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2 years ago

Please answer quickly :)

Physics

1 answer:

Elena L [17]2 years ago

6 0

Answer:

idk butttttt.... i had fun talking to u yesterday and the other conversation is getting a little....uhm....how do i say this.....weird.....everyone is friends now¯\_('-')_/¯

Explanation:

You might be interested in

A large lightning bolt is observed to have a 19500 A current and move 36 C of charge. What was its duration?

Lynna [10]

Answer:

Its duration is 1.85*10⁻³ s or 1.85 ms

Explanation:

The intensity of electric current I is defined as the amount of electric charge Q (measured in Coulombs) that passes through a section of a conductor in each unit of time. The letter I is used to name the Intensity and its unit is the Ampere (A).

The intensity of electric current is expressed as:

$I=\frac{Q}{t}$

where:

I: Intensity expressed in Amps (A)

Q: Electric charge expressed in Coulombs (C)

t: Time expressed in seconds (s)

Being:

I= 19500 A
Q=36 C
t=?

Replacing:

$19500 A=\frac{36 C}{t}$

Solving:

19500 A*t= 36 C

$t=\frac{36 C}{19500 A}$

t= 1.85*10⁻³ s= 1.85 ms (being 1 s= 1,000 ms)

<u><em>Its duration is 1.85*10⁻³ s or 1.85 ms</em></u>

8 0

2 years ago

If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc

prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

4 0

2 years ago

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur

sp2606 [1]

Answer:

a) q = -9.23 cm, b) h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

1 / f = 0.6 / 4 = 0.15

f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

1 / q = 1 / f - 1 / p

1 / q = 1 / 6.67 - 1/24

1 / q = 0.15 - 0.04167 = 0.10833

q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

M = h ’/ h = - q / p

h’= - q / p h

h’= - (-9.23) / 24.0 0.150

h’= 0.05759 cm

h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0

3 years ago

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago

Which is an example of a screw? Lightbulb,ladder,screwdriver,flag pole

Andrei [34K]

screwdriver :)

hope you'll give a good review!

4 0

3 years ago