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Alja [10]
3 years ago
12

A snowball that will be used to build a snowman is at the top of a only hill. If the

Physics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

h = 24.11 m

Explanation:

Given that,

The potential energy of the snowball is 520 J

The mass of the snowball is 2.2 kg

We need to find the height of the hill. The potential energy of an object is given by the formula as follows :

E=mgh

g is acceleration due to gravity

h is height of the hill

h=\dfrac{E}{mg}\\\\h=\dfrac{520}{2.2\times 9.8}\\\\h=24.11\ m

So, the height of the hill is 24.11 m.

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Two forces are acting on a wheelbarrow. One force is pushing to the right and an equal force is pushing to the left. What can yo
andreev551 [17]

-- As far as we know, the forces on the wheelbarrow are balanced.

-- That tells us that the net force on the wheelbarrow is zero, just
as if there were no forces acting on it at all.

-- That tells us that the wheelbarrow's acceleration is zero ... its
speed and direction of motion are not changing.

-- That tells us that the wheelbarrow is moving in a straight line
at a constant speed.  It's very possible that relative to us, the speed
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6 0
3 years ago
How to measure the volume of a baseball bat ( need answers ASAP )
vaieri [72.5K]

<em>Measure the amount of water it displaces.</em>

This won't be easy, because the bat floats in water.  But I think you can get around that little problem like this:

-- Get some kind of a tank or tub that's big enough to hold the whole bat under water.

-- Get a heavy weight, like a big wrench or a small rock.  

-- Fill the tub almost to the tippy top with water.

-- Slip the heavy weight into the tub, slowly.  Some water will run over the top and out of the tub.  That's OK ... it's exactly what you want.  If NO water runs over the top, pour some more in, until it runs out and then stops.  You want the tub full to the brimmy rim with the rock at the bottom of it.

-- Take the heavy weight out of the tub.

-- Now set the tub into a bigger tub or a deep pan.  The next time it overflows and some water runs out of it, you'll need to catch that water and measure it.

-- Get a short piece of heavy string.  Tie the heavy weight to somewhere near the middle of the bat.

-- Slowly slide the bat into the water, with the rock tied to it.  The bat needs to go complete underwater.

-- Some more water will run over the top and out of the tub, and INTO the lower tub.  Wait until the overflow stops and everything settles down again.

-- Take the bat (tied to the weight) out of the tub.  Slowly and carefully, so that your hand or your arm doesn't make any MORE water run over and out.

-- Lift the upper tub out of the lower tub.

-- Take the lower tub, with the overflow water in it.  Using a kitchen measuring cup, or a saucepan or a bottle, or anything else with liquid amounts marked on it, measure how much water overflowed into the lower tub.

THAT amount is the volume of the bat.

You may have to do some units conversions.  Like if you need the volume of the bat in cm³ and you used measuring vessels marked in fluid ounces.  But you can find all those conversion factors with a search on Floogle.

8 0
3 years ago
Find the sum of the vectors: 40m/s2 Northeast, 10 m/s2 Northeast
podryga [215]

Answer:

Explanation:

Since both vectors are pointing on the same direction (Northeast), the sum of them will point in that same direction, and its magnitud will be the sum of the magnitudes of each vector (40m/s2+10m/s2). This problem is just a problem in one dimension. The sum of the vectors is then 50m/s2 Northeast.

8 0
3 years ago
A coastal location in Florida experiences very strong winds throughout the year. What would the coastal community in the locatio
Zina [86]

Answer:

install wind turbines to produce

6 0
3 years ago
The error in the measurement of the radius of a sphere is 2%. What will be the error in the calculation of its volume?
BlackZzzverrR [31]

To solve this problem we will apply the geometric concepts of the Volume based on the consideration made of the radius measurement. The Volume must be written in differential terms of the radius and from the formula of the margin of error the respective response will be obtained.

The error in radius of sphere is not exceeding 2%

\frac{dr}{r} = \pm 0.02

The objective is to find the percentage error in the volume.

The volume can be defined as

V = \frac{4}{3} \pi r^3

Differentiate with respect the radius we have,

\frac{dV}{dr} = 4\pi r^2

dV = 4\pi r^2 \times dr

dV = 4\pi r^2 (\pm 0.02r)

dV = \pm 4\times 0.02 \times \pi r^3

The percentage change in the volume is as follows

\% change = \frac{dV}{V} \times 100

\% change = \frac{\pm 4 \times 0.02 \times \pi r^3 \times 3}{4\pi r^3}\times 100

\% change = \pm 6\%

Therefore the percentage change in volume is \pm 6\%

3 0
3 years ago
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