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icang [17]
3 years ago
8

The bowl now contains . Milliters of oil

Physics
2 answers:
Sergeeva-Olga [200]3 years ago
7 0
Data????????????????????????????
erik [133]3 years ago
4 0
Post the rest of the question, I think you missed some of the data. Then I will try to help
You might be interested in
What is the kinetic energy of a 50-kg child running to catch the school bus at
Vinvika [58]

Answer:

Option C

100 J

Explanation:

Kinetic energy, KE is given by

KE=0.5mv^{2} where m is the mass and v is the velocity

Substituting 50 Kg for mass, m and 2 m/s for velocity v then we obtain

KE=0.5*50*2^{2}=100 J

Therefore, the child's kinetic energy is equivalent to 100 J

6 0
3 years ago
Read 2 more answers
A ball is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the wa
vovangra [49]

Answer:

a) 48.5 ft/s

b) 36.5 ft

c) -80.3 ft/s

Explanation:

a)

The equation of motion of the ball is :

y(t) = -16.1 ft/s^2 * t^2 + Vo*t

Where Vo is the initial velocity

If y(5s) = - 160 ft:

-160 ft = -16.1 ft/s^2 * (5 s)^2 + Vo*(5s)

Solving for Vo

Vo  = (16.1*25- 160) ft / 5s = 48.5 ft/s

b)

To answer this question we must first know when the velocity became zero, at this time is when the ball was at its highest point.

v(t) = -32.2 ft/s^2 * t + Vo

t = Vo/32.2ft/s^2 = 1.5 s

And now, the highest point which the ball reached is given by:

y(1.5s) = -16.1 ft/s^2 * (1.5)^2 + Vo*(1.5s)

y(1.5s) = 36.52 ft

c)

We now need the time at which y(t') = -64 ft

-64 = -16.1*t'^2 + 48.5*t'

By means of the quadratic formula, we find that

t' = 4.00498 s ≈ 4 s

And the velocity at t = 4s is:

v(4s) = -32.2 ft/s^2 * 4s +48.5 ft/s = -80.3 ft/s

3 0
3 years ago
An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0
sleet_krkn [62]

Answer:

22.26 years

, 15.585 light years  , 11.13 light years

Explanation:

a)

t' = t/(\sqrt{1-(v/(c*v)/c)}

= 15.9/\sqrt{(1-0.7*0.7)}

= 22.26 years

b)

0.7*c*22.26 years

=15.585 light years  

c)

0.7*c*15.9

=11.13 light years

3 0
3 years ago
A 5kg ball is on top of the school building at a height of 40m above the ground.
mojhsa [17]

Answer:

A-Caclcuate the potential energy of the ball at that height

Explanation:

(a). Mass of the Body = 10 kg.

Height = 10 m.

Acceleration due to gravity = 9.8 m/s².

Using the Formula,Potential Energy = mgh

= 10 × 9.8 × 10 = 980 J.

(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.

∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.

∴ Kinetic Energy = 980 J.

(c). Kinetic Energy = 980 J.

Mass of the ball = 10 kg.

∵ K.E. = 1/2 × mv²

∴ 980 = 1/2 × 10 × v²

∴ v² = 980/5

⇒ v² = 196

∴ v = 14 m/s.

3 0
2 years ago
What is the average power consumption in watts of an appliance that uses 5.00 kWh of energy per day? How many joules of energy d
denpristay [2]

Answer:

(A)  power  = 0.208 kW = 208 watts

(B)  energy = 6.6 x 10^{9} joules

Explanation:

energy consumed per day = 5 kWh

(a) find the power consumed in a day

         1 day = 24 hours

        power = \frac{energy}{time}

        power = \frac{5}{24}

          power  = 0.208 kW = 208 watts

         

(b) find the energy consumed in a year

    assuming it is not a leap year and number of days = 365 days

     1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds

            energy = power x time

            energy = 208 x 31,536,000

            energy = 6.6 x 10^{9} joules

5 0
3 years ago
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