Ask question

Ask question

All categories

2 years ago

12

Have you ever had a dream that, that, um, that you had, uh, that you had to, you could, you do, you wit, you wa, you could do so

, you do you could, you want, you wanted him to do you so much you could do anything?

2 answers:

Ivenika [448]2 years ago

4 0

DNDUDHHDHEUDIEIDID DUDE

Murljashka [212]2 years ago

3 0

Answer:

I love that video it’s so funny!!!

Explanation:

You might be interested in

Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of

GalinKa [24]

Answer:

D) equal to the flux of electric field through the Gaussian surface B.

Explanation:

Flux through S(A) = Flux through S (B ) = Charge inside/ ∈₀

4 0

2 years ago

6. A satellite is orbiting Earth just above its surface. The centripetal force making the satellite follow a circular trajectory

Svetllana [295]

Answer:

Engular velocity: $w=1,24*10^{-3}/s$

Linear velocity: $V=7905 m/s$

The time it takes:

$t=5060s=84min$

Explanation:

The magnitude of the centripetal acceleration can be related to the angular velocity and radius as:

(1) $a=r*w^{2}$

Solving for w:

(2) $w=\sqrt{\frac{a}{r} }$

Replacing a=9,8m/s2 and r=6,375,000m:

(3) $w=\sqrt{\frac{9,8m/s^{2}}{6375000m} }=1,24*10^{-3}/s$

And the angular velocity relates to the linear velocity:

$V=w*r=1,24*10^{-3}/s*6375000m=7905 m/s$

The perimeter of the orbit is:

$P=\pi *2*r=\pi *2*6375000m=40.05*10^{6}m$

The time it takes:

$t=\frac{P}{V} =\frac{40.05*10^{6}m}{7905 m/s}=5060s=84min$

5 0

2 years ago

Hardness and density are physical properties.<br> a. True<br> b. False

antiseptic1488 [7]

a . true hardness and density are physical properties

7 0

2 years ago

Read 2 more answers

The third one pls help

Nadya [2.5K]

Answer:

20 ms¯¹

Explanation:

3. Determination of the final velocity

From the question given above, the following data were obtained:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

Acceleration is simply defined as the change in velocity per unit time.

Mathematically, it can be expressed as:

Acceleration (a) = final velocity – Initial velocity / time

a = v – u / t

With the above formula, we can obtain the final velocity of the car as follow:

Time (t) = 4 s

Acceleration (a) = 5 ms¯²

Initial velocity (u) = 0 ms¯¹

Final velocity (v) =?

a = v – u / t

5 = v – 0 / 4

5 = v / 4

Cross multiply

v = 5 × 4

v = 20 ms¯¹

Thus, the final velocity of the car is 20 ms¯¹

7 0

2 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

2 years ago