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jeka57 [31]
3 years ago
7

A weight lifter struggles but manages to keep a heavy barbell abovehis head. Occasionally he slips and the barbell starts to fal

l downward, but he always recovers. In each case, compare the force exerted by the weight lifter on the barbell to that exerted by the barbell on the weight lifter.
A)- With the barbell at rest.
B)- With the barbel moving upward.
C)- With the barbel moving downward.
Physics
2 answers:
shutvik [7]3 years ago
4 0
A) The lifter's and the barbell's force are equal and opposite in direction
B) The lifter's force is greater than that of the barbell and in the upward direction
C) The barbell's force is greater than that of the lifter and in the downward direction
lakkis [162]3 years ago
3 0

Explanation:

(A)

-With the barbell at rest- Both forces are equal

When the barbell is at rest then in this condition both forces are equal. The force exerted by the weight lifter on the barbell is equal to the force exerted by the barbell on the weight lifter.

(B)

-With the barbell moving upward- Forces are equal

When the barbell is moved upward by the weight lifter then the force exerted on the barbell by the weight lifter is greater than the gravitational force exerted on the barbell.

(C)

-With the barbell moving downward-Forces are equal

When the barbell is moved downward by the weight lifter then the force exerted on the barbell due to gravity is greater than the force exerted by the weight lifter on the barbell.

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Answer:

Waves

Explanation:

Subatomic particles are particles that are of the subdivision of an atom and are smaller that atom, they are also described as waves in quantum mechanics.

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3 years ago
Why is it important to have a closed circuit?
yaroslaw [1]

D. It allows the flow of an electric current in the circuit.

<h3>Why is it important to have a closed circuit?</h3>

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2 years ago
a rocket is launched with a constant acceleration straight up. exactly 4.00 seconds after lift off, a bolt falls off the side of
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The initial speed of the bolt is not 58.86 m/s.  

Let a be the acceleration of the rocket.  

During the 4 sec lift off, the rocket has reached a height of  

h = (1/2)*a*t^2  

with t=4,  

h = (1/2)*a^16  

h = 8*a  

Its velocity at 4 sec is  

v = t*a  

v = 4*a  

The initial velocity of the bolt is thus 4*a.  

During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,  

h = (1/2)*g*t^2 + V0*t  

Substituting h0=8*a, t=6 and V0=-4*a into it,  

8*a = (1/2)*g*36 - 4*a*6  

Solving for a  

a = 5.52 m/s^2

6 0
3 years ago
What is the science principle that explains magnetic fields? I need this fast, please!
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3 0
2 years ago
g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k
Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s

Now, we will apply the law of conservation of momentum:

m_1v_1 = m_2v_2

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s

Now, we again use the third equation of motion for the upward motion of the ball:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\

<u>h = 3.5 m</u>

6 0
2 years ago
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