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3 years ago

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Match the block to its correct density. 1. Block A 0.64 Kg/L 2. Block B 0.917 kg/L 3. Block C 0.70 Kg/L 19.3 kg/L 4. Block D 3.5

3 Kg/L 5. Block E

1 answer:

Sedbober [7]3 years ago

4 0

Answer:

quit school aghahahha

Explanation:

xD

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A conducting sphere with radius R is charged until the magnitude of the electric field just outside its surface is E. The electr

dusya [7]

Answer:

he correct answer is V = ER

Explanation:

In this exercise they give us the electric field on the surface of the sphere and ask us about the electric potential, the two quantities are related

ΔV = ∫ E.ds

where E is the elective field and normal displacement vector.

Since E is radial in a spray the displacement vector is also radial, the dot product e reduces to the algebraic product.

ΔV = ∫ E ds

ΔV = E s

since s is in the direction of the radii its value on the surface of the spheres s = R

ΔV = E R

checking the correct answer is V = ER

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4 years ago

Sophie [7]

Answer:

Figure xaina ta

Explanation:

Figure samet pathau

5 0

3 years ago

9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku

Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately <u>33.23 m/s</u> in a direction from the North of

approximately <u>9.18°</u>.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

v₁ = -( $\frac{\sqrt{2} }{2}$ × 3.5)·i + ( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

v₂ = ( $\frac{\sqrt{2} }{2}$ × 4.0)·i - ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-( $\frac{\sqrt{2} }{2}$ × 3.5) - ( $\frac{\sqrt{2} }{2}$ × 4.0))·i + (( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5) + ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15))·j

v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

Red Skull's velocity relative to Captain America, v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

Red Skull appears to be moving West at <u>5.3 m/s</u> and North at <u>32.8 m/s</u>

The direction is $arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}$

Therefore;

Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

$|v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23$ ·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ <u>33.23 m/s</u>

Learn more here:

brainly.com/question/24430414

6 0

3 years ago

a car traveling at a velocity of 2 m/s undergoes an acceleration of 4.5 m/s^2 over a distance of 340 m. How fast will it be goin

ra1l [238]

Vi = 2m/s
a= 4.5 m/s
d= 340 m
vf= ?

use this equation ... vf^2=vi<span>^2+2ad

you should get vf = 55.3
hope this helps </span>

3 0

3 years ago

4. A billiard ball rolls from rest down a smooth ramp that is 8.0 m long. The

Arisa [49]

Answer:

Explanation:

The formula S=(at^2)/2 will be used during the entire explanation.

1. 4 = (2t^2)/2

t = 2 s.

V = at = 2 * 2 = 4 m/s

2. 8 = (2t^2)/2

t = 2.8 s

iii. 2s

iv. 2.8 - 2 = 0.8s

Hope you understand)

8 0

3 years ago