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frutty [35]
4 years ago
7

I need help on the following questions.

Physics
1 answer:
dusya [7]4 years ago
6 0
What is the question you need help with
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Isostatic rebound
svp [43]
I think it is B the rise of the land
4 0
3 years ago
It is weigh-in time for the local under 85 kg rugby team. The bathroom scale used to assess eligibilty can be described by Hooke
Varvara68 [4.7K]
<span>k = 1.7 x 10^5 kg/s^2 Player mass = 69 kg Hooke's law states F = kX where F = Force k = spring constant X = deflection So let's solve for k, the substitute the known values and calculate. Don't forget the local gravitational acceleration. F = kX F/X = k 115 kg* 9.8 m/s^2 / 0.65 cm = 115 kg* 9.8 m/s^2 / 0.0065 m = 1127 kg*m/s^2 / 0.0065 m = 173384.6154 kg/s^2 Rounding to 2 significant figures gives 1.7 x 10^5 kg/s^2 Since Hooke's law is a linear relationship, we could either use the calculated value of the spring constant along with the local gravitational acceleration, or we can simply take advantage of the ratio. The ratio will be both easier and more accurate. So X/0.39 cm = 115 kg/0.65 cm X = 44.85 kg/0.65 X = 69 kg The player masses 69 kg.</span>
4 0
4 years ago
Read 2 more answers
A measure of the average kinetic energy of the individual particles in an object is called
svlad2 [7]
<h2>Answer: Temperature </h2>

Temperature is a physical quantity that reflects the amount of heat in a body or medium. This amount of heat is related to the internal energy of a system (thermodynamically speaking), <u>according to the movement (speed) of each of the particles that compose it</u>, this means that it is related to its kinetic energy.

Therefore, the higher the kinetic energy, the higher the thermal energy in the system and the higher the temperature.

7 0
4 years ago
Chapter 05, Problem 15 Multiple-Concept Example 7 and Concept Simulation 5.2 review the concepts that play a role in this proble
Llana [10]

Answer:

Explanation:

The question relates to motion on a circular path .

Let the radius of the circular path be R .

The centripetal force for circular motion is provided by frictional force

frictional force is equal to μmg , where μ is coefficient of friction and mg is weight

Equating cenrtipetal force and frictionl force in the case of car A

mv² / R = μmg

R = v² /μg

= 26.8 x 26.8 / .335 x 9.8

= 218.77 m

In case of moton of car B

mv² / R = μmg

v²  = μRg

= .683  x 218.77x 9.8

= 1464.35

v = 38.26 m /s .

3 0
3 years ago
What would happen if you took a really strong metal box and you started teleporting in hundreds and hundreds of small cotton bal
mojhsa [17]
It will not explode since the mass of the cotton balls is so low but rather will most likely break the lock and hinges and come out.
4 0
3 years ago
Read 2 more answers
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