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lapo4ka [179]
3 years ago
14

Pablo preto lift a barbell. During which stage of the lift does Pablo do work on the barbell

Physics
2 answers:
Arlecino [84]3 years ago
6 0
While he is lifting the dumbell as the definition of work done = moving a mass through a distance = F x d
____ [38]3 years ago
3 0

this is the original question:

Pablo prepares to lift a barbell. During which stage of the lift does Pablo do work on the barbell?

when he holds the barbell up above his head

when he applies force to the barbell but has not moved it

when he lifts the barbell from the floor to his chest

when he lets go of the barbell and allows it fall to the ground

that is the answer-> when he lifts the barbell from the floor to his chest

hope this helps!

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lyudmila [28]

Answer:

C

Explanation:

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3 years ago
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A person is lying on a diving board 2.00 m above the surface of the water in a swimming pool. She looks at a penny that is on th
wlad13 [49]

Answer:

7.98 m

Explanation:

In the given question,

distance above surface= 2 m

Distance penny from person = 8 m

Since the swimming pool is filled with water and atmosphere has air therefore the refractive index phenomenon will occur.

The refractive index of water: air is 4/3 (1.33).

Using the formula, 4/3 = real depth, apparent depth

real depth= 4/3 x apparent depth

Now, calculating apparent depth = 8 - 2

= 6 m

therefore, real depth =  4/3 x apparent depth

= 1.33 x 6

= 7.98

thus, 7.98 m is the real depth of water.

8 0
3 years ago
A motorcycle moving at a constant velcoity suddenly accelerates at a rate of 4.0 m/s/s to a speed of 35 m/s in 5.0 s. What was t
oee [108]

Answer:

15 m/s

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v = u+ at

35 = u + 20

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u= 15 m/s

7 0
3 years ago
A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre
Alisiya [41]

Answer:

0.546 \hat k

Explanation:

From the given information:

The force on a given current-carrying conductor is:

F = I ( \L  \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})

where the length usually in negative (x) direction can be computed as

\L ^ {\to }  = -x\hat i \\dL\limits ^ {\to }- dx\hat i

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})

F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)

F = I \int^3_1  - 9.0x^2 \ dx \hat k

F = I  (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k

F = I  (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ]  \hat k

where;

current I = 7.0 A

F = (7.0 \ A)  (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ]  \hat k

F = (7.0 \ A)  (9.0) \bigg [\dfrac{26}{3} \bigg ]  \hat k

F = 546 × 10⁻³ T/mT \hat k

F = 0.546 \hat k

4 0
3 years ago
With all his gear, Neil Armstrong weighed 360 pounds on Earth. When he landed on the Moon, he weighed 60 pounds. Why?
mina [271]

Answer:C: The gravity on the moon is less than the gravity on Earth.

Explanation:

4 0
2 years ago
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