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Neporo4naja [7]
3 years ago
12

The table below shows the difference between the average temperatures in April and October in four locations.

Physics
1 answer:
nevsk [136]3 years ago
6 0

i dont know if im right but

b. 1.7

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A transformer has an output coil with 6 times as many turns as the input coil. If the input frequency is f then the output frequ
sveticcg [70]

Answer:

d. f

Explanation:

Based on the principle of operation of transformer, when a.c voltage is applied at the terminal of the primary coil (input coil), an alternating magnetic flux is produced in the iron core which links the secondary coil (output coil).

Also, an alternating e.m.f of <em>the same frequency</em> as that of primary coil (input coil) <u>will be induced in the secondary coil (output coil) by mutual inductance.</u>

Thus, If the input frequency is "f" then the output frequency will be "f" (since the frequency will be the same).

The correct option is "D"

5 0
3 years ago
Two parallel-plate capacitors, identical except that one has twice the plate separation of the other, are charged by the same vo
erik [133]

Answer:

The capacitor having less distance of separation has a stronger electric field.

Explanation:

The capacitors are identical and only difference between them is that one has twice the plate separation of the other. Therefore, capacitance of the given capacitors C1 and C2 is,

C1= Aε/d  and C2=Aε/2d

The charges Q1 and Q2 on the capacitors of capacitance C1 and C2 respectively, is then given by the equation,

Q1=VC1

Q1=VAε/d

Q2=VC2

Q2=VAε/2d

Therefore, the surface charge density σ1 and σ2 for the capacitors is,

σ1=Q1/A

σ1=VAε/(d*A)

σ1=Vε/d

Similarly,

σ2=Q2/A

σ2=Vε/2d

The electric field between the plates is directly proportional to the surface charge density. And so electric field is inversely proportional to the distance of separation. Therefore the capacitor whose distance of separation is less has a stronger electric field.

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8 0
3 years ago
Which type of reaction is shown in this energy diagram?
valentinak56 [21]

Answer:

Option C

Explanation:

The graph shows endothermic reaction because the reactants are lower in energy and the products are higher is energy. Endothermic reactions absorb energy having products with higher energy.

5 0
4 years ago
Read 2 more answers
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
Huck Finn walks at a speed of 0.70 m/sm/s across his raft (that is, he walks perpendicular to the raft's motion relative to the
exis [7]

Answer:

Explanation:

Given

Velocity of Huck w.r.t to raft v_{H,raft}=0.7\ m/s

Perpendicular to the motion of raft

Velocity of Raft in the river v_{raft,river}=1.6\ m/s

As Huck is traveling Perpendicular to the raft so he possess two velocities i.e. vertical velocity and horizontal velocity of River when observed from bank

v_{Huck,river\ bank}=0.7\hat{j}+1.6\hat{i}

So magnitude of velocity is given by

|v|=\sqrt{0.7^2+1.6^2}

|v|=\sqrt{0.49+2.56}

|v|=\sqrt{3.05}

|v|=1.74\ m/s

For direction \tan =\frac{0.7}{1.6}=0.4375

\theta =23.63^{\circ} w.r.t river bank

                       

4 0
3 years ago
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