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Alex Ar [27]
3 years ago
14

A tiger runs one kilometer. The tiger does 15,000 J of work and has a total power output of 500 W. How many minutes does it take

the tiger to run the distance? (Points : 3)
0.30 sec
30 sec
3.0 sec
300 sec
Physics
1 answer:
defon3 years ago
3 0
Power is the amount of energy consumed per unit time. Having no direction, it is a scalar quantity. <span>As is implied by the equation for </span>power<span>, a unit of </span>power <span>is equivalent to a unit of work divided by a unit of time. The formula would be as follows:

P = W/t 

We calculate as follows:

500 W = 15000 J / t
t = 30 s</span>
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A 1.2 kg block of wood hangs motionless from strings. A 50 gram bullet, traveling horzontally, strikes the block and becomes emb
Firdavs [7]

Answer:

speed of the bullet before it hit the block is 200 m/s

Explanation:

given data

mass of block m1 = 1.2 kg

mass of bullet m2 = 50 gram = 0.05 kg

combine speed V= 8.0 m/s

to find out

speed of the bullet before it hit the block

solution

we will apply here conservation of momentum that is

m1 × v1 + m2 × v2 = M × V    .............1

here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet

put all value in equation 1

m1 × v1 + m2 × v2 = M × V

1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8

solve it we get

v2 = 200 m/s

so speed of the bullet before it hit the block is 200 m/s

8 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
Without an atmosphere, the equatorial curve would show minimum daily values on the solstices in June when the sub-solar point is
lapo4ka [179]

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Explanation:

At the sub-solar point, the sun strikes directly at the surface with an angle of 90 degrees at a given point.

Solistice refers to that point in time when the sun’s zenith is located at the farthest point from the equator.  

During summer solistice on June 21, the sun’s zenith reaches northernmost point, sub-solar point is fixed at 23.5°S Tropic of Cancer making the earth tilt 23.4 degrees

During winter soliscitse on December 21, the sub-solar point is fixed at)  Tropic of Capricorn.

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It says find the slope for each line I'm stuck on number one can you help me
Allushta [10]
\text{slope}=\frac{y_2-y_1}{x_2-x_1}

(-2, -1)(3, 1)

Therefore,

\text{slope}=\frac{1-(-1)}{3-(-2)}=\frac{1+1}{3+2}=\frac{2}{5}

7 0
1 year ago
Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how
zimovet [89]

Look at the title of the graph, in small print under it.

Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !

3 0
3 years ago
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