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3 years ago

A planet orbits a sun in a clockwise elliptical orbit as shown in the diagram below

Physics

1 answer:

STatiana [176]3 years ago

3 0

Answer:

Greatest gravitational energy is at "C".

The planet has to do work "against" the field to get to "C".

Also, if m v R (angular momentum) is constant then as R increases v must decrease for this term to be constant and KE = 1/2 M v^2 must decrease also to get to point C.

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When 224-nm light falls on a metal, the current through a photoelectric circuit is brought to zero at a stopping voltage of 1.84

OLga [1]

Answer:

3.71 eV

Explanation:

λ = Wavelength of light = 224 nm = 224 x 10⁻⁹ m

c = speed of electromagnetic wave = 3 x 10⁸ m/s

V₀ = stopping potential = 1.84 volts

W₀ = Work function of the metal = ?

Using the equation

$\frac{hc}{\lambda } = eV_{o} + W_{o}$

$\frac{(6.63\times 10^{-34})(3\times 10^{8})}{224\times 10^{-9} } = (1.6\times 10^{-19})(1.84) + W_{o}$

$W_{o}$ = 5.94 x 10⁻¹⁹

$W_{o}$ = 3.71 eV

7 0

3 years ago

When waves strike an object and bounce off, what has occured

Vesna [10]

The answer is reflection

5 0

3 years ago

A 20" round duct is 275' in length. The duct is carrying 2,900 CFM at a friction loss of 0.12 inches WG per 100 feet.

ivolga24 [154]

50500000 FACTION : uo/80

3 0

3 years ago

A 600-g block is dropped onto a relaxed vertical spring that has a spring constant k =190.0 N/m. The block becomes attached to t

charle [14.2K]

Answer:

Work done will be 2.205 j

Explanation:

We have given that the spring is compressed b 37.5 cm

So d = 0.375 m

Mass of the block m = 600 gram = 0.6 kg

Acceleration due to gravity $g=9.8m/sec^2$

Gravitational force on the block $F=mg=0.6\times 9.8=5.88N$

Now we know that work done is give by $W=Fd=5.88\times 0.375=2.205J$

5 0

3 years ago

The three forces shown act on a particle. what is the direction of the resultant of these three forces?

melisa1 [442]

Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png

Solution:
We need to find the magnitude of the resultant on both x- and y-axis.

x-axis) The resultant on the x-axis is
$F_x = 65 N\cdot cos 30^{\circ} - 30 N - 20 N\cdot sin 20^{\circ} = 19.45 N$
in the positive direction.

y-axis) The resultant on the y-axis is
$F_y = 65 N \cdot sin 30^{\circ} - 20 N \cdot cos 20^{\circ} = 13.70 N$
in the positive direction.

Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
$\tan \alpha = \frac{F_y}{F_x} = \frac{13.70 N}{19.45 N}=0.7$
from which we find
$\alpha=35^{\circ}$

7 0

3 years ago