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marysya [2.9K]
3 years ago
11

The rate at which a force displaces a mass a horizontal distance is measured in?

Physics
1 answer:
kodGreya [7K]3 years ago
6 0

Answer:

c. joules

Explanation:

The rate at which a force displaces a mass a horizontal distance is measured in joules. In science, this phenomenon is known as work done.

Work done can be defined as the rate at which a force acting on an object or a body causes it to experience a displacement. The work done is a scalar quantity and is measured in joules (J).

Mathematically, work done is given by the formula;

Work done = force * distance

W = F * d

Where,

  • W is the work done
  • F represents the force acting on a body.
  • d represents the distance covered by the body.

<em>For example, a bull pulling a plough through a farm, a girl pushing a shopping cart down the aisle of a supermarket etc. </em>

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A slice of bread contains about 100 kcal. If specific heat of a person were 1.00 kcal/kg·°C, by how many °C would the temperatur
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Answer:

(c) 1.43°C

Explanation:

If the energy in the bread are converted to heat.

Then, The heat transferred from the bread to person = 100 kcal.

From specific heat capacity,

Q = cmΔT............................ equation 1

Where Q = quantity of heat, m = mass of the person, c = specific heat capacity of the person, Δ = increase in temperature.

Making ΔT the subject the equation 1,

ΔT = Q/cm........................ equation 2

Where Q = 100 kcal, c= 1.00 kcal/kg.°C, m = 70.0 kg

Substituting these values into equation 2,

ΔT = 100/(1×70)

ΔT = 100/70

ΔT = 1.428

ΔT ≈ 1.43°C

The increase in temperature of the body is = 1.43°C

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If 2.2 lbs = 1.0 kg, and Megan Progress weighs 130 lbs, what is her weight in newtons? W = N (Round your answer to nearest whole
lord [1]

as it is given that

2.2 lbs = 1 kg

here we know that Megan Progress weighs 130 lbs

so its mass in kg is given as

m = 130 * \frac{1}{2.2} = 59.1 kg

now to find the weight in Newton unit we can say

W = mg

W = 59.1* 9.8

W = 579 N

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3 years ago
A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo
Orlov [11]

Explanation:

(a)   Net force acting on the block is as follows.

           F_{net} = -mg Sin (\theta)

or,           ma = -mg Sin (\theta)[/tex]

                 a = -g Sin (\theta)

                    = -9.8 \times Sin (20^{o})

                    = -3.35 m/s^{2}

According to the kinematic equation of motion,

             v^{2} - v^{2}_{o} = 2as

Distance traveled by the block before stopping is as follows.

     s = \frac{v^{2} - v^{2}_{o}}{2a}

        = \frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}

        = 21.5 m

According to the kinematic equation of motion,

               v = v_{o} + at

      0 = 12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t

   t_{1} = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b)    When the block is moving down the inline then net force acting on the block is as follows.

                 F_{net} = -mg Sin (\theta)

                ma = mg Sin (\theta)

                    a = g Sin (\theta)

                       = 9.8 m/s^{2} \times Sin (20^{o})

                       = 3.35 m/s^{2}

Kinematics equation of the motion is as follows.

                   s = v_{o}t + \frac{1}{2}at^{2}

      21.5 m = 0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}

     t_{2} = \sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}

             = 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

               t = t_{1} + t_{2}

                 = 7.16 sec + 3.58 sec

                 = 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

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