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notsponge [240]
3 years ago
13

The sum of two component vectors is referred to as the

Physics
1 answer:
blondinia [14]3 years ago
8 0
Resultant is the correct answer!
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What is the force of an object with a mass of 65 kg and an an unknown acceleration?
Alik [6]

We know, F = m.a

F = 65 * a

Where, F = force

a = unknown acceleration

8 0
3 years ago
An unknown substance has a mass of 15 g and a volume of 4 cm3. Calculate the density of the unknown substance.
myrzilka [38]

Answer:

Density is 3.75g/cm3

Explanation:

Density = mass ÷ volume

= 15g ÷ 4cm3

= 3.75g/cm3

5 0
3 years ago
A 95.0kg weightlifter pushes on a 800,000 kg wall for 400s but it does not move. How much work does he do on the wall?
Soloha48 [4]
<span>The weightlifter does no work. Although he has exerted force, work is the product of force over distance. Since he has not moved the wall he has done no work.</span>
7 0
3 years ago
What is an independent variable
Iteru [2.4K]

Answer:

An independent variable is a variable that is manipulated to determine the value of a dependent variable. The dependent variable is what is being measured in an experiment or evaluated in a mathematical equation and the independent variables are the inputs to that measurement.

Explanation:

8 0
3 years ago
Read 2 more answers
child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s
Strike441 [17]

Answer:

θ = 13.7º

Explanation:

  • According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
  • The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
  • As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

       \Delta K + \Delta U = W_{nc} (1)

  • ΔK, is the change in kinetic energy, as follows:

       \Delta K = \frac{1}{2}* m* (v_{f} ^{2}  - v_{0} ^{2}) (2)

  • ΔU, is the change in the gravitational potential energy.
  • If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

        \Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)

  • Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

        W_{nc} = F_{f} * d * cos 180\º \\\\  = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)

  • Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

      \frac{1}{2}* (v_{f} ^{2}  - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d

  • Replacing by the givens and the knowns, we can solve for sin θ, as follows:              \frac{1}{2}*( (4.30 m/s) ^{2}  - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2}  = 0.236⇒ θ = sin⁻¹ (0.236) = 13.7º
8 0
3 years ago
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