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fredd [130]
3 years ago
14

An object is dropped from rest from the top of a 400 m cliff on Earth. If air resistance is negligible, what is the distance the

object travels during the first 6 s of its fall?
Physics
1 answer:
IrinaVladis [17]3 years ago
8 0

Answer:

176.58 m

Explanation:

t = Time taken = 6 seconds

u = Initial velocity = 0

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 6^2\\\Rightarrow s=176.58\ m

The object travels 176.58 m from the cliff in 6 seconds.

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If the efficiency and mechanical advantage of a certain machine are given as 65 % and 3 respectively.What is the velocity ratio
Vinvika [58]

Answer:

b. 4.6 %

Explanation:

From the question,

E = M.A/V.R................ Equation 1

Where E = percentage Efficiency of the machine, M.A = machanical accurancy of the machine, V.R = Velocity ratio of the machine

Make V.R the subject of the equation

V.R = M.A/E

Given: M.A = 3, E = 65% = 0.65

Substitute this values into equation 2

V.R = 3/0.65

V.R = 4.6

Hence the right option is b. 4.6 $

8 0
3 years ago
Does the surface tension of water affect the rate of evaporation? If so, would adding a surfactant speed up evaporation?
Goshia [24]

Answer:

The evaporation time gradually increased with the increase in surfactant concentration, i.e., from water to the concentration level of 0.005%. Furthermore, the evaporation time is significantly reduced, even lower than that of water containing relatively high concentrations from 0.01% to 0.1%.

6 0
2 years ago
Two floors in a building are separated by 4.1 m. People move between the two floors on a set of stairs. (a) Determine the change
Ket [755]

Answer:

a) The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

Explanation:

Let consider the bottom of the first floor in a building as the zero reference (z = 0\,m). The change in potential energy experimented by a particle (\Delta U_{g}), measured in joules, is:

\Delta U_{g} = m\cdot g\cdot (z_{f}-z_{o}) (1)

Where:

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

z_{o}, z_{f} - Initial and final height with respect to zero reference, measured in meters.

Please notice that m\cdot g is the weight of the particle, measured in newtons.

a) If we know that m = 3\,kg, g = 9.807\,\frac{m}{s^{2}}, z_{o} = 0\,m and z_{f} = 4.1\,m, then the change in potential energy is:

\Delta U_{g} = (3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.1\,m-0\,m)

\Delta U_{g} = 120.626\,J

The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) If we know that m\cdot g = 650\,N, z_{o} = 4.1\,m and z_{f} = 0\,m, then the change in potential energy is:

\Delta U_{g} = (650\,N)\cdot (0\,m-4.1\,m)

\Delta U_{g} = -2665\,J

The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

5 0
2 years ago
A force of 5.25 newtons acts on an object of a mass 25.5 kilograms. How far is the object from the center of Earth? (The value o
nexus9112 [7]

Answer:

The force between two objects is calculated through the equation,

                       F = Gm₁m₂/d²

where m₁ and m₂ are the masses of the objects. In this case, an unknown mass and Earth. d is the distance between them and G is the universal gravitation constant.

In the second case, if the force is to become 2.5 times the original and all the variables are constant except d then,

                      2.5F = Gm₁m₂ / (D²)

                               D = 0.623d

Subsituting the known value of d,

                               D = 0.623(6.9 x 10^8) = 4.298 x 10^8 m

                         

5 0
1 year ago
Why are objects that fall near earth’s surface rarely in free fall?
Sloan [31]

Answer:

Because of the presence of air resistance

Explanation:

When an object is in free fall, ideally there is only one force acting on it:

- The force of gravity, W = mg, that pushes the object downward (m= mass of the object, g = acceleration of gravity)

However, this is true only in absence of air (so, in a vacuum). When air is present, it exerts a frictional force on the object (called air resistance) with upward direction (opposite to the motion of free fall) and whose magnitude is proportional to the speed of the object.

Therefore, it turns out that as the object falls, its speed increases, and therefore the air resistance acting against it increases too; as a result, the at some point the air resistance becomes equal (in magnitude) to the force of gravity: when this happens, the net acceleration of the object becomes zero, and so the speed of the object does not increase anymore. This speed reached by the object is called terminal velocity.

3 0
3 years ago
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