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2 years ago

PLEASE HELP THE question on picture

Physics

1 answer:

Gnoma [55]2 years ago

7 0

Vertical:

(20 m/s) sin(25º) ≈ 8.45 m/s

Horizontal:

(20 m/s) cos(25º) ≈ 18.1 m/s

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Why do our eyes see the color red when we look at a tomato?

forsale [732]

Answer:

B. Tomatos reflect red light

Explanation:

The only reason colors exist is because the objects with color reflect all other light except for what they are portrayed as. White reflects all colors, and black absorbs all colors.

If you have any questions feel free to ask :)

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3 years ago

Read 2 more answers

imagine you cut something in half. Then, you cut each half in half and continue doing so. Could you keep cutting the pieces in h

algol [13]

You can't keep cutting because the object will be gone

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2 years ago

Tom Cruise jumped from one building to other building while filming the roof chase scene in Mission: Impossible - Fallout. He di

OLga [1]

Answer:

$v_0=9.9\ m.s^{-1}$

Explanation:

Given:

angle of launch of projectile from horizontal, $\theta=15^{\circ}$

range of projectile, $R=5\ m$

<u>We have formula for the range of projectile:</u>

$R=\frac{v_0^2\times sin\ 2\theta}{g}$

putting the respective values

$5=\frac{v_0^2\times sin\ 30^{\circ}}{9.8}$

$v_0=9.9\ m.s^{-1}$ is the velocity with which Tom should jump to land on the other roof.

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2 years ago

You are on the south bank of the river in your canoe you need to reach the north bank you know that you can row your canoe at 2

Anna71 [15]

Answer:

(θ) = 60°

Explanation:

Given:

Speed of canoe Vc = 2 m/s

Speed of River Vr = 1 m/s

Computation:

Vc (Cosθ) = Vr

2 (Cosθ) = 1

(Cosθ) = 1 / 2

(Cosθ) = (Cos60)

(θ) = 60°

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2 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

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3 years ago