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Semenov [28]
3 years ago
8

How to measure voltage across an ungrounded component with a two channel oscilloscope?

Physics
1 answer:
N76 [4]3 years ago
8 0
If you cannot ground it, then what you should do is measure each channel and then subtract them and you'll see what the voltage is. If it's grounded, the way it works is different.
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An object is projected at 25m/s from the top of a building of height 50m. At the same instant,another object is projected from t
docker41 [41]

A) The objects have the same vertical position after 2 seconds

B) The objects have same vertical position at y = 30.4 m (but they do not collide since they have different x-position)

Explanation:

The motion of the first object along the vertical direction is a uniformly accelerated motion, so we can write its position at time t using the following equation:

y_1(t)=h+u_1 t + \frac{1}{2}gt^2

where:

h = 50 m is the initial height

u_1=0 is the initial vertical velocity (the object is projected horizontally, so the vertical velocity is zero at the beginning)

g=-9.8 m/s^2 is the acceleration of gravity

So, its vertical position can be rewritten as

y_1(t)=50-4.9t^2

The position of object 2 instead can be written as

y_2(t)=(u_2 sin \theta)t + \frac{1}{2}gt^2

where

u_2 sin \theta is the initial vertical velocity, where

u_2 = 50 m/s is the initial velocity

\theta=30^{\circ} is the angle of projection

Substituting, we get:

y_2(t)=(50)(sin 30^{\circ})t+\frac{1}{2}(-9.8)t^2=25t-4.9t^2

The two objects collide when their vertical position is the same, so:

y_1(t)=y_2(t)\\50-4.9t^2 = 25t-4.9t^2

And solving for t, we find:

50=25t\\t= 2 s

Note that this means that the two object at t = 2 s have the  same vertical position: however, this is not true for the horizontal position.

B)

In order to find the point where they collide, we have to substitute the time of the collision that we found in part A into one of the expressions of the vertical position.

Substituting into the expression of object 2, we find:

y_2(t) = 25t-4.9t^2=25(2.0)-4.9(2.0)^2=30.4 m

We can verify that at the same time, the vertical position of object 1 is the same:

y_1(t)=50-4.9t^2=50-4.9(2.0)^2=30.4 m

This means that the two objects have the same vertical position at 30.4 m.

However, in reality, the two objects do not collide. In fact, object 1 is moving in the horizontal direction with constant velocity

v_{1x}=25 m/s

So its horizontal position at t = 2.0 s is

x_1(2.0)=v_{1x}t=(25)(2.0)=50 m

While object 2 is moving in the horizontal plane with velocity

v_{2x}=u_2 cos \theta=(50)(cos 30^{\circ})=43.3 m/s

So its horizontal position  at t = 2.0 s is

x_2(2.0)=v_{2x}t=(43.3)(2.0)=86.6 m

So in reality, the two objects do not collide, if they start from the same x-position.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0
3 years ago
An object has an initial velocity v and uniform acceleration a. If it covers a distance d, then its final velocity, v, is given
viktelen [127]

Answer:

.

Explanation:

5 0
3 years ago
A cue ball, moving with 9.0 N·s of momentum strikes the nine-ball at rest. The nine-ball moves off with 2.0 N·s in the original
lisov135 [29]

Answer:

P = 7.28 N.s

Explanation:

given,

initial momentum of cue ball in x- direction,P₁ = 9 N.s

momentum of nine ball in  x-  direction, P₂ = 2 N.s

momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s

momentum of the cue after collision = ?

using conservation of momentum

in x- direction

P₁ + p = x  + P₂

p is the initial momentum of the nine balls which is equal to zero.

9 + 0  = x  + 2

x = 7 N.s

momentum in x-direction.

equating along y-direction

P'₁ + p = y + P'₂

0 + 0 = y + 2

y = -2 N.s

the momentum of the cue ball after collision is equal to resultant of the momentum .

P = \sqrt{x^2+y^2}

P = \sqrt{7^2+(-2)^2}

      P = 7.28 N.s

the momentum of the cue ball after collision is equal to P = 7.28 N.s

7 0
3 years ago
Find the wavelength of the third line in the lyman series, and identify the type of em radiation.
Natalija [7]

The wavelength of the third line in the Lyman series, and identify the type of EM radiation

In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.

1 / lambda = R(h)* ( \frac{1}{(n1)^{2} } -   \frac{1}{(n2)^{2} })

                 = 109678 ( \frac{1}{1^{2} } -  \frac{1}{3^{2} } )

                 = 109678 (8/9)

   Lambda = 9 / (109678 * 8 )

                  = 102.6 * 10^{-9} m = 102.6 nm

To learn more about Lyman series here

brainly.com/question/5762197

#SPJ4

8 0
2 years ago
What is the density of a iphone with a mass of 200g and a volume of 40cm3
Leviafan [203]
Answer: 5 gm/cc

Explanation:

200 gm/40 cc
= 5 gm/cc
8 0
3 years ago
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