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3 years ago

6

A 2.0 m conductor is formed into a square and placed in the horizontal xy-plane. A magnetic field is oriented 30.0° above the ho

rizontal with a strength of 1.0 T. What is the magnetic flux through the conductor?

1 answer:

Gnesinka [82]3 years ago

5 0

Answer:

$\phi_B = 0.216 T m^2$

Explanation:

As we know that the length of the conductor is given as

$L = 2 m$

now if it is converted into a square then we have

$L = 4a$

$a = \frac{L}{4} = 0.5 m$

now the are of the loop will be

$A = a^2 = 0.5(0.5) = 0.25 m^2$

now the magnetic flux is defined as

$\phi_B = BAcos\theta$

here we know

B = 1.0 T

$\theta = 30.0^o$

$\phi_B = (1.0 T)(0.25 m^2)(cos30)$

$\phi_B = 0.216 T m^2$

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Use the worked example above to help you solve this problem. An Alaskan rescue plane drops a package of emergency rations to str

Vlad [161]

Answer:

(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact is 77.19°

Explanation:

The parameters given are;

Velocity of the plane, vₓ = 39.0 m/s

Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m

(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;

h = u·t + 1/2×g×t²

Where:

g = Acceleration due to gravity = 9.81 m/s²

u = Initial vertical velocity = 0 m/s

Hence;

1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²

∴ t = √(1500/4.905) = 17.49 s

The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m

The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The vertical velocity, $v_y$ , of the package just before landing is given by the relation;

$v_y$ ² = u² + 2·g·h

u = 0 m/s

∴ $v_y$ ² = 0 + 2×9.81×1500 = 29430 m²/s²

$v_y$ = √29430 = 171.55 m/s

Hence the horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact, θ, is given as follows;

$tan \theta = \dfrac{v_y}{v_x} = \dfrac{171.55}{39.0 } = 4.4$

∴ θ = tan⁻¹(4.4) = 77.19°.

6 0

4 years ago

Which has the most kinetic energy?

lara31 [8.8K]

Answer:

I think A golf ball shot out of a small cannon

Explanation:

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2 years ago

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

7 0

4 years ago

Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?

eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = $\frac{1}{2}$ m $v^{2}$

Plug in the numbers = $\frac{1}{2}$ (4.0)( $8^{2}$ )

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

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3 years ago

When the temperature of the air is 25 degress C, the velocity of a sound wave traveling through the air is approximately

dusya [7]

Assuming an ideal gas, the speed of sound depends on temperature
only. Air is almost an ideal gas.

Assuming the temperature of 25°C in a "standard atmosphere", the
density of air is 1.1644 kg/m3, and the speed of sound is 346.13 m/s.

The velocity can't be specified, since the question gives no information
regarding the direction of the sound.

6 0

3 years ago