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Paul [167]
2 years ago
14

Convert 5.6 bushels to hectoliters

Physics
1 answer:
Pavel [41]2 years ago
4 0

Answer:

Explanation:

5.6 bushel(1 hectoliter/2.8 bushel) = 2.0 hectoliters

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A 10.0kg water balloon is dropped from a height of 12.0m. Calculate the speed of the balloon just before it hits the ground
kolbaska11 [484]

Answer:

15.5 m/s.

Explanation:

Potential energy of the balloon has been converted to kinetic energy.

potential energy = kinetic energy.

mgh = ½mv².

10* 10* 12= ½ *10 *v²

1200 = 5v²

v²=1200÷5

v=√240

v= 15.49~15.5 m/s.

5 0
2 years ago
What is the available energy content, in joules, of a battery that operates a 2.00-W electric clock for 18 months
iren2701 [21]

Answer:

93312000 Joules

Explanation:

Energy can be explained as the capacity for doing work. And can be expressed as

E= pt....................eqn(1)

Where

E= Energy in Joules

P= power in Watt= 2.00W

t= time in seconds

Time= 18 months, which can be converted to "seconds" as;

18 months = [ 18 monthsx 30 days x 24 hours x 60 min x 60 sec] = 46656000

Then

time= 46656000 seconds

Power=2 watt

If we substitute into equation (1) we have

Energy = 2 watt × 46656000 seconds

Energy= 93312000 Joules

7 0
2 years ago
Earth's rising atmospheric temperature is MOSTLY caused by unnatural
nataly862011 [7]
Greenhouse gases
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2 years ago
The magnetic field of a magnet is strongest
Talja [164]
The answer is C !!!!
4 0
2 years ago
A 1.2 W point source emits sound waves isotropically. Assuming that the energy of the waves is conserved, find the intensity (a)
pshichka [43]

Answer:

(a) I = 3.7\times 10^{-2}\ W/m^{2}

(b) I = 1.9\times 10^{-2}\ W/m^{2}

Explanation:

Given:

Point source P_{b}=1.2\ W

Distance from source Part a L = 1.6\ m

Distance from source Part b L = 2.2\ m

Solution:

Part (a)

Using intensity formula at distance L from an isotropic point.

I = \frac{P_{b}}{4\pi(L)^{2}}  -------------------(1)

Substitute L = 1.6\ m and P_{b}=1.2\ W in equation 1..

I = \frac{1.2}{4\times 3.14(1.6)^{2}}

I = \frac{1.2}{32.15}

I = 0.037\ W/m^{2}

I = 3.7\times 10^{-2}\ W/m^{2}

Part (b)

Substitute L = 2.2\ m and P_{b}=1.2\ W in equation 1.

I = \frac{1.2}{4\times 3.14(2.2)^{2}}

I = \frac{1.2}{60.79}

I = 0.019\ W/m^{2}

I = 1.9\times 10^{-2}\ W/m^{2}

Therefore, the intensity at distance 1.6 m from the source:

I = 1.9\times 10^{-2}\ W/m^{2}.

And, the intensity at distance 2.2 m from the source:

I = 1.9\times 10^{-2}\ W/m^{2}.

7 0
3 years ago
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