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3 years ago

15

This is my first question don’t really know how to use this app yet lol but somebody answer it for me pls!! Seded the corredare

1 answer:

miskamm [114]3 years ago

7 0

3x6

(This just for extras )

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You have discovered a planet that is one-quarter the radius of Earth (Rp = 1/4R⊕) and one-half as massive (Mp = 1/2M⊕). How does

ExtremeBDS [4]

It's 8 times as much as before.

4 0

2 years ago

A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re

lbvjy [14]

Answer:

The beam of light is moving at the peed of:

$\frac{dy}{dt} = \frac{80\pi}{3}$ km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, $\omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi$ (1)

Now, in the right angle in the given fig.:

$tan\theta' = \frac{y}{3}$

Now, differentiating both the sides w.r.t t:

$\frac{dtan\theta'}{dt} = \frac{dy}{3dt}$

Applying chain rule:

$\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}$

$sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}$

Now, using $tan\theta = \frac{1}{m}$ and y = 1 in the above eqn, we get:

$(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}$

Also, using eqn (1),

$8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}$

$\frac{dy}{dt} = \frac{80\pi}{3}$

7 0

2 years ago

If you could help me out that would be fantastic.?!!!

Volgvan

It's either C. Or B.

3 0

3 years ago

White raven [17]

Answer:

The direction of defliection of the site to the left I think ..

4 0

2 years ago

3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a

serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as $\mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}$ .

Where, ${V}_{i}$ is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

${V}_{i}$ =3 m/s, V = 0 m/s, and ∆s = 2 m

Substitute the values in the above formula,

$0=3^{2}-2 \times 2 \times a$

0 = 9 - 4a

4a = 9

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration.

Calculating Coefficient of friction:

$\mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}$

Compare the above equation

$\mu \times m \times g=m \times a$

Cancel "m" common term in both L.H.S and R.H.S

$\text { Equation becomes, } \mu \times g=a$

$\text { Coefficient of friction } \mu=\frac{a}{g}$

$\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}$

$\mu=\frac{2.25}{9.8}$

$\mu=0.229$

Hence coefficient of friction is 0.229.

calculating force:

$\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })$

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0

2 years ago