1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Amanda [17]
3 years ago
12

Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin

g when determining the work done on the object?
A. Only the component of the force perpendicular to the motion is used to calculate the work.

B. If the force acts in the same direction as the motion, then no work is done.

C. When there is an angle between the two directions, the cosine of the angle must be considered.

D. A force at a right angle to the motion requires the use of the sine of the angle.
Physics
1 answer:
ELEN [110]3 years ago
3 0
C is correct.  The work-force relation is given by W=F·d, where F is force vector, and d is the displacement vector.  The dot is the dot product, which is a measure of how parallel the two vectors are.  It can be restated as the product of two vector magnitudes times the cosine of the angle between them.  Therefore work is a scalar, not a vector, since the dot product returns a scalar.  
W=Fdcos(\theta)
You might be interested in
A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
3 years ago
What is the resultant of a pair of forces, 100N, upward and 75N, downward?
soldi70 [24.7K]

Answer:

25N

Explanation:

100 - 75 = 25

That should be right if im not dumb...

3 0
3 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
4 years ago
Question #4
IrinaVladis [17]
2) transverse



hope this helped:)
3 0
3 years ago
Read 2 more answers
Which description best explains the distortion of color at the bottom of the leaves in the image?
Vilka [71]

Answer:

A.

Explanation:

When the light travels through the lenses and disperses it can create other colors around objects that aren't there.

6 0
3 years ago
Other questions:
  • What is a discussion?
    14·2 answers
  • A survey team is trying to estimate the height of a mountain above a level plain. from one point on the plain, they observe that
    10·1 answer
  • What is the greenhouse effect?
    9·2 answers
  • A positive charge Q is distributed uniformly along the positive y-axis between y=0 and y=a. A negative point charge, -q, lies on
    6·1 answer
  • Suppose we hang a heavy ball with a mass 13 kg (so the weight is ) from a steel wire 3.9 m long that is 3.1 mm in diameter (radi
    8·1 answer
  • - A person is on an elevator that moves
    8·1 answer
  • 5. Define joint<br><br> answer asap plz and ty
    6·2 answers
  • Two horses are side by side on a carousel. Which has a greater tangential speed the one closer to the center or the one farther
    15·1 answer
  • The wind blows a lawn chair that weighs 4 kg into a fence with a force of 8 N. How much reaction force does the fence exert on t
    10·1 answer
  • Which type of wave can only be transmitted through matter?
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!