Answer:F(of gravity) = MA
F(normal force) = MA * cos(angle)
F = 72 * 9.81 * cos28
Don't have a calculator, so can't really do all the math right there. So just plug that in
Explanation:
i dont really know
The distance traveled by the particle at the given time interval is 0.28 m.
<h3>
Position of the particle at time, t = 0</h3>
The position of the particle at the given time is calculated as follows;
x = 2 sin2(t)
y = 2 cos2(t)
x(0) = 2 sin2(0) = 0
y(0) = 2 cos2(0) = 2(1) = 2
<h3>
Position of the particle at time, t = 4</h3>
x = 2 sin2(t)
y = 2 cos2(t)
x(4) = 2 sin2(4) = 0.28
y(4) = 2 cos2(4) = 2(1) = 1.98
<h3>Distance traveled by the particle at the given time interval</h3>
d = √[(x₄ - x₀)² + (y₄ - y₀)²]
d = √[(0.28 - 0)² + (1.98 - 2)²]
d = 0.28 m
Thus, the distance traveled by the particle at the given time interval is 0.28 m.
Learn more about distance here: brainly.com/question/23848540
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Answer:
The temperature is 
Explanation:
From the question ewe are told that
The rate of heat transferred is 
The surface area is 
The emissivity of its surface is 
Generally, the rate of heat transfer is mathematically represented as

=> ![T = \sqrt[4]{\frac{P}{e* \sigma } }](https://tex.z-dn.net/?f=T%20%20%3D%20%20%5Csqrt%5B4%5D%7B%5Cfrac%7BP%7D%7Be%2A%20%5Csigma%20%7D%20%7D)
where
is the Boltzmann constant with value 
substituting value
![T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }](https://tex.z-dn.net/?f=T%20%20%3D%20%20%5Csqrt%5B4%5D%7B%5Cfrac%7B13.1%7D%7B%200.287%2A%205.67%20%2A10%5E%7B-8%7D%20%7D%20%7D)

its C because i just anwserd it & it was right
The electron is accelerated through a potential difference of

, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:

where
m is the electron mass
v is the final speed of the electron
e is the electron charge

is the potential difference
Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:

where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B: