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kolbaska11 [484]

3 years ago

10

An object’s position vs. time graph is a straight line with positive slope, draw the velocity vs. time graph that represents thi

s motion

1 answer:

maksim [4K]3 years ago

8 0

Answer:

Explanation:

its going to be a straight line with increasing acceleration know that slop of velocity -tine graph is acceleration

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A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe

andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

4 0

3 years ago

Read 2 more answers

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

3 years ago

Fill-in-the-blank test questions measure ________; matching concepts with their definitions measures ________.

Ulleksa [173]

Test questions measure recall; matching concepts with their definitions measures recognition.

<u>Explanation: </u>

According to Psychology our brain remembers everything what we learn but the understanding and remembering the right answer for the right question needs training and understanding ability. So in order to enhance the ability of recalling and recognizing among the students, the concept of test questions and matching with definitions are used in curricular activities.

As the students will be learning different terms, definitions, methods and different subjects, they should be able to distinguish among different definitions as well as they should recall the things they have learnt. So the answers for the test questions will help to recall the topics learnt by the students while the matching concept will help the students to recognize each definition with their terms.

6 0

3 years ago

In a compression wave, particles in the medium move

MissTica

<span>The particles through which compressional waves travel move in the same direction as the wave. This may be observed by fixing one end of a large spring and then compressing and extending the other end. The wave travels from one end to the other and the spring's parts move in the same direction.</span>

7 0

3 years ago

Read 2 more answers

How many newtons of force are represented by the following amount: 3 kg·m/sec^2?

Jet001 [13]

1N=1kg•m/s^2 so the answer is 3N

8 0

3 years ago