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3 years ago

5

If a certain silver wire has a resistance of 3.00 Ï at 11.0Â°c, what resistance will it have at 25.0Â°c?

1 answer:

Romashka [77]3 years ago

7 0

11.30 times the nuber of toys

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Two cars travel westward along a straight highway, one at a constant velocity of 85 km/h, and the other at a constant velocity o

spayn [35]

To solve letter a:

d1 = 85t1 = 16 km,
85t1 = 16,
t1 = 16 / 85 = 0.1882 h = 11.29 min.

d2 = 115t2 = 16 km,
115t2 = 16,
t2 = 16 / 115 = 0.139 h = 8.35 min.

t1 - t2 = 11.29 - 8.35 = 2.94 min.
Car #2 arrives 2.94 minutes sooner.

To solve letter b:

15 min = 1/4 h = 0.25 h.
d1 = d2,
115t = 85(t + 0.25),
115t = 85t + 21.25,
115t - 85t = 21.25,
30t = 21.25,
t = 21.25 / 30 = 0.71 h,

d = 115 * 0.71 = 81.65 km.

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3 years ago

Which statement about electromagnetic waves is true?

Alborosie

Answer:

B

Good luck! Have a great day!

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3 years ago

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A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 7.0 s

marta [7]

Answer

given,

initial speed of merry-go-round = 0 rad/s

final speed of merry-go-round = 1.5 rad/s

time = 7 s

Radius of the disk = 6 m

Mass of the merry-go-round = 25000 Kg

Moment of inertia of the disk

$I = \dfrac{1}{2}MR^2$

$I = \dfrac{1}{2}\times 25000\times 6^2$

I = 450000 kg.m²

angular acceleration

$\alpha = \dfrac{\omega_f-\omega_0}{t}$

$\alpha = \dfrac{1.5-0}{7}$

$\alpha =0.214\ rad/s^2$

we know,

$\tau= I \alpha$

$\tau= 450000\times 0.214$

$\tau=96300\ N.m$

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4 years ago

In industry _____.

Alchen [17]

Answer:

B.useful products

Explanation:

industry is a sector that produces goods or services within an economy

6 0

3 years ago

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A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.

SashulF [63]

Answer:

Part a)

$F = 0$

Part b)

$F = 0.25 N$

Part c)

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

Explanation:

As we know that force on a current carrying wire is given as

$\vec F = i(\vec L \times \vec B)$

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

$F = i(\vec L \times \vec B)$

here we know that L and B is parallel to each other so

$F = 0$

Part b)

For 68.1 cm length wire we have

$F = iLB sin\theta$

here we know that

$cos\theta = \frac{68.1}{166}$

$\theta = 65.8$

so we have

$F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8$

$F = 0.25 N$

Part c)

For 151 cm length wire we have

$F = iLB sin\phi$

here we know that

$cos\phi = \frac{151}{166}$

$\theta = 24.5$

so we have

$F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5$

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

4 0

3 years ago