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Alexeev081 [22]

3 years ago

12

A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'

s speed at the bottom is 4.22 m/s and is fast enough so that the marble never loses contact with the pipe. What is its speed at the top?

1 answer:

Leviafan [203]3 years ago

7 0

Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom $v=4.22 m/s$

Diameter of loop $d=28.6 cm$

As Energy is conserved therefore Energy at top is equal to energy at bottom

$E_T=E_B$

$\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}$ ,where $v_0$ is the velocity at bottom

$\frac{v^2}{2}+gh=\frac{v_0^2}{2}$

$v_0^2=v^2+2gh$

$v^2=v_0^2-2gh$

$v=\sqrt{v_0^2-2gh}$

$v=\sqrt{4.22^2-2\times 9.8\times 0.286}$

$v=\sqrt{17.8084-5.6056}$

$v=3.49 m/s$

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Suppose you are climbing a hill whose shape is given by the equation z = 2000 − 0.005x2 − 0.01y2, where x, y, and z are measured

Ierofanga [76]

Answer:

(a) Ascend at 0.8 vertical meter/meter

(b) Descend at -0.2·√2 vertical meter/meter

(c) In the (-0.6, -0.8) direction. The path begins at 45° to the horizontal

Explanation:

The given equation of the shape of the hill is z = 2000 - 0.005·x² - 0.01·y²

The current location = (60, 40, 1966)

The direction of the positive x-axis = east

The direction of the positive y-axis = north

(a) Walking due south = Reducing the y-value 40

From the equation, the elevation varies inversely with the motion towards the north

Therefore, walking south increases the elevation, and we ascend

The rate is given by the partial derivative at in the -j direction, which is 0.02

The rate is therefore 40 × 0.02 = 0.8

(b)The unit vector in the northwest direction u = 1/√2·(-1, 1)

∴ The rate = (-0.01(60), -0.02(40))·u = (-0.6, -0.8)·1/√2·(-1, 1) = -0.2·√2

Therefore we descend

(c) The slope is largest in the grad of the function at the point (60, 40) which is given as follows;

d(2000 - 0.005·x² - 0.01·y² )/dx, d(2000 - 0.005·x² - 0.01·y² )/dy = (-0.6, -0.8)

Therefore, the direction is tan⁻¹(-0.8/-0.6) ≈ S 36.87° W

The slope =(√((-0.4)² + (-0.8)²) = 1

Therefore, the angle is 45° to the horizontal.

4 0

3 years ago

Can someone help me a bit on this? Will mark brainliest. ( no physical science option soooo)

Advocard [28]

Answer:

When a light wave goes through a slit, it is diffracted, which means the slit opening acts as a new source of waves. How much a light wave diffracts<em> (how much it fans out)</em> depends on the wavelength of the incident light. The wavelength must be larger than the width of the slit for the maximum diffraction. Thus, for a given slit, red light, because it has a longer wavelength, diffracts more than the blue light.

The corresponding relation for diffraction is

$d sin(\theta) \approx \lambda$ ,

where $\lambda$ is the wavelength of light, $d$ is the slit width, and $\theta$ is the diffraction angle.

From this relation we clearly see that the diffraction angle $\theta$ is directly proportional to the wavelength $\lambda$ of light—longer the wavelength larger the diffraction angle.

7 0

3 years ago

Which change would decrease the total current, I, flowing through this circuit?

iren [92.7K]

Answer:

A. Increasing the voltage of the battery

Explanation:

The relationship between voltage, V, current, I and resistance, R, is given as follows;

V = I × R

∴ I = V/R

From the above relationship, the current flowing in the circuit is directly proportional to the voltage of the battery, and inversely proportional to the resistance, 'R', of the circuit

Therefore, increasing the voltage, 'V', of the battery, increases the total current, 'I', flowing in the circuit.

7 0

3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

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3 years ago

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Katena32 [7]

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