Answers
Answer:
F = M2 ω^2 R centripetal force of sun on planet
ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P where P is the period
P = 2 pi (M2 * R / F)^1/2
F = G M1 M2 / R^2 gravitational force on planet
P = 2 pi {R^3 / (G M1)]^1/2
P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2
P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec
1 yr = 3600 * 24 * 365 = 3.15E7 sec
P = 3.9 / 3.2 = 1.2 years
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A baseball player throws a ball into the stands at 15.0 m/s and at an angle 45.0° above the horizontal. On its way down, the ball is caught by a spectator 4.10 m above the point where the ball was thrown. How much time did it take for the ball to reach the fan in the stands?
Which statement would most likely be found in an advertisement from a cell phone provider
A wire with mass 90.0 g is stretched so that its ends are tied down at points 98.0 cm apart. The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude of 0.300 cm at the antinodes. Part A What is the speed of propagation of transverse waves in the wire
Answers
The force between objects that are any distance apart is expressed as
According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,
M and m are the masses
r is the distance between the masses
If the force between objects that are 10 meters apart, hence;
To find the force between objects that are any distance apart, we will use the same formula above to have;
Substitute the result above into the expression to have:
Hence the force between objects that are any distance apart is expressed as
Learn more on gravitational law here: brainly.com/question/11760568
Answer:
F' = 100 F/r²
Explanation:
The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:
F = Gm₁m₂/r²
where,
F = Force between objects
G = Universal Gravitational Constant
m₁ = mass of first object
m₂ = mass of second object
r = distance between objects = 10 m
Therefore,
F = Gm₁m₂/10²
Gm₁m₂ = 100F --------------------- equation (1)
Now, we consider these objects at any distance r apart. So, the force becomes:
F' = Gm₁m₂/r²
using equation (1), we get:
F' = 100 F/r²
So, if the force (F) between objects 10 m apart is known, we can find it at any distance from the above formula.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225 $193,800
Par value at maturity 570,000
Total repaid 763,800
Less amount borrowed 645 669
Total bond interest expense $118.131
2. Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Period End Unamortized Discount Carrying Value
01/01/2019
06/30/2019
12/31/2019
06/30/2020
12/31/2020
3. Record the interest payment and amortization on June 30. Note:
Date General Journal Debit Credit
June 30
4. Record the interest payment and amortization on December 31.
Date General Journal Debit Credit
December 31
Answers
Answer:
1) Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225: $193,800
Par value at maturity: $570,000
Total repaid: $763800 (193,800 + 570,000)
Less amount borrowed: $508050
Total bond interest expense: $255750 (763800 - 508,050)
2)Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Interest Period End; Unamortized Discount; Carrying Value
01/01/2019 61,950 508,050
06/30/2019 54,206 515,794
12/31/2019 46,462 523,538
06/30/2020 38,718 531,282
12/31/2020 30,974 539,026
3) Record the interest payment and amortization on June 30:
June 30 Bond interest expense, dr 31969
Discount on bonds payable, Cr (61950/8) 7743.75
Cash, Cr ( 570000*8.5%/2) 24225
4) Record the interest payment and amortization on December 31:
Dec 31 Bond interest expense, Dr 31969
Discount on bonds payable, Cr 7744
Cash, Cr 24225
1 pound
1 kilometer
1 gram
Answers
Answer:
it's answer is 1 newton
Answers
Answer:
a) The fly is 2.24 m from the origin.
b) In polar coordinates, the position of the fly is (2.24 m, 26.7°).
Explanation:
Hi there!
The position vector of the fly is r = (2.00, 1.00)m. The distance from that point to the origin is the magnitude of the vector "r" (see figure).
a) Notice in the attached figure that the distance from the origin to the point where the fly is located is the hypotenuse of the triangle formed by r, the x-component of r (2.00 m) and the y-component ( 1.00 m). Then:
r² = (2.00 m)² + (1.00 m)²
r² = 5.00 m²
r = 2.24 m
The fly is 2.24 m from the origin.
b) To find the angle θ (see figure) we can use trigonometry:
cos θ = adjacent / hypotenuse
cos θ = 2.00 m / √5 m
θ = 26.7°
The same will be obtained if we use sin θ:
sin θ = opposite / hypotenuse
sin θ = 1.00 m / √5 m
θ = 26.7°
In polar coordinates, the position of the fly is (2.24 m, 26.7°).
Answers
Answer:
d' = 75.1 cm
Explanation:
It is given that,
The actual depth of a shallow pool is, d = 1 m
We need to find the apparent depth of the water in the pool. Let it is equal to d'.
We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,
or
d' = 75.1 cm
So, the apparent depth is 75.1 cm.
The apparent depth of a 1.00-meter-deep pool, when viewed from above, is around 75.2 centimeters. This difference is due to light refraction in water, causing optical distortion.
When observing a shallow pool of 1.00 meter depth from above, the apparent depth is altered by the phenomenon of light refraction in water. Light bends as it passes from air into water, affecting the way objects are perceived underwater.
The apparent depth is less than the actual depth due to this bending of light. To calculate the apparent depth, one can use the Snell's Law formula, which relates the angles of incidence and refraction to the refractive indices of the two media.
However, a simplified formula for the apparent depth (d') in terms of the actual depth (d) is given by d' = d/n, where 'n' is the refractive index of water (approximately 1.33). Therefore, in this case, the pool's apparent depth, when viewed from above, will be approximately 75.2 centimeters, making it shallower than it appears at first glance due to the optical effects caused by light traveling through water.
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Complete question below:
"What is the apparent depth, in centimeters, when looking straight down at a shallow pool that is 1.00 meter deep? Note that the apparent depth is different from the actual depth due to the refraction of light in water."
Answers
Work = Force times Distance
Work = 200 x 30
Work = 6000
The work done by a force of 200N on a body that moved 30m is 6000J or 6000 Joules.