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3 years ago

List several examples of applied force, normal force, and friction that you’ve observed in your life.

Physics

1 answer:

grandymaker [24]3 years ago

3 0

applied forces would be push for example.

normal forces would seem to be a force such as gravity.

friction for example when you try to slide on carpet but the fabric or whatever its made of stops you.

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What is the speed of the tip of the minute hand of a clock where the hand is of length 7cm?

azamat

Explanation:

Q1) What is the speed of the tip of the minute hand of a clock where the hand is of length 7cm?

Ans1) speed, v=st=2πrT=2×227×7×10-260×60=119×10-4=1.22×10-4m/s

<h2><em><u>Hope it helps</u></em></h2>

5 0

3 years ago

4. If the Kitty Hawk Flyer has 12 HP, do some research to find out how to express this power in the SI system and report the num

lesya [120]

Product of my research: 1 HP = 746 watts .

12 HP = (12 x 746 W) = 8,952 W

8,952 W = 8.952 kW

8 0

3 years ago

Read 2 more answers

the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t

givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

$a_{t} = v = 0.8 m/s^{2}$

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

= 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

= 0.8

at x= 5 Km

dy/dx = 0.8(5)

= 4

$p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }$

now insert the values,

$p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km$

→ Now the normal component of acceleration is given by

$a_{n} = \frac{v^{2} }{p}$

= (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

$a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}$

$a = [(0.8)^{2} + (0.457)^{2}]^{0.5}$

a = 0.921 m/s²

4 0

2 years ago

The view that time is a scarce resource that must be rationed and controlled through the use of schedules and appointments, and

Diano4ka-milaya [45]

Answer:

Monochronic.

Explanation:

This view is called as Monochronic.

The dictionary definition of Mono-chronic is that Monochronic people just like to do one thing at a time. They respect a certain uniformity and sense of being a suitable place and time for different activities. They don't esteem interruptions. It has been generally observe that their focus and productivity remains higher than the polychronic.( Those we are comfortable with performing multiple task at a time)

6 0

3 years ago

How much heat is required to convert 2.55g of water at 28 degrees c to steam?

pantera1 [17]

There are two different processes here:
1) we must add heat in order to bring the temperature of the water from $28^{\circ}C$ to $100^{\circ}C$ (the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates

1) The heat needed for process 1) is
$Q_1=m C_s \Delta T$
where
$m=2.55 g$ is the water mass
$C_s = 4.18 g/J^{\circ}C$ is the water specific heat
$\Delta T=100^{\circ}C-28^{\circ}C=72^{\circ}C$ is the variation of temperature of the water
If we plug the numbers into the equation, we find
$Q_1 = (2.55 g)(4.18 J/g^{\circ}C)(72^{\circ}C)=767.4 J$

2) The heat needed for process 2) is
$Q=m L_e$
where
$m=2.55 g$ is the water mass
$L_e = 2264.7 J/g$ is the latent heat of evaporation of water
If we plug the numbers into the equation, we find
$Q_2=(2.55 g)(2264.7 J/g)=5775.0 J$

So, the total heat needed for the whole process is
$Q=Q_1+Q_2=767.4 J+5775.0 J=6542.4 J$

4 0

3 years ago