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marshall27 [118]
2 years ago
7

If an egg person starts from rest then falls directly downward and hits the ground with a velocity of 12 m/s but

Physics
2 answers:
Anni [7]2 years ago
6 0

Answer:

Explanation: Determine the gravitational acceleration. ...

Decide whether the object has an initial velocity. ...

Choose how long the object is falling. ...

Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt

vfiekz [6]2 years ago
4 0

Answer:

14

Explanation:

this is totally my work :)

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What do you call the material or "stuff" that a wave travels through
frutty [35]
C I think if not it’s B
8 0
2 years ago
Read 2 more answers
Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),
yuradex [85]

Answer:

v=3.564\ m.s^{-1}

\Delta v =2.16\ m.s^{-1}

Explanation:

Given:

  • mass of John, m_J=30\ kg
  • mass of William, m_W=30\ kg
  • length of slide, l=3\ m

(A)

height between John and William, h=1.8\ m

<u>Using the equation of motion:</u>

v_J^2=u_J^2+2 (g.sin\theta).l

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3

v_J=5.94\ m.s^{-1}

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

m_J.v_J+m_w.v_w=(m_J+m_w).v

where: v = velocity with which they move together after collision

30\times 5.94+0=(30+20)v

v=3.564\ m.s^{-1} is the velocity with which they leave the slide.

(B)

  • frictional force due to mud, f=105\ N

<u>Now we find the force along the slide due to the body weight:</u>

F=m_J.g.sin\theta

F=30\times 9.8\times \frac{1.8}{3}

F=176.4\ N

<em><u>Hence the net force along the slide:</u></em>

F_R=71.4\ N

<em>Now the acceleration of John:</em>

a_j=\frac{F_R}{m_J}

a_j=\frac{71.4}{30}

a_j=2.38\ m.s^{-2}

<u>Now the new velocity:</u>

v_J_n^2=u_J^2+2.(a_j).l

v_J_n^2=0^2+2\times 2.38\times 3

v_J_n=3.78\ m.s^{-1}

Hence the new velocity is slower by

\Delta v =(v_J-v_J_n)

\Delta v =5.94-3.78= 2.16\ m.s^{-1}

8 0
3 years ago
PLEASE ANSWER ASAP
neonofarm [45]

Answer:

2.55sec

Explanation:

time = distance/speed = (87 m)/(34 m/s) = 2.55sec

6 0
2 years ago
the resistor of values 6 ohm,6 ohm are connected in series and 12 ohm are connected in parallel. the equivalent resistance of th
andrezito [222]

Answer:

The equivalent or total resistance of the circuit is 6

Explanation:

6 &6 are in series

6+6=r

r= 12

1/Rtotal= 1/12+1/2

1/Rt=2/12=1/6

Rt=6

6 0
3 years ago
John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2
Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function cos(\omega t +\phi) is defined between -1 and 1, So it is not possible obtain a value 2\pi greater.  

In addition, if you  move the function cosine a T Value, and T is the Period,  the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have f=1+cos(\omega t +\phi), the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the f=1+cos(\omega t +\phi), and its mean that the function f=cos(\omega t +\phi), in general, is not greater than cos(\omega t +\phi).

3 0
3 years ago
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