Answer:
6.0 m/s vertical and 9.0 m/s horizontal
Explanation:
For the vertical component, we use the formula:
- Sin(34°) = <em>y</em> / 10.8
Then we <u>solve for </u><u><em>y</em></u>:
- 0.559 = <em>y</em> / 10.8
And for the horizontal component, we use the formula:
- Cos(34°) = <em>x</em> / 10.8
Then we <u>solve for </u><u><em>x</em></u><u>:</u>
- 0.829 = <em>x</em> / 10.8
So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".
By using drift velocity of the electron, the current flow is 7.20 ampere.
We need to know about drift velocity of electrons to solve this problem. The drift velocity can be determined as
v = I / (n . A . q)
where v is drift velocity, I is current, n is atom number density, A is surface area and q is the charge.
From the question above, we know that
d = 2.097 mm
r = (0.002097 / 2) m
v = 1.54 mm/s = 0.00154 m/s
ρ = 8.92 x 10³ kg/m³
q = e = 1.6 x 10¯¹⁹C
Find the atom density
n = Na x ρ / Mr
where Na is Avogadro's number (6.022 x 10²³), Mr is the atomic weight of copper (63.5 g/mol = 0.635 kg/mol).
n = 6.022 x 10²³ x 8.92 x 10³ / 0.635
n = 8.46 x 10²⁷ /m³
Find the current flows
v = I / (n . A . q)
0.00154 = I / (8.46 x 10²⁷ . πr² . 1.6 x 10¯¹⁹)
0.00154 = I / (8.46 x 10²⁷ . π(0.002097 / 2)² . 1.6 x 10¯¹⁹)
I = 7.20 ampere
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Given what we know, despite not having the figure attached to the question, we can still confirm that the magnitude for the acceleration of the dancer will be zero.
<h3>Why is the dancer's acceleration equal to zero?</h3>
This has to do with how the question clarifies the speed of the dancer. Though it does not give us an exact value, we are told that the speed is constant. This is an indicator that the acceleration is zero because with any other value for acceleration the speed <u>cannot remain</u> constant.
Therefore, given that any value for acceleration will increase or decrease the speed of the dancer, but we are told that the dancer's speed is constant throughout the trip, we can confirm that the magnitude for the acceleration of the dancer is zero.
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A 1-kg mass at the earth's surface weighs about C. 10N
The third planet from the Sun is the Earth. It is the seventh largest in terms of size and weighs roughly 5.98 1024 kg. The inherent quality of mass is unaffected by the environment of the object or the technique employed to quantify it.
Newton's law of gravitation can be used to estimate the mass of the Earth. This is set to the fundamental formula, which reads: force (F) = mass (m) times acceleration. Gravitational acceleration (G) is equal to 9.8 m/s2, the Earth's radius is 6.37 106 m, and the gravitational constant (G) is 6.673 1011 Nm2/kg2. The Earth has a mass of 5.96 1024 kg after rearranging the equation and entering all the numbers.
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