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3 years ago

I really need help with the graphs

Physics

1 answer:

nata0808 [166]3 years ago

3 0

Answer:

t = 2s

Explanation:

When you're looking for instantaneous portions of a graph, of any sort really, it means you're observing a rate at a single point in time [or possibly some other variable]. It's sorta like a snapshot of a rate as opposed to an average rate over an interval. After choosing this rate we'll typically draw a straight, tangent line through it to indicate it's slope. (Tangent lines are just lines that only touch a single point on a graph or shape.)

Another thing to take note of are the values of the graph's major axes. The "y-axis" corresponds to velocity in meters per second, while the "x-axis" corresponds to time in seconds. Normally when relating the two we put "y" over the "x" and say that at any point there are "y[units]" per "x[units]". Though with instantaneous rates, we say the value of "x" is "1"; for reasons I can try to further explain later if you'd like.

With the above information in mind we can turn our attention to your graph. You're told to find the point on this graph where the instantaneous rate of acceleration is -2 m/s². The only place where the graph reflects an instantaneous rate of -2m/s² is at t = 2s. At t = 2, the rate comes out to (2[m/s]/1s), which simplifies to 2m/s². If you then draw the tangent line through the point, you'll find that the line is decreasing (going down from left to right) which means that the instantaneous rate is negative.

So at t = 2s, we have an instantaneous acceleration of -2m/s².

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An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min. (a) Calculate the magnitude of the centripe

jonny [76]

To solve this problem it is necessary to apply the concepts related to the relationship between tangential velocity and centripetal velocity, as well as the kinematic equations of angular motion. By definition we know that the direction of centripetal acceleration is perpendicular to the direction of tangential velocity, therefore:

$a_c= \frac{v^2}{r}=r\omega^2$

Where,

V = the linear speed

r = Radius

$\omega =$ Angular speed

The angular speed is given by

$\omega =6500\frac{rev}{min}(\frac{2\pi rad}{1 rev})(\frac{1 min}{60s})$

$\omega = 680.6784 rad/s.$

Replacing at our first equation we have that the centripetal acceleration would be

$a_c = r\omega^2$

$a_c = (0.0750)(680.6784)^2$

$a_c = 34 749.23 m/s^2$

To transform it into multiples of the earth's gravity which is given as $9.8m / s$ the equivalent of 1g.

$a_c =34749.23 \frac{m}{s^2} (1\frac{g}{9.8m/s^2})$

$a_c = 3545.84g$

PART B) Now the linear speed would be subject to:

$v = \omega r$

$v= (680.6784)(0.0750)$

$v=51.05 m/s$

Therefore the linear speed of a point on its edge is 51.05m/s

8 0

3 years ago

A 4.0-kg block is pushed up a 36 incline by a force of magnitude P applied parallel to the incline. When P is 31 N, it is obser

denpristay [2]

Answer:

15.1 N

Explanation:

mass of block (m) = 4 kg

angle of inclination = 36 degrees

applied force (P) = 31 N

acceleration due to gravity (g) = 9.8 m/s^{2}

since the block is moving at a constant speed, it means the acceleration is 0 and therefore the summation of all the forces acting on the body is 0

therefore

P - f - mgsinθ = 0

where

P = applied force
f = frictional force
m = mass
g = acceleration due to gravity

θ = angle of inclination

when P = 31 N and the block is pushed upward

31 - f - (4 x 9.8 x sin 36) = 0

f = 7.96 N

now that we have the value of the frictional force we can find P required to lower the block, our equation becomes p + f - mgsinθ = 0 since the block is to be lowered

P + f - mgsinθ = 0

P = mgsinθ - f

P = (4 x 9.8 x sin 36) - 7.96 = 15.1 N

5 0

3 years ago

GRAPH INCLUDED PLEASE HELP

swat32

Answer:

10 days

Explanation:

The half-life of a radioactive sample is the time taken for half of the sample to decay.

In the diagram, the half-life corresponds to the time after which the % of cobalt-57 has halved. We can observe the following:

At t=10 days, the % of Co remaining is approximately 45%

At t=20 days, the % of Co remaining is approximately 22%

This means that the sample of cobalt-57 has halved in 10 days, so the half-life of cobalt-57 is 10 days.

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4 years ago

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3 years ago

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