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ozzi
4 years ago
9

A transformer is used to change: 1) power 2) frequency 3) capacitance 4) none of the above

Physics
2 answers:
tangare [24]4 years ago
8 0

4)none of the above.

A transformer is an electrical device  that allows to increase or decrease from one voltage level to another in an AC circuit while conserving the power.  

It is basically composed of two coils magnetically coupled in a core of ferromagnetic material increasing its permeability and effectiveness.

kicyunya [14]4 years ago
8 0

Answer: None of the above

Explanation:

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What is the product of an object’s mass and velocity? momentum net force kinetic energy
kaheart [24]

<u>Answer:</u>

The correct answer option is momentum.

<u>Explanation:</u>

Momentum is the product of an object's mass and velocity.

It is a term which describes the relationship between the mass of an object and the velocity of an object.

Momentum can be represented in the form of an equation as:

P = mv

where P is the momentum,

m is the mass of the object; and v is the velocity of that object.

6 0
3 years ago
Number 10 and an explaination would be fabulous. thanks!
Sati [7]
Linear momentum has to be conserved. It was zero before the thread eas burned ... when nothing was moving ... so the momentum of the masses moving in opposite directions has to add up to zero. ... Momentum = mass times speed. ... In one direction, you have 5 kg times 1/5 m/s= 1 kg-m/s. ... We need 1 kg-m/s in the other direction. ... 7 kg times speed = 1 kg-m/s. ... Can you finish it from here ?
3 0
3 years ago
Read 2 more answers
A car's velocity as a function of time is given by vx(t)=α+βt2, where α=3.00m/s and β=0.100m/s3.
Mice21 [21]

1) Analyze the equation

Vₓ = α + β t²

Vₓ = 3.00 + 0.100 t²

That is a quadratic equation, so the graph is a parabola.

2) Therefore, take into account that the main points to draw a parabola are:

i) vertex

ii) concavity

iii) y-intercepts

iv) x - intercepts

Also, for all graphs you need the domain and the range.

3) Find the y-intercept (t = 0)

t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00

4) Find the x-intercepts (Vₓ = 0)

Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²

⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.

5) Concavity

Since, the coefficient of t² is positive, the parabola open upwards.

6) Vertex

It is the local minimum of the equation. You can find it by the first derivative

Vₓ' = 2×0.100 t = 0 ⇒ t = 0

Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s

⇒ vertex = (0,3.00)

7) The domain is given t ∈ [0,5.00]

8) You can also build a table with several points in the domain

t =0; Vₓ = 3.00 + 0.100 (0)² = 0

t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10

t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40

t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90

t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60

t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50

9) Range: Vₓ ∈ [ 3.00, 5.50]

10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.

4 0
3 years ago
Read 2 more answers
Give. 1325419200 seconds , convert to years.
GaryK [48]

Answer:

42 years

Explanation:

1325419200/3.154e+7

8 0
3 years ago
The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the
Ivanshal [37]

Answer:

(a) \alpha = - 1.32\ rev/m^{2}

(b) \theta = 13674\ rev

(c) \alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}

(d) a = 22.458\ m/s^{2}

Solution:

As per the question:

Angular velocity, \omega = 190\ rev/min

Time taken by the wheel to stop, t = 2.4 h = 2.4\times 60 = 144\ min

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

\omega' = \omega + \alpha t

omega' = final angular velocity

\omega = initial angular velocity

\alpha = angular acceleration

Now,

0 = 190 + \alpha \times 144

\alpha = - 1.32\ rev/m^{2}

Now,

(b) The no. of revolutions is given by:

\omega'^{2} = \omega^{2} + 2\alpha \theta

0 = 190^{2} + 2\times (- 1.32) \theta

\theta = 13674\ rev

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}

(d) The radial acceleration is given by:

\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s

Linear acceleration is given by:

a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}

a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}

5 0
3 years ago
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