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3 years ago

I need help pleaseee

Physics

1 answer:

Ludmilka [50]3 years ago

3 0

Answer:

8) 709.8875 J

9) The object is at 7.24375 m from the ground

10) Kinetic energy increases as the object falls.

Explanation:

We use the expression for the displacement h(t) as a function of time of an object experiencing free fall:

h(t) = hi - (g/2) t^2

hi being the initial position of the object (10m) above ground, g the acceleration of gravity (9.8 m/s^2), and t the time (in our case 0.75 seconds):

h(0.75) = 10 - 4/9 (0.75)^2 = 7.24375 m

This is the position of the 10 kg object after 0.75 seconds (answer for part 9)

Knowing this position we can calculate the potential energy of the object when it is at this height, using the formula:

U = m g h = 10kg * 9.8 (m/s^2) * 7.24375 m = 709.8875 J (answer for part 8)

Part 10)

the kinetic energy of the object increases as it gets closer to ground, since its velocity is increasing in magnitude because is being accelerated in its motion downwards.

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Calculate the mass (in g) of 346 cm³ of polythene. The density of polythene is 0.95 g/cm³. Give your answer to 2 decimal places.

IceJOKER [234]

Answer:

M = 328.70g

Explanation:

From the given values:

V = 346 cm³

M of 1 cm³ of Polythene = 0.95g or 95/100g

Solve:

M = (95×346)

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2 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

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R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

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3 years ago

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mixas84 [53]

Answer:

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3 years ago

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storchak [24]

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<h3>Radius of the prop blade</h3>

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a = v²/r

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Thus, the radius of the prop blade of an airplane is determined as 4.25 m.

Learn more about centripetal acceleration here: brainly.com/question/79801

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