Answer:
25.08m/s
Explanation:
mgh1 + 0.5mv1² = mgh2 + 0.5mv2²
h1 = 0m
v1 = u
h2 = 5m
v2 = 23m/s
putting the values into the formula above;
m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)
0 + 0.5mu² = 50m + 264.5m
0.5mu² = 314.5m
dividing through by m
0.5u² = 314.5
u² = 629
u = <u>2</u><u>5</u><u>.</u><u>0</u><u>8</u><u>m</u><u>/</u><u>s</u>
<u>Theref</u><u>ore</u><u>,</u><u> </u><u>the</u><u> </u><u>init</u><u>ial</u><u> </u><u>speed</u><u> </u><u>"</u><u>u</u><u>"</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>m</u><u>/</u><u>s</u>
Answer:
The acceleration is equal to the net force divided by the mass. If the net force acting on an object doubles, its acceleration is doubled. If the mass is doubled, then acceleration will be halved. If both the net force and the mass are doubled, the acceleration will be unchanged.
People, and cars
is the right answer!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
7.78x10^-8T
Explanation:
The Pointing Vector S is
S = (1/μ0) E × B
at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,
S = (1/μ0) E B
where S, E and B are magnitudes. The average value of the Pointing Vector is
<S> = [1/(2 μ0)] E0 B0
where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)
Also at any instant,
E = c B
where E and B are magnitudes, so it must also be true at the instant of peak values
E0 = c B0
Substituting for E0,
<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²
Solve for B0.
Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)
= 7.79 x10 ^-8 T