ounces is In the apothecary system of measurement, equal to one apothecary pound. a) eight b) 16 c) 12 d) four

What is the peak emf generated (in V) by rotating a 1000 turn, 42.0 cm diameter coil in the Earth's 5.00 ✕ 10−5 T magnetic field

ELEN [110]

Answer:

5.77 Volt

Explanation:

N = 1000, Diameter = 42 cm = 0.42 m, t = 12 ms = 12 x 10^-3 s

Change in magnetic field, B = 5 x 10^-5 T

The peak value of emf is given by

e = N x dФ / dt

e = N x A x dB/dt

e = (1000 x 3.14 x 0.21 x 0.21 x 5 x 10^-5) / (1.2 x 10^-3)

e = 5.77 Volt

3 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

On a hot summer day several swimmers decide to jump into a river. They step off the bridge and strike the water at 1.5s later. I

Lilit [14]

The only equation you need is the Δy = v₀t + 1/2at², and in this case, v₀t = 0 because the swimmers are starting from rest. Therefore, Δy = 1/2at², and that is 1/2 x (10m/s²) x (1.5)² = 11.25m.
With sigfigs it is 11m.

6 0

3 years ago

Projectile A is launched horizontally at a speed of 20 meters per second from the top of a cliff and strikes a level surface bel

pogonyaev

Answer: 3 seconds

Explanation:

Initial velocity(u) of projectile A in vertical direction = 0m/s

acceleration due to gravity a=g=9.81m/s^2

Time taken(t) of projectile A = 3s

Initial velocity of projectile B = 0m/s(vertical direction)

We can get height of cliff using parameters of projectile A since it's the same location.

Height(S) = u×t + 0.5×a×t^2

u =0

S= 0.5×9.81×3^2 = 44.145m

Time taken for projectile B to reach the ground:

S = u×t + 0.5×a×t^2

u =0, S=44.145m, a=9.81m/s^2

44.145 = 0.5×9.81×t^2

44.145 = 4.905×t^2

44.145 ÷ 4.905 = t^2

9 = t^2

t = sqrt(9)

t = 3seconds

5 0

4 years ago

Environment A is warm and gets lots of rain. Environment B is warmer and gets more rain year round. Identify Environment A.(2 po

zlopas [31]

The Swamp
There are many websites that say the rainforest but the rainforest is warmer and gets rain year round and swamps are warm and gets lots of rain but not year round.

5 0

2 years ago