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3 years ago

14

a father and his son are racing down a mountain. if they're traveling at the same velocity who has more kinetic energy

1 answer:

lakkis [162]3 years ago

8 0

Answer:

I think it's the son

Explanation:

bcs he might love it

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How a measuring scale measure our weight

Aloiza [94]

Scales measure how much something weighs—and they do it by measuring how much force exists between the object you're weighing and planet Earth. Although scales measure force, they give you measurements of mass in kilograms, grams, pounds, or whatever.

5 0

3 years ago

Determine the gravitational potential energy, in kJ, of 3 m3 of liquid water at an elevation of 40 m above the surface of Earth.

melomori [17]

Explanation:

We will calculate the gravitational potential energy as follows.

$P.E_{1} = mgz_{1}$

$P.E_{1} = (\rho V)gz_{1}$

= $1000 kg/m^{3} \times 3 m^{3} \times 9.7 \times 40 m$

= 1164000 J

or, = 1164 kJ (as 1 kJ = 1000 J)

Now, we will calculate the change in potential energy as follows.

$\Delta P.E = mg(z_{2} - z_{1})$

= $\rho \times V \times g (z_{2} - z_{1})$

= $1000 \times 3 \times 9.7 (10 - 40)m$

= -873000 J

or, = -873 kJ

Thus, we can conclude that change in gravitational potential energy is -873 kJ.

4 0

3 years ago

You lift a stuffed toy up above the ground (over your head).

BigorU [14]

Answer:

Gravitational energy

Explanation:

when over head , when it hits the floor thermal energy

7 0

3 years ago

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

Electro negativity increases when atoms __. (Apex)

abruzzese [7]

C. I took the test...........

3 0

3 years ago

Read 2 more answers