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harina [27]
3 years ago
14

in 1994, Vladimir Kurlivich , from Belarus, set the world record as the world's strongest weight lifter. he did this by lifting

and holding above his head a barbell whise mass was 253kg. Kurlovich's mass at the time was 133 kg. caculate the normal force exerted on each of his feet during the time he. was having ldi g the barbell
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0

The free-body diagram of your question is; 2 downward forces (253 kg mass of barbell & 133 kg body mass of Kurlovich) acting together on a point supported by 2 upward forces as normal forces exerted by Kurlovich's feet.  

Solving the normal forces exerted by 2 feet :  

Summation of Forces Vertical = 0  

2 Dowwnard Forces = 2 Upward Forces (2F)  

253 + 133 = 2F  

2F = 386 Kgs  

F = 386 / 2  

F = 193 Kgs (Normal Force Exerted by Each Foot)

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Serga [27]

Answer:

"For towing, a tow chain should be of a length that keeps both vehicles within the maximum 4.5 meter distance, also  tow chains an be any length 20 foot chains are often chosen"  

Explanation:

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3 0
3 years ago
The velocity of a particle is given by v=25t2 -80t-200, where velocity is meter per second and time is seconds. Determine the ve
jarptica [38.1K]

Answer:

 v = 220 m / s

Explanation:

This is a kinematics exercise, the expression for velocity is

          v = 25 t² - 80 t - 200

asks the velocity for time t = 6 s.

let's calculate

        v = 25 6² - 80 6 - 200

        v = 220 m / s

3 0
3 years ago
what force is needed to give a 0.25-kg arrow an acceleration of 196 m/s^2A. 0.25 NB. 49 NC. 748 ND. 196 N
svp [43]
We Know, F = m*a
Here, m = 0.25 Kg
a = 196 m/s²

Substitute it into the expression, 
F = 0.25*196
F = 49 N

So, option B is your final answer.

Hope this helps!
8 0
4 years ago
Read 2 more answers
A motorcyclist started a race from START line and from rest, and accelerated with the acceleration of 2 m/s2 for 12 seconds. The
ElenaW [278]

Answer:

The total distance is 280 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

v_{f} =v_{o} +(a*t)

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 2 [m/s²]

t = time = 12 [s]

Vf = 0 + (2*12)

Vf = 24 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

24² = 0 + (2*2*x)

x = 576/(4)

x = 144 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

v_{f}= v_{i} +(a*t)

Vf = final velocity [m/s]

Vi = initial velocity = 24 [m/s]

a = desacceleration = 5 [m/s²]

t = time = 4 [s]

Vf = 24 + (4*5)

Vf = 44 [m/s]

With this velocity, we can calculate the displacement using the following expression.

v_{f} ^{2} =v_{o} ^{2} +2*a*x

where

x = distance traveled [m]

44² = 24² + (2*5*x)

x = 1360/(10)

x = 136 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 144 + 136 = 280 [m]

7 0
3 years ago
An object accelerates from rest to 85m/s over a distance of 36m. What acceleration did it experience?
Marizza181 [45]
Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s 
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2
6 0
3 years ago
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