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In-s [12.5K]
2 years ago
5

If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?

Physics
1 answer:
steposvetlana [31]2 years ago
3 0

Hi there!

\large\boxed{\text{B) 20 meters}}

We know that:

E_T = U + K

U = Potential Energy (J)

K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

U + K = 50 + 50 = 100 J

To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:

100J = mgh

We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².

50J = 10(10)m

m = 1/2 kg

Now, solve for height given that E = 100 J:

100J = 1/2(10)h

100J = 5h

<u>h = 20 meters</u>

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While participating in a blood drive at school, Keona learns that blood has a density of 1.06 g/mL. She donates one pint of bloo
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Answer:

m≈501.57 g

Explanation:

The density formula is:

d=m/v

Let’s rearrange the formula for m. m is being divided by v. The inverse of division is multiplication, so multiply both aides by v.

d*v= m/v*v

d*v=m

The mass can be found by multiply the density and the volume.

m=d*v

The density is 1.06 grams per milliliter and the volume is 473.176 milliliters.

d= 1.06 g/mL

v= 473.176 mL

Substitute the values into the formula.

m= 1.06 g/mL * 473.176 mL

Multiply. When multiplying, the mL will cancel out.

m= 501.56656 g

Let’s round to the nearest hundredth. The 6 in the thousandth place tells us to round the 6 to a 7 in the hundredth place.

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7 0
3 years ago
Read 2 more answers
A spring with spring constant of 34 N/m is stretched 0.12 m from its equilibrium position. How much work must be done to stretch
Nesterboy [21]

Answer:0.253Joules

Explanation:

First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.

F = ke where;

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e is the extension = 0.12m

F = 34× 0.12 = 4.08N

To get work done,

Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.

Work done = Force × Distance

Since F = 4.08m, distance = 0.062m

Work done = 4.08 × 0.062

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Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules

8 0
4 years ago
A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener
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Answer:

Explanation:

From the given information:

The initial PE (PE)_i = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = P.E_f -P.E_i

ΔP.E = 0 - (PE)_i

ΔP.E = -P.E_i

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

\Delta P.E + \Delta K.E + \Delta U = 0

\Delta U = -\Delta P.E - \Delta K.E

this can be re-written as:

\Delta U =- (-\Delta P.E_i) - \Delta K.E

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

\Delta U =(70\%) \Delta P.E_i-0

\Delta U =(0.70) (490.5)

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

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These are called neap tides!

Hope this helps :)
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