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3 years ago

Which statement should she place in the region marked X?

Physics

2 answers:

Sidana [21]3 years ago

6 0

Answer:

DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

Explanation:

Orlov [11]3 years ago

3 0

I think the answer is D

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A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

Drupady [299]

Answer:

$4.5\times 10^{-5} T$

Explanation:

We are given that

Current in wire=40 A

Magnetic field= $B_1=3.5\times 10^{-5}$ T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

$B_{wire}=B_2=\frac{\mu_0I}{2\pi R}$

We have R=29 cm= $\frac{29}{100}=0.29 m$

1 m=100 cm

Substitute the values in the given formula

$B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T$

The resultant magnetic field is given by

$B=\sqrt{B^2_1+B^2_2}$

Substitute the values then we get

$B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}$

$B=4.5\times 10^{-5} T$

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire= $4.5\times 10^{-5} T$

Hence, the resultant magnitude of the magnetic field 29 cm above the wire= $4.5\times 10^{-5} T$

7 0

3 years ago

Vector question,university physics zemanski

just olya [345]

Answer:

yes This is correct Answer

3 0

3 years ago

one horsepower is a unit of power equal to 746w. how much energy can a 150-horsepower engine transform in 10.0s?

Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

(150 horsepower) x (746 watt/HP) x (1 joule/sec / watt) x (10 sec)

= (150 x 746 x 1 x 10) joule = 1,119,000 joules .
if correct plz mark brainly

8 0

3 years ago

A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/L, where x is measured from the

marissa [1.9K]

Answer:

Explanation:

λ(x) = λo x/ L

(a) The total charge is Q.

$Q=\int_{0}^{L}dq$