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Step2247 [10]
1 year ago
11

You will now use Graphical Analysis to calculate the slope of the line. Choose Analyze from the menu. Select Automatic Curve Fit

. Select the proper stock function from the list. (Hint: Your answers from question 1 should help you make the proper selection.) Just below the graph you will see the slope value. Remember that the symbol for slope is "M." When you choose OK to get back to the regular graph screen, you will see a little box on the graph. This box also shows the slope value. (Remember that slope is "m.") If you are unable to use Graphical Analysis, you can also graph your data on paper to determine the slope. Hint: to determine the slope, use the formula m = (y2 - y1) ÷ (x2 - x1). What is the numerical value of the slope of the line? Show your work.
Physics
1 answer:
Vlad1618 [11]1 year ago
3 0
The value for the slope is <span>M=1.13</span>
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An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st
RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

s=ut+\frac{1}{2}at^2

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2

Now we can apply Newton's second law to find the net force on the object:

F=ma=(24 kg)(4 m/s^2)=96 N

This net force is the resultant of the applied force forward (F_a) and the component of the weight acting backward (mg sin \theta), so we can find what is the applied force:

F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N

where

m = 24 kg is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

W=F_a s = (237.6)(18)=4276 J

where s = 18 m is the displacement.

Therefore the average power needed is:

P=\frac{W}{t}=\frac{4276}{3}=1425 W

b)

The instantaneous power at any point of the motion is given by

P=F_av

where

F_a is the force applied

v is the velocity of the object

We already calculated the applied force:

F_a=237.5 N

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

v=u+at=0+(4)(3.0)=12.0 m/s

And therefore, the instantaeous power at 3.0 sec is:

P=Fv=(237.5)(12)=2850 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
2 years ago
A 50 kg rocket generates 990 N of thrust. What will be its acceleration if it is launched straight up?
Ilia_Sergeevich [38]

Answer:

The acceleration of the rocket is 10 m/s².

Explanation:

Let the acceleration of the rocket be a m/s².

Given:

Mass of the rocket is, m=50\ kg

Thrust force acting upward is, F_{th}=990\ N

Acceleration due to gravity is, g=9.8\ m/s^2

Now, force acting in the downward direction is due to the weight of the rocket and is given as:

W=mg=50\times 9.8=490\ N

Now, net force acting on the rocket in upward direction is given as:

F_{net}=F_{th}-W\\F_{net}=990-490=500\ N

Therefore, from Newton's second law, net force acting on the rocket is equal to the product of mass and acceleration.

F_{net}=ma\\500=50a\\a=\frac{500}{50}=10\ m/s^2

Therefore, the acceleration of the rocket is 10 m/s².

4 0
1 year ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
2 years ago
calculate the work done to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it?
Charra [1.4K]
100N is how much work is needed 
4 0
2 years ago
Describe how private citizens have a voice in which projects the federal government will fund
erastovalidia [21]
<span>Two main ways, free elections, casting your vote for the candidate supporting your goals. Second, Lobbying the sitting gov.</span>
3 0
1 year ago
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