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kap26 [50]
3 years ago
13

Please Help! When drawing free-body diagrams, does the label "centripetal force" get used? Explain why.

Physics
1 answer:
Yakvenalex [24]3 years ago
4 0

Answer:

When drawing free-body diagrams, the label "centripetal force" does not get used.

Explanation:

The label does not get used because it is not one of the forces acting on the object.

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A sprinter has a mass of 80 kg and a kinetic energy of 2500 J. What is the sprinter's speed?
nata0808 [166]
Using KE = 1/2mv^2
m = mass (80kg)
v = velocity (?)
KE = kinetic energy (2500J)
2500/((1/2) x (80)) = v^2
Square root the answer to get: 7.91 m/s
3 0
2 years ago
Which of the following is an example of applying a force?
maw [93]
The correct response would be C. Carpenter hammering a nail. The carpenter is applying a force as he or she is hitting the surface of the nail with the hammer.
4 0
3 years ago
PLEASE HELP WILL GIVE BRAINLEST!!!
Aleks04 [339]

Answer:

huh what do u want tho its nothing bad

Explanation:

3 0
3 years ago
The graph below shows the variation with distance r from the nucleus of the square of the wave function, Ψ^2, of a hydrogen atom
Ludmilka [50]

The region a represents the distance of the electron from the nucleus.

According to the wave mechanical model of the atom, the probability of finding an electron within a given volume element (representing the atom) is the square of the wave function psi.

Since a is the region in space where there is the greatest probability of finding the electron in the atom, it follows that distance of the electron form the atom is always a.

Learn more about the wave mechanical model: brainly.com/question/1382157

3 0
2 years ago
How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C ? The specific heat of ice is 2090
cupoosta [38]

Answer:209.98 kJ

Explanation:

mass of water m=456 gm

Initial Temperature of Water T_i=25^{\circ}C

Final Temperature of water T_f=-10^{\circ}C

specific heat of ice c=2090 J/kg-K

Latent heat L=33.5\times 10^4 J/kg

specific heat of water c_{water}=4.184 KJ/kg-K

Heat require to convert water at T=25^{\circ}C to T=0^{\circ}C

Q_1=0.456\times 4.184\times (25-0)=47.69 kJ

Heat require to convert water at T=0^{\circ} to ice at T=0^{\circ}

Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ

heat require to convert ice at T=0^{\circ} C\ to\ T=-10^{\circ} C

Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ

Total heat Q=Q_1+Q_2+Q_3

Q=47.69+152.76+9.53=209.98 kJ

7 0
3 years ago
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