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Ymorist [56]
3 years ago
7

A weightlifter presses a 400 n weight 0.5 m over his head in 2 seconds, what is the power of the weight lifter

Physics
1 answer:
Scrat [10]3 years ago
3 0
There are two ways to solve this problem. First we write the given.

Given: Force F = 400 N;  Height h = 0.5 m;  Time t = 2 s

Formula: P = W/t;  but Work W = Force x distance or W = f x d
 
Weight is also a Force, therefore:  W = mg, solve for Mass m = ?

m = w/g  m = 400 N/9.8 m/s²  m = 40.82 Kg

P = W/t = F x d/t  = mgh/t  P = (40.82 Kg)(9.8 m/s²)/2 s  

P = 100 J/s or 100 Watts

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Why does the band of stability curve upward at high atomic numbers? excess neutrons are required due to the repulsion between th
ella [17]
The band of stability curves upward at high atomic numbers due to the fact that excess of neutrons are required due to the repulsion between protons.

Atomic number is the number of protons. As the number of protons (atomic number) increase, the electrical repulsion force, due to the same sign of the protons inside the nucleus, becomes more dominant compared to the nuclear force attractions, then the nucleus needs more neutrons to gain stability.The presence of more neutrons decrease the density of protons which decreases the repulsion among them.


3 0
3 years ago
An increase in temperature will ___________ the speed of a sound wave?
lbvjy [14]
I think it’s cause no change i think
3 0
3 years ago
Read 2 more answers
Scientists often work on projects for a long time and fail to see sources of error in their research. Which process allows an ou
Marina CMI [18]

Answer:

Peer Review

Explanation: I took the test

6 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
4 years ago
Find the velociity of a car which travels 35 m to the right over a period of 40 seconds
GalinKa [24]

Answer:

the velocity of the car is 0.875 m/s

Explanation:

v = d \div t \\ v = 35m \div 40sec = 0.875 \: m \: per \: sec

therefore the V of car is 0.875 m

8 0
3 years ago
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