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hjlf
3 years ago
15

How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec^2?

Physics
2 answers:
DiKsa [7]3 years ago
4 0
<span>Answer: Force = 81.6 N

Explanation:
According to Newton's Second law:
F = ma --- (1)

Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2

Plug in the values in (1):
(1) => F = 68 * 1.2
F = 81.6 N (The force needed to accelerate the skier at a rate of 1.2 m/s^2)</span>
Alex73 [517]3 years ago
3 0

Answer:

The force needed to accelerate a 68 kilogram-skier at a rate of 1.2ms2 is 81.6 Net forces

Explanation:

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An astronaut circling the earth at an altitude of 400 km is horrified to discover that a cloud of space debris is moving in the
elena-14-01-66 [18.8K]

One of the essential concepts to solve this problem is the utilization of the equations of centripetal and gravitational force.

From them it will be possible to find the speed of the body with which the estimated time can be calculated through the kinematic equations of motion. At the same time for the calculation of this speed it is necessary to clarify that this will remain twice the ship, because as we know by relativity, when moving in the same magnitude but in the opposite direction, with respect to the ship the debris will be double speed.

By equilibrium the centrifugal force and the gravitational force are equal therefore

F_c = F_g

\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}

Where

m = mass spacecraft

v = velocity

G = Gravitational Universal Constant

M = Mass of earth

r \rightarrow R+h \Rightarrow Radius of earth and orbit

Re-arrange to find the velocity

\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}

\frac{v^2_{orbit}}{r} = \frac{GM}{r^2}

v^2_{orbit}=\frac{GM}{r}

v_{orbit} = \sqrt{\frac{GM}{r}}

v_{orbit} = \sqrt{\frac{GM}{R+h}}

Replacing with our values we have

v_{orbit} = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{6.37*10^6+0.4*10^6}}

v_{orbit} = 7676m/s

From the cinematic equations of motion we have to

t = \frac{d}{2v_{orbit}} \rightarrow Remember that the speed is double for the counter-direction of the trajectories.

Replacing

t = \frac{29000m}{7676m/s}

t = 3.778s

Therefore the time required is 3.778s

4 0
3 years ago
Select all the correct answers. a leaf falls through the air, floating as it falls. air resistance is visibly affecting the leaf
musickatia [10]

Surface area and Mass

When a leaf falls, it is being accelerated by gravity to the ground but opposed by air resistance also the drag. The net force on a leaf will therefore be calculated by subtracting its weight of the leaf from its drag.

<h3>What is Air resistance ?</h3>

Air exerts a force known as air resistance. When an object is travelling through the air, the force works in the opposite direction.

  • While a sports vehicle with a streamlined design will encounter reduced air resistance and experience less drag, the automobile will be able to move more quickly than a truck with a flat front.
  • The speed, area, and shape of the object passing through the air all affect air resistance. Air density and resistance are affected by altitude, temperature, and humidity. The resistance increases with speed and area, respectively.

Learn more about Air resistance here:

brainly.com/question/27965545

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5 0
2 years ago
A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?
allsm [11]

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side.  See more details further below.

Explanation:

Start with the formula for pressure

(pressure P) = (Force F) / (Area A)

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

F_g = m \cdot g

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3

The density of water is 997 kg/m^3, so the mass of the barrel is:

m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg

and so the weight is

F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N

and so the pressure is

P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.  

7 0
3 years ago
the wattage ratting of a light bulb is the power it sonsumers when it is connected accross a 120 V potential difference what is
igor_vitrenko [27]

Answer:

240 Ω

Explanation:

Resistance: This can be defined as the opposition to the flow of current in an electric field. The S.I unit of resistance is ohms (Ω).

The expression for resistance power and voltage is give as,

P = V²/R.......................... Equation 1

Where P = Power, V = Voltage, R = Resistance

Making R the subject of the equation,

R = V²/P.................... Equation 2

Given: V = 120 V, P = 60 W.

Substitute into equation 2

R = 120²/60

R = 240 Ω

Hence the resistance of the bulb = 240 Ω

8 0
4 years ago
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