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AveGali
1 month ago
13

4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity

.]
Physics
1 answer:
Yuliya22 [3.3K]1 month ago
8 0

Answer:

w_f= 22.41 rad/s

Explanation:

To begin with, we have:

a = 4 rad/s²

S = 10 rev = 62.83 rad

Now we understand that:

w_f^2-w_i^2=2aS

where w_f denotes the final angular velocity, w_i the initial angular velocity, a represents angular acceleration, and S symbolizes radians.

<pplugging in="" the="" values="" we="" obtain:="">

w_f^2-(0)^2=2(4)(62.83)

Ultimately, solving for w_f gives:

w_f= 22.41 rad/s

</pplugging>
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We fully submerge an irregular lump of material in a certain fluid. The fluid that would have been in the space now occupied by
Ostrovityanka [3204]

Answer:

the lump descends

Explanation:

The full articulation of the query is

Upon fully submerging a 3 kg lump of material in a certain fluid, the fluid that would have occupied the space now filled by the lump weighs 2 kg. (a) When released, does the lump float up, sink, or remain steady

(a)

F_{b} = Buoyant force acting upward on the lump

M = mass of irregular lump = 3 kg

m = mass of fluid displaced = 2 kg

The upward buoyant force on the lump is given by the weight of the displaced fluid, thus

F_{b} = mg = (2) (9.8) = 19.6 N

the weight of the irregular lump of material is represented as

W = mg\\W = (3) (9.8)\\W = 29.4 N

Given that the weight of the lump downward exceeds the upward buoyant force, the lump will indeed descend

4 0
12 days ago
A sinusoidal electromagnetic wave of frequency 6.10×1014hz travels in vacuum in the +x direction. the magnetic field is parallel
Yuliya22 [3333]
Part a) The connection between the electric field and the magnetic field in an electromagnetic wave is
E=cB
where
E signifies the strength of the electric field
B indicates the strength of the magnetic field
c represents the speed of light
Using the equation, we determine:
E=cB=(3 \cdot 10^8 m/s)(5.80 \cdot 10^{-4} T)=1.74 \cdot 10^5 N/C

Part b) The text does not clarify the orientation of the magnetic field on the y-axis: I speculate it points in the y+ direction.
The direction of the electric field can be established using the right-hand rule, which states:
- the index finger shows the direction of E
- the middle finger indicates the orientation of B
- the thumb reveals the propagation direction of the wave
Because the wave propagates in the x+ direction, and the magnetic field in the y+ direction, we conclude that the electric field direction (index finger) must be z-.
7 0
29 days ago
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
inna [3103]

Answer:

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu ≈ 3077.34

Explanation:

To calculate the flux of F (vector field) across surface S, where

F(x,y,z) = y i − x j + z^{2} k

and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We will evaluate the following integral:

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + v^{2}k

The normal vector N is computed as

N = S_{u}XS_{v}

Where:

S_{u} = =

S_{v} = =

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, v^{2}>. < 2v sin v, -2v cos v, u >

F(S(u,v)).N = 2uv + uv^{2}

Thus

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {2uv}+u v^{2}\,dvdu ≈ 3077.34

8 0
22 days ago
A student is trying to classify an unidentified, solid gray material as a metal or a nonmetal. Which question will best help the
kicyunya [3294]
The most effective question for the student to determine whether the substance is metal or nonmetal would be option C.
8 0
16 days ago
Read 2 more answers
Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat
Yuliya22 [3333]

Answer:

Explanation:

The distance between the electrodes is denoted as d.

The kinetic energy of the electron is represented as Ek when the electrodes are positioned at a distance of "d" apart.

Our goal is to determine the kinetic energy when they are separated by a distance of d/3.

K.E = ½mv²

It’s important to note that the mass remains constant; only velocity varies.

Additionally,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Assuming constant acceleration

Hence, m and a are fixed,

therefore,

K.E is directly related to d

Thus, as d increases, K.E increases, and conversely, when d decreases, K.E decreases.

Consequently,

K.E_1 / d_1 = K.E_2 / d_2

With K.E_1 equating to E_k

and d_1 being d

while d_2 is represented as d/3

This leads to K.E_2 = K.E_1 / d_1 × d_2

Thus, K.E_2 = E_k × ⅓d / d

Finally,

K.E_2 = ⅓E_k

Therefore, the resultant kinetic energy is one third of the original E_k

7 0
16 days ago
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