The vertical component is B) 7.9 m/s
Explanation:
A vector can be resolved into its horizontal and vertical components. The horizontal and the vertical components form the sides of a right triangle, in which the resultant corresponds to the hypothenuse, so we can use Pythagorean's theorem:
where
R is the resultant vector
Rx is the horizontal component
Ry is the vertical component
In this problem, we have:
R = 12 m/s is the resultant vector
Rx = 9 m/s is the horizontal component
Solving the formula for Ry, we find the vertical component:
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The answers is
D. The acid creates cracks in the rocks, which
allow air to circulate through the rock,
causing it to weather
Answer:
Explanation:
2 )
power of an electric device = V² / R where V is volts and R is resistance
putting given data
power = 9²/ 5
= 16.2 J/s
energy produced in 7 minutes
= 16.2 x 7 x 60
= 6804J .
3 ) Power of an electrical device
= V² / R
= V X I where I is current
= 4.5 x .5
= 2.25 W or J/s
4 )
energy used in 3 minutes with power of 2.25 W
= 2.25 x 3 x 60
= 405 J .
7 )
power of a electrical device
= V x I
IR x I where R is resistance .
= I²R
putting given data
power = .005² x 50
= 1.25 x 10⁻³ W .
8 )
Energy used up by a 60 W bulb in 2.5 hours
= 60 x 2.5 x 60 x 60
= 5.4 x 10⁵ J .
Students and scientists construct scale models of natural phenomena to better observe patterns and relationships.
What is natural phenomena?
Things that happen naturally—phenomena—occur or show up without human intervention. Gravity, tides, moons, planets, volcanic lightning, starling swarms, ant armies, sandstorms, biological processes and oscillation, among countless more events, are examples of natural phenomena.
Scientists and students use models to help them visualize their current understanding of a system in the condition that it is in right now.
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Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
