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vaieri [72.5K]
3 years ago
5

Could someone explain to me how to got the answer B, thank you very much​

Physics
1 answer:
Mnenie [13.5K]3 years ago
4 0

Answer:

since -6 lasted for 5 seconds, multiplying both would result in -30

3 lasted for 10 seconds, so multiplying both would give +30

average = ( 30 + (-30) ) / 2

30 -30 is already equal to zero, so the answer should be 0

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A system is a group of objects that’s analyzed as one unit. Consider a car moving along a road that has a flat section and a hil
vladimir1956 [14]

Answer:

a) Em= K +U,  b) Em= K

Explanation:

The system in this case is formed by the mobilizes and the hill.

Let's write the expressions correctly and completely.

a) When the car moves in the path, the mechanical energy is the siua of the kinetic energy of the car and the potential energy of the car when going up the hill.

              Em = K + U

be) when the car moves in the flat part all the mechanical energy is formed by its kinetic energy that is calculated with the mass and speed of the car

             Em = K

c) When the car goes up the hill the energy the mechanical energy is conserved, but part of the kinetic energy is transformed into potential energy.

8 0
3 years ago
PLEASE HURRY, WILL GIVE BRAINLIEST!!!
Brilliant_brown [7]

Answer:

1. Convection (Moving Water)

2. Radiation (Sunlight)

3. Conduction (Direct Contact)

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5. Convection (Moving Air)

6. Radiation (Feeling Heat)

Explanation:

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3 0
2 years ago
In a 200-turn automobile alternator, the magnetic flux in each turn is ΦB = 2.50 10-4 cos ωt, where ΦB is in webers, ω is the an
n200080 [17]

Answer:

10.63952sin(209.43951t)

10.63952 V

Explanation:

N_t = Number of turns = 200

\phi_B = Magnetic flux = 2.5\times 10^{-4}cos(\omega t)

\omega_e = Engine angular speed = 1\times 10^{3}\ rpm

Alternator angular speed is given by

N_a=2\times \omega_e\\\Rightarrow N_a=2\times 1\times 10^{3}\\\Rightarrow N_a=2\times 10^{3}\ rpm

\omega=N_a\dfrac{2\pi}{60}\\\Rightarrow \omega=2\times 10^{3}\dfrac{2\pi}{60}\\\Rightarrow \omega=209.43951\ rad/s

Induced emf is given by

\epsilon=-N_t\dfrac{d\phi_B}{dt}\\\Rightarrow \epsilon=-200\dfrac{d}{dt}2.54\times 10^{-4}cos(209.43951 t)\\\Rightarrow \epsilon=200\times 2.54\times 10^{-4}\times 209.43951 sin(209.43951t)\\\Rightarrow \epsilon=10.63952sin(209.43951t)

The function is 10.63952sin(209.43951t)

The induced maximum emf is 10.63952 V

5 0
3 years ago
How is friction reduced between an air hockey puck and the table?
irakobra [83]

Cause it's Corona Time

3 0
3 years ago
a rectangular coil of 25 loops is suspended in a field of 0.20wb/m2.the plane of coil is parallel to the direction of the field
statuscvo [17]

Answer:

The current in the coil is 60 Ampere.

Explanation:

Given:

Number of turns in the coil is N = 25

Dimension of the coil = 15cm X 12cm

magnitude of magnetic field = 0.20T

angle in the xy plane is θ = 0 degree

torque τ = 5.4 N-m

To find:

current in the coil is i = ?

Solution:

The torque acting on the coil is given by

=> \tau = NiAB cos\theta

Converting cm to m

12 cm = 0.12 m

15 cm = 0.15 m

The area of the coil is  

A = 0.12 X 0.15

A = 0.018 m^2

Substituting the values

=>5.4 = 25\times i \times 0.018 \times 0.20 \times cos\theta

=>5.4 = 25\times i \times 0.018 \times 0.20 \times cos(0)

=>5.4 = 25\times i \times 0.018 \times 0.20 \times 1

=>5.4 = 25\times i \times 0.018 \times 0.20 \times 1

=>5.4 = 0.09\times i

=>i = \frac{5.4}{0.09}

=> i = 60 A

3 0
3 years ago
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