3 years ago

What is the kinetic energy of a 0.5 kg soccer ball that is traveling at a

Physics

1 answer:

Charra [1.4K]3 years ago

3 0

Answer:

kinetic energy = 1/2mv²

k.E =1/2 X 0.5 X 3²

k.e = 9/4

K.E = 2.25j

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sp2606 [1]

True
False
False.
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True
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4 0

3 years ago

A book starts on a shelf 1 m off of the ground. You then move the book to a higher shelf that is 3 m off of the ground. If the s

JulsSmile [24]

Answer:

Using g = 9.8: 1.02 kg, Using g = 10: 1 kg

Explanation:

E = mgh

20 = m(9.8)(3 - 1)

20 = 9.8m(2)

20 = 19.6m

m = 1.02 kg

I'm now assuming you may be using a g constant of 10, thus the close integer result, in which case the mass would be exactly 1 kilogram.

4 0

3 years ago

Which statement bet explains the relationship between the electric force between two charged objects and the distance between th

Nataliya [291]

Coulomb's Law: Force = k x q1x q2 divided distance square
where k=9x10^9 , q1 and q2 are the charge
So if you distance is halved, your force is stronger by 4 times
and if you distance is doubled, your force is 1/4
Ask me again if you aren't clear :)

4 0

3 years ago

Estimate the volume of a typical house (2050 feet squared in size and 10 feet tall) answer in units of meters squared

Aloiza [94]

Answer:

$Volume = 6248.48 m^{3}$

Explanation:

Given:

The area of the house $A = 2050\ ft^{2}$

The height of the house $h=10\ ft$

We need to find the volume of a typical house.

Solution:

We find the volume of the house by multiplying the area of the house and height of the house.

$Volume = Area\times height$

$Volume = A\times h$

Area and height of the house are known, so we substitute these value in above equation.

$Volume = 2050\times 10$

$Volume = 20500\ ft^{3}$

Now we convert the unit from feet to meter.

Divide the volume by 3.2808 for $m^{3}$

$Volume = \frac{20500}{3.2808}$

$Volume = 6248.48\ m^{3}$

Therefore, the volume of the house is $6248.48 m^{3}$

8 0

3 years ago

A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 781 K, and r

andreev551 [17]

Answer:2.517 J/K

Explanation:

Given

Reservoir 1 Temperature $T_1=781 K$

Reservoir 2 Temperature $T_2=335 K$

Let Q is the amount of heat Flows i.e. $Q=1477 J$

thus change in Entropy is given by $\frac{\sum Q}{T}$

$\Delta S=\frac{\sum Q}{T}=-\frac{Q}{T_1}+\frac{Q}{T_2}$

$\Delta S=\frac{\sum Q}{T}=-\frac{1477}{781}+\frac{1477}{335}$

$\Delta S=\frac{\sum Q}{T}=-1.891+4.4089$

$\Delta S=\frac{\sum Q}{T}=2.517 J/K$

6 0

3 years ago