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Mekhanik [1.2K]
3 years ago
13

When an athlete holds a barbell overhead, the reaction force is the weight of the barbell on his hand. how does this force vary

for the case where the barbell is decelerated upward?
Physics
1 answer:
Sonja [21]3 years ago
3 0
The force applied by the competitor is littler than the heaviness of the barbell. At the point when the barbell quickens upward, the power applied by the competitor is more prominent than the heaviness of the barbell. When it decelerates upward, the power applied by the competitor is littler than the heaviness of the barbell.
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What potential difference is needed to give a helium nucleus (q=2e) 85.0 kev of kinetic energy?
Vilka [71]
The kinetic energy K given to the helium nucleus is equal to its potential energy, which is 
E=q \Delta V
where q=2e is the charge of the helium nucleus, and \Delta V is the potential difference applied to it.
Since we know the kinetic energy, we have
E=K=85~keV=q \Delta V
and from this we can find the potential difference:
\Delta V =  \frac{K}{q}= \frac{85~keV}{2e}=42.5~kV

6 0
3 years ago
A 90. 0-kg ice hockey player hits a 0. 150-kg puck, giving the puck a velocity of 45. 0 m/s. If both are initially at rest and i
Mice21 [21]

The distance traveled by the hockey player is 0.025 m.

<h3>The principle of conservation of linear momentum;</h3>
  • The principle of conservation of linear momentum states that, the total momentum of an isolated system is always conserved.

The final velocity of the hockey play is calculated by applying the principle of conservation of linear momentum;

m_1v_1 = m_2 v_2\\\\v_1 = \frac{m_2 v_2}{m_1} \\\\v_1 = \frac{0.150 \times 45}{90} \\\\v_1 = 0.075 \ m/s

The time taken for the puck to reach 15 m is calculated as follows;

t = \frac{d}{v} \\\\t = \frac{15\ m}{45 \ m/s} \\\\t = 0.33 \ s

The distance traveled by the hockey player at the calculated time is;

d = vt\\\\d = 0.075 \ m/s \ \times 0.33 \ s\\\\d = 0.025 \ m

Learn more about conservation of linear momentum here: brainly.com/question/7538238

4 0
2 years ago
PLZ HELP NOW
ZanzabumX [31]

Answer: mass x height x gravitational field strength (g)

note: gravitational field strength (g) = 10 N/Kg

55 x 15 x 10 = 8250

gpe = 8250j

Explanation:

4 0
3 years ago
By what factors does the speed of the elctron exceed that of the proton?<br>​
____ [38]

Answer:

» An electron is lighter than a proton.

<u>explanation</u><u>:</u>

{  =  \: \sf{an \: electron \: has \: formula \:  \: }}{ \bf{ {}^{0}_{  - 1}e }}

hence it's mass number is zero

{  =  \: \sf{an \: electron \: is \: helium \: particle \:  \: }}{ \bf{ {}^{4} _{2}He  }}

hence it's mass number is 4

<u>Therefore</u><u>,</u><u> </u><u>proton</u><u> </u><u>is</u><u> </u><u>heavier</u><u> </u><u>than</u><u> </u><u>electron</u>

» An electron has a small charge magnitude than a proton.

<u>Explanation</u><u>:</u>

An electron has charge of -1 while proton has charge of +2, therefore electron is less deflected by any energetic fields than a proton

8 0
3 years ago
A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collis
Elza [17]

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2               ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1                  ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

4 0
3 years ago
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