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USPshnik [31]
3 years ago
6

You need to push a heavy box across a rough floor, and you want to minimize the average force applied to the box during the time

the box is being pushed. Which method for pushing results in the minimum average force being applied to the box? a. Keep pushing the box forward at a steady speed, b. Push the box forward a short distance, rest, then repeat until finished, c. Push the box so that it accelerates forward at a constant rate.
Physics
1 answer:
7nadin3 [17]3 years ago
7 0

When pushing the body it is necessary to break the frictional force generated by the floor. Once this frictional force is overcome, the body will begin to move. Ideally, if a constant velocity is maintained or close to this value, the acceleration that will be exerted will tend to be zero and therefore, by Newton's second law the value of the Force will also tend to minimum values.

Remember that this law tells us that

F= ma

F= m \frac{\Delta v}{t}

Therefore the best strategy is A. keep pushing the box forward at a steady speed

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3 years ago
A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.
aleksandrvk [35]

Answer:

ΔH°comb=-5899.5 kJ/mol

Explanation:

First, consider the energy balance:

m_{c} *Cp*(T_{2}-T_{1})=-n_{s} *H_{c} Where m_{c} is the calorimeter mass and n_{s} is the number of moles of the samples; H_{c} is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.

The product m_{c} *Cp is the heat capacity, so the balance equation is:

6.21\frac{kJ}{K}*(75-25)=-8.00g*\frac{mol}{152.0g}*H_{c}

So, the enthalpy of combustion can be calculated:

H_{c}=-5899.5\frac{kJ}{mol}

I will be happy to solve any doubt you have.

4 0
3 years ago
A car travels at a constant velocity of 70 mph for one hour. At the second hour, the car’s velocity was 60 mph. At the third hou
katovenus [111]

When the velocity increases, then the acceleration will be positive and when the velocity decrease then the acceleration will be negative.

During the first hour, the velocity was 70 mph and during the seconds hour the velocity was 60 mph. Hence, the velocity decrease in the seconds hour. So, the acceleration will be negative during the second hour.

Now, during the third hour the velocity increases as it is 80 mph. Hence, the acceleration will be positive during the third hour.


8 0
3 years ago
Read 2 more answers
A car is moving at 20 m/s. The driver steps on the gas pedal and accelerates the car for 4 seconds at an acceleration of 4 m/s2.
polet [3.4K]

Answer:

The final velocity of the car is 36 m/s.

Explanation:

Given;

initial velocity of the car, u = 20 m/s

time of the car acceleration, t = 4 s

acceleration of the car, a = 4 m/s²

the final velocity of the car is calculated as;

v = u + at

where;

v is the final velocity of the car

v = 20 + (4 x 4)

v = 36 m/s

Therefore, the final velocity of the car is 36 m/s.

6 0
2 years ago
(a) As you ride on a Ferris wheel, your apparent weight is different at the top and at the bottom. Explain. (b) Calculate your a
otez555 [7]

Answer:

a. The component of the net force which make up the apparent weight are added to each other at the bottom and subtracted (the centripetal force from the weight) at the top)

b. Apparent weight at the top is approximately 519.06 N

Apparent weight at the bottom is approximately 558.94 N

Explanation:

a. The apparent weight at the top is different from the apparent weight at the bottom of a moving Ferris wheel because of the opposite direction in which the centripetal force acts at the top and the bottom, which are upwards and downwards respectively, while the weight acts downwards constantly

b. The given parameters are

The radius of the Ferris wheel, r = 7.2 m

The period for one complete revolution, t = 28 seconds

The angle covered in one revolution, θ = 2·π radian

The mass of the person riding on the Ferris wheel, the passenger  = 55 kg

Therefore, we have;

The angular speed, ω = Δθ/Δt = 2·π/(28)

From which we have;

Centripetal force, F_c = m × ω² × r

Substituting the known values, we have F_c = 55 kg × (2·π/(28 s))² × 7.2 m ≈ 19.94 N

The centripetal force, F_c = 19.94 N always acting outward from the center

Weight = Mass × Acceleration due to gravity

The weight of the passenger = 55 kg × 9.8 m/s² = 539 N

The weight of the passenger = 539 N always acting downwards

At the top of the Ferris wheel the the centripetal force is acting upwards and the weight is acting downwards

Therefore;

The net force, which is the apparent weight of the passenger at the top F_{NET_{Top}} = 539 N - 19.94 N ≈ 519.06 N

Apparent weight at the top ≈ 519.06 N

At the bottom of the Ferris wheel the weight is acting downwards and the centripetal force is also acting downwards

Therefore;

The net force at the bottom, which is the apparent weight of the passenger at the bottom F_{NET_{bottom}} = 539 N + 19.94 N ≈ 558.94 N

Apparent weight at the bottom ≈ 558.94 N.

5 0
3 years ago
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