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3 years ago

Compare a wedge and a screw with an inclined plane

Physics

1 answer:

Jlenok [28]3 years ago

3 0

a wedge is a small inclined plane

a screw is an inclined lane wkich goes round a centrral axis ... like a 'spiral' or helical staircase

easier per step ... more steps

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A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a sta

creativ13 [48]

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

$F_c=F_e=ma_c$ (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

$F_e=k\frac{qq'}{r^2}$ (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

$a_c=\frac{v^2}{r}$ (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

$k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}$

Finally, you replace the values of all parameters in the previous expression:

$r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m$

The radius of the circular trajectory is 0.22m

5 0

3 years ago

A light bulb is rated at 25 w when operated at 110 v. how much charge enters (and leaves) the light bulb in 1.0 hour?

DedPeter [7]

According to the current, the charge is 818.18 coulombs.

We need to know about current to solve this problem. Current can be defined as the flow rate of charge through the medium. It can be determined as

I = Q / t

where I is current, Q is charge and t is time.

From the question above, we know that

P = 25 W

V = 110 V

h = 1 hour = 3600 second

Calculate the current

P = V . I

25 = 110 . I

I = 0.23 A

By substituting the given parameters, we can calculate the charge

I = Q / t

0.23 = Q / 3600

Q = 818.18 coulombs

Hence, the charge is 818.18 coulombs.

Find more on current at: brainly.com/question/24858512

#SPJ4

4 0

1 year ago

An object is propelled straight up from ground level with an initial velocity of 48 feet per second. Its height at time t is mod

Ket [755]

Answer:

Explanation:

For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:

v₀ = 48 ft/s

a = -32 ft/s/s

v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:

v² = v₀² + 2aΔx and filling in:

$0=48^2+2(-32)$ Δx and

0 = 2300 - 64Δx and

-2300 = -64Δs so

Δx = 36 feet.

Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:

v = v₀ + at and filling in:

0 = 48 - 32t and

-48 = -32t so

t = 1.5 sec. That's part a. Onto part b:

The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in:

$0=48t+\frac{1}{2}(-32)t^2$ and

$0=48t-16t^2$ and

0 = 16t(3 - t) so

t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:

We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:

$48t+\frac{1}{2}(-32)t^2 >32$ and get everything on one side and factor it again:

$-16t^2+48t-32>0$ and we find that

1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.

8 0

3 years ago

When a source of dim orange light shines on a photosensitive metal, no photoelectrons are ejected from its surface. what could b

padilas [110]

Answer: Replace the orange light source with a higher frequency light source

Explanation:

To expel electrons from a piece of metal, the incoming light must have a minimum frequency to cause a photoelectric effect, i.e., the ejection of photoelectrons from a metal surface, which is also known as the metal's threshold frequency.

If v = frequency of incident photon and vth= threshold frequency, then,

For v < vth, there will be no ejection of photoelectron.

For v = vth, photoelectrons are just ejected from the metal surface, in this case, the kinetic energy of the electron ejected is zero

For v > vth, then photoelectrons will come out of the surface along with kinetic energy

Therefore, we would have to increase the frequency of incident light so that it becomes greater than the threshold frequency of that surface, and consequently a photoelectric process takes place.

5 0

2 years ago

if we're interested in knowing the rate at which light energy is received by a unit of area on a particular surface we're really

klasskru [66]

Illuminance of a surface<span>
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3 0

3 years ago

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