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Irina-Kira [14]
3 years ago
14

In which situation is work being done?

Physics
1 answer:
Keith_Richards [23]3 years ago
4 0

Answer: D. The force and displacement are in the same direction.

Explanation:

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path with distance d.  

Work is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy).

Now, when the applied force is constant and the direction of the force and the direction of the displacement are <u>parallel</u>, the equation to calculate it is:  

W=(F)(d) (1)

When they are not parallel, both directions form an angle, let's call it \alpha. In that case the expression to calculate the Work is:  

W=Fdcos{\alpha} (2)

When the force and displacement are perpendicular to each other,\alpha=90\° and <u>no work is done</u>.

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A dumbell has a mass of 95 kg. What force must be applied to accelerate it upward at 2.2 m/s2?
Sveta_85 [38]
A :-) F = ma
Given - m = 95 kg
a = 2.2 m/s^2
Solution -
F = ma
F = 95 x 2.2
F = 209

.:. The force is 209 N
5 0
3 years ago
A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi
Ahat [919]
<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = ?

        Time, t = 6.8 s    

        Displacement, s = 1/4 mi =    400 meters

     Substituting

                      s = ut + 0.5 at²

                      400 = 0 x 6.8 + 0.5 x a x 6.8²

                      a = 17.30 m/s²

Now we have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = ?

     Time, t = 6.8 s

      Acceleration, a = 17.30 m/s²

     Substituting

                      v = u + at  

                      v = 0 + 17.30 x 6.8

                      v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

6 0
3 years ago
Physics double pivot question​
andriy [413]

Explanation:

Assuming the wall is frictionless, there are four forces acting on the ladder.

Weight pulling down at the center of the ladder (mg).

Reaction force pushing to the left at the wall (Rw).

Reaction force pushing up at the foot of the ladder (Rf).

Friction force pushing to the right at the foot of the ladder (Ff).

(a) Calculate the reaction force at the wall.

Take the sum of the moments about the foot of the ladder.

∑τ = Iα

Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0

Rw (3.0 sin 60°) = mg (1.5 cos 60°)

Rw = mg / (2 tan 60°)

Rw = (10 kg) (9.8 m/s²) / (2√3)

Rw = 28 N

(b) State the friction at the foot of the ladder.

Take the sum of the forces in the x direction.

∑F = ma

Ff − Rw = 0

Ff = Rw

Ff = 28 N

(c) State the reaction at the foot of the ladder.

Take the sum of the forces in the y direction.

∑F = ma

Rf − mg = 0

Rf = mg

Rf = 98 N

3 0
3 years ago
List the jovian planets in order of increasing distance from the sun?
trapecia [35]
Here are the planets listed in order of their distance from the Sun: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune. An easy mnemonic for remembering the order is “My Very Educated Mother Just Served Us Noodles

Hope this helps.
5 0
3 years ago
The near point of an eye is 48.5 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan
Masteriza [31]

Answer:

The focal length of the lens should be -51.5 cm (a concave lens).

Explanation:

The purpose of the lens is to make objects at 48.5 cm appear at the healthy near point. The healthy near point is 25.0 cm.

We use the lens formula

\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}

where <em>f</em> = focal length, <em>u</em> = object distance and <em>v</em> = image distance.

In this case, <em>u</em> = 48.5 cm and <em>v</em> = -25.0 cm.

<em>v</em> is negative because the image is virtual an not real. (Here, we are using the real-is-positive sign convention)

\dfrac{1}{f} = \dfrac{1}{48.5} + \dfrac{1}{-25.0} = -\dfrac{23.5}{1212.5}

f = -51.5

The negative sign indicates the lens is concave.

3 0
3 years ago
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