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exis [7]
2 years ago
10

All pressure topics in physics​

Physics
1 answer:
AfilCa [17]2 years ago
5 0
Absolute, Atmospheric, Differential, and Gauge Pressure
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If a ball is given a push so that it has an initial velocity of 5 m/s down a certain inclined plane, then the distance it has ro
iren [92.7K]

Answer:

v= 13 m/s

Explanation:

Velocity is defined as the derivative of displacement with respect to time

v= ds/dt

Known data

s(t) = 5t + 2t²  : distance that the ball has rolled after t seconds

vi= 5 m/s : initial velocity

t= 2 s

Problem develoment

s(t) = 5t + 2t²

v= ds/dt= 5 + 4t : velocity of the ball in function of the time

We replace t =2 s in the equation of velocity

v= 5 + 4(2)

v= 13 m/s : velocity after 2 seconds

4 0
3 years ago
A single loop of wire with an area of 0.0900 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic
erica [24]

Answer:

(a) 0.0171 V

Explanation:

A = 0.09 m^2, dB/dt = 0.190 T/s

(a) According to the law of electromagntic induction

e = dФ / dt

e = A dB / dt

e = 0.09 x 0.190 = 0.0171 V

(b)

as we know

i = e / R

we can find induced current by dividing induced emf by resistance

5 0
3 years ago
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field
lawyer [7]

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = \frac{1}{2} mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = \frac{1}{2}  \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             K_e = K_p

              K_p = ½ m v_e²

              K_p = \frac{1}{2}  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = \frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = \frac{7 \ 10^4 }{ 3 \ 10^8}

        v/c= 2.33 10⁻⁴

8 0
3 years ago
Arigid body must rotate about an axis in order for it to have angular momentum about that axis. True False
kompoz [17]

Answer:

False

Explanation:

Let's consider the definition of the angular momentum,

\vec{L} = I \vec{\omega}

where I = \int\limits_m r^2 dm = \lim_{n \to \infty} \sum\limits_{i=1}^n m_i r_i^2 is the moment of inertia for a rigid body. Now, this moment of inertia could change if we change the axis of rotation, because "r" is defined as the distance between the puntual mass and the nearest point on the axis of rotation, but still it's going to have some value. On the other hand,

\vec{\omega} = \frac{\vec{r} \times \vec{v}}{r^2} so \vec{\omega} \neq 0 unless \vec{r} ║  \vec{v}.

In conclusion, a rigid body could rotate about certain axis, generating an angular momentum, but if you choose another axis, there could be some parts of the rigid body rotating around the new axis, especially if there is a projection of the old axis in the new one.

7 0
3 years ago
Which statement best describes electrons?
crimeas [40]
They are positive and remain inside the nucleus.
4 0
2 years ago
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