How much time does it take for the water to go from the top to bottom 4.5 m/s 53m?

A 20 kg sled is pulled up a 10m tall hill. What work is done against gravity?

Digiron [165]

Answer:

1962

Explanation:

w = f × d

= 20×9.81 × 10

= 1962

5 0

3 years ago

What increases the work output of a machine

Ilia_Sergeevich [38]

There are only two things you can do to increase the work output of a machine:

1). Increase the work INput to the machine.

2). Make the machine more efficient ... do things like lubricating it better to eliminate some internal friction.

4 0

3 years ago

A rock is thrown upward with a velocity of 22 meters per second from the top of a 25 meter high cliff, and it misses the cliff o

Mice21 [21]

Answer:

The rock will reach 9 m from the ground at eaxactly 5.06 s after it was initially thrown upwards.

Explanation:

We will use the equations of motion for this.

u = initial velocity of the rock = 22 m/s

g = acceleration due to gravity = -9.8 m/s²

y = vertical position of the rock at a time t = 9 m

y₀ = initial height of the rock = 25 m

t = time it takes for the rock to reach height of 9 m.

(y-y₀) = ut + 0.5gt²

(9 - 25) = 22t + 0.5(-9.8)t²

- 14 = 22t - 4.9t²

4.9t² - 22t - 14 = 0

solving this quadratic equation,

t = 5.055 s or - 0.565 s

Since time cannot be negative,

t = 5.055 s = 5.06 s

Hope this Helps!!!

6 0

3 years ago

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gwendolyn has a mass of 17 kg. she is riding her scooter at 1.15 m/s to the right when she runs into maya. maya has a mass of 13

kykrilka [37]

Inelastic collision happens when two objects joined and move together after the collision

7 0

3 years ago

A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .

Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}$

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
$B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}$

Taking out constants from the integral:
$B =\frac{\mu_0 i}{4\pi R^2} \int ds$

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

$B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds$

Evaluate:
$B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}$

Plugging in our givens to solve for the magnetic field strength of one loop:

$B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T$

Multiply by the number of loops to find the total magnetic field:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

b)

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}$

Using the diagram, if 'z' is the point's height from the center:

$r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}$

Substituting this into our expression:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }$

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds$

Evaluate:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Multiplying by the number of loops:
$B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Plug in the given values:
$B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}$

5 0

2 years ago

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