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olasank [31]
3 years ago
11

A cat climbs 10 m directly up a tree.

Physics
1 answer:
Harman [31]3 years ago
6 0
 <span>there is no horizontal displacement if he went straight up 

straight up means vertical, so his vertical displacment is 20 m</span>
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The diagram shows a measuring cylinder containing a liquid,and at thay same measuring cylinder when it is empty . What is the de
KatRina [158]

Answer:

A measuring cylinder contains 30 cm :))

Explanation: Hope this helps !!!!

3 0
2 years ago
The student quickly discovers that placing the marshmallow over the flame is more effective than heating the marshmallow on the
ANEK [815]

Answer:

The answer is convection.

Explanation:

There are three types of heat transfer: conduction, convection and radiation.

  • Conduction occurs when two objects touch each other and transfer heat.
  • Convection occurs when an object heats its surrounding fluid (like air, or water) and, since the hot fluids are less dense than the cold ones, they go up. So convection is a type of heat transfer that usually goes from down to up.
  • Radiation occurs when objects emanate heat in the form of electromagnetic waves that propagates in all directions.

So in this case, when the marshmallow is above the fire, it is exposed to convection, which does not occur when it is on the side of the fire.

4 0
4 years ago
I will give more points <br>for this thank you I u helped<br>me ​
scZoUnD [109]

Answer:

weight = mass×gravity

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<h2>stay safe healthy and happy.</h2>
6 0
3 years ago
When a baseball is thrown straight upward, what is its acceleration at its greatest height? Please explain your answer.
anyanavicka [17]

Answer: A

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6 0
3 years ago
An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC charge uniformly distributed throughout its interior volume.
Kobotan [32]

Explanation:

(a)   Formula to calculate the density is as follows.

            \rho = \frac{Q}{\frac{4}{3}\pi a^{3}}

                       = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}

                     = 2.42 \times 10^{-2} C/m^{3}

Now, calculate the charge as follows.

            q_{in} = \rho(\frac{4}{3} \pi r^{3})

                      = 2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}

                      = 10.106 \times 10^{-8} C

or,                   = 101.06 nC

(b)  For r = 6.50 cm, the value of charge will be calculated as follows.

                q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}

                          = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}

                          = 7.454 \mu C

7 0
3 years ago
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