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Ksivusya [100]
2 years ago
5

A carrot has a volume of 180 cubic centimeters and a mass of 81 grams . Calculate its destiny

Physics
2 answers:
Tomtit [17]2 years ago
7 0
I have no idea I am sorry someone will help you soon
Zepler [3.9K]2 years ago
4 0

Answer:

0.45 grams per cu centimeter

Explanation:

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A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h
hram777 [196]

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

8 0
3 years ago
Can someone please answer this, ill give you brainliest Would be very appreciated.
marshall27 [118]

Answer:

cohesive properties

Explanation:

The property of cohesion allows liquid water to have <u>no tension on the surface</u>.

7 0
2 years ago
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Can work done=mass*acceleration*displacement(work=m*a*s)
Airida [17]

no, work is = force * distance or displacement


5 0
3 years ago
What is the kinetic energy of a an 80kg football player running at 8 m/s?
Ugo [173]
In the question it is already given that the football player is 80 kg.
Then the mass of the football player = 80 kg
Velocity at which the football player is running = 8 m/s
<span>Kinetic Energy = 0.5 • mass • square of velocity
Now we have to put the known data in this equation to find the actual velocity of the footballer.
</span> <span></span>So
Kinetic Energy of the footballer = 0.5 * 80 * (8 * 8)
                                                 = 0.5 * 80 * 64
                                                 = 2560
So the Kinetic energy of the footballer is 2560 joules


4 0
3 years ago
Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of
Lana71 [14]

Electric field due to a point charge is given as

E = \frac{kq}{r^2}

here we know that

q = 1.6 \times 10^{-19} C

also the distance is given as

r = 5.29 \times 10^{-11} m

now we will have

E = \frac{(9\times 10^9)(1.6 \times 10^{-19})}{(5.29 \times 10^{-11})^2}

so we will have

E = 5.14 \times 10^{11} N/C

so above is the electric field due to proton

5 0
3 years ago
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