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Natalka [10]
3 years ago
13

In which of the following situations has work been done? Select one: a. A weightlifter holds a 50kg barbell over his head for 3

minutes. b. A yoga instructor holds a handstand for 1 minute. c. A man pulls against a stationary fire truck for 5 minutes. d. A child pushes a 2kg toy 10m across the floor.
Physics
1 answer:
Arlecino [84]3 years ago
5 0
Work done = Force * distance --- In other words, for work to be done, a force should be applied such that there is motion.

Therefore, the only situation meeting the criteria is situation d. where a force is applied resulting to motion.
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) In the figure below, three charges K, L and M are placed at the corners of an equilateral triangle. If K and L are fixed, in w
abruzzese [7]

Answer:

B)

Explanation:

Negative (-) charge M will not move towards negative (-) charge K because, same charges will not attract each other in the given case

Negative (-) charge at the M tends to move towards positive (+) charge L in the direction of B) because opposite charges attract each other.

4 0
3 years ago
a child's toy consists of a piece of plastic attached to a spring. the spring is compressed against the floor a distance of 2.0
Nat2105 [25]

Answer:

1.7N

Explanation:

Force = kx

Where x = spring compression and

K = spring constant

K =85N/m

x = 2.0cm / 100

= 0.02m

Force = 85 x 0.02

= 1.7N

5 0
3 years ago
Gas
Brut [27]
A: what is it called when a solid jumps straight to a gas?
B: what is it called when a liquid becomes a gas?
C: what is it called when a solid changes into a liquid?

Im sure you can at least figure out the last one! (hint: this happens when ice becomes water)
3 0
3 years ago
Read 2 more answers
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3
Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC       Q₃ = 51 nC        Q₂ = 30.5 nC

x₁ = - 1.715m         x₃ = - 1.085m     x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0
3 years ago
You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3
jeka57 [31]

Answer:

W = 0J

Explanation:

The work done by the dresser is described as

W = f d (cos θ)

F has been given as the weight of this dresser. And it is 3500 N

d = 0 m

When you put these values into the equation

W = 3500 x 0 x cosθ

W = 0 J

This value tells us that the work done on this dresser is zero. No work has been done. Therefore the last option answers the question.

3 0
3 years ago
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