Explanation:
Find the time it takes for the roadrunner to land.
Given (in the y direction):
Δy = 0 m
v₀ = v sin 10°
a = -9.81 m/s²
Find: t as a function of v
Δy = v₀ t + ½ at²
(0 m) = (v sin 10°) t + ½ (-9.81 m/s²) t²
t = (v sin 10°) / 4.905
Given (in the x direction):
Δx = 20.5 m
v₀ = v cos 10°
a = 0 m/s²
Find: t as a function of v
Δx = v₀ t + ½ at²
(20.5 m) = (v cos 10°) t + ½ (0 m/s²) t²
t = 20.5 / (v cos 10°)
Set equal and solve for v:
(v sin 10°) / 4.905 = 20.5 / (v cos 10°)
v² sin 10° cos 10° = 100.5525
v = 24.2485398301588
Graph:
desmos.com/calculator/x4b2zf1hxj
None of the options shown are correct.
The coefficient of expansion is 13 * 10^-6 m per meter length.per oK
The temperature difference = 42 - - 8 = 50 oC
delta T = (42 + 273) - (-8 + 273) = 50 oK
delta L = L * 13* 10^6 m/oK
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m
Answer: A changing magnetic flux establishes a current in a circuit.
Faraday's Law of Electromagnetic Induction was formulated from the experiments made by him and states that:
<em>The voltage induced in a closed circuit is directly proportional to the speed with which the magnetic flux that crosses any surface with the circuit as edge changes in time</em>
where:
is the Electric Field
is the infinitesimal element of the C contour
is the magnetic field density
is an arbitrary surface, whose edge is C
The negative sign indicates the direction of the induced current and refers to the opposition between the fields induced by the magnetic flux and the electromotive force.
So, it is the change in the magnetic flux that establishes a voltage in the circuit, hence current.
Answer:
The strength of magnetic field is 0.25 T
Explanation:
Time period sec
Mass of proton kg
Charge of proton C
Here proton moves in circular path
Velocity of proton is given by,
Put the value of velocity in above equation,
Now magnetic field is given by,
T
Therefore, the strength of magnetic field is 0.25 T
Work=applied Force x distance
= 1275 x 26
=33150 Joules