s

Therefore we use the formula,

s = ut + 1/2 at^2

*where s(the distance travelled)...u(the initial velocity)...t(the time period)

; s = 6.6(7.9) + 1/2 (2.8)(7.9)^2

; s = 52.14 + 87.374

The distance moved by the truck = 139.514m

8 0
3 years ago
A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas
Radda [10]

Answer:

change in momentum, \Delta p=7.65 \,kg.m.s^{-1}

  • \Delta p_x= 6.6251 \,kg.m.s^{-1}
  • \Delta p_y= 3.825 \,kg.m.s^{-1}

Average Force, F=144.3396\,N

  • F_x=125.0018\,N
  • F_y=72.1698\,N

Explanation:

Given:

angle of kicking from the horizon, \theta= 30^{\circ}

velocity of the ball after being kicked, v=18 m.s^{-1}

mass of the ball, m=0.425\, kg

time of application of force, t=5.3\times 10^{-2}\,s

We know, since body is starting from the rest

\Delta p=m.v.....................(1)

\Delta p=0.425\times 18

\Delta p=7.65 \,kg.m.s^{-1}

Now the components:

\Delta p_x= 7.65\times cos 30^{\circ}

\Delta p_x= 6.6251 \,kg.m.s^{-1}

similarly

\Delta p_y= 7.65\times sin 30^{\circ}

\Delta p_y= 3.825 \,kg.m.s^{-1}

also, impulse

I=F\times t.........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

F\times t=m.v

F\times 5.3\times 10^{-2}= 7.65

F=144.3396\,N

Now, the components

F_x=144.3396\times cos 30^{\circ}

F_x=125.0018\,N

&

F_y=144.3396\times sin 30^{\circ}

F_y=72.1698\,N

6 0
3 years ago
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