The speed of Matt is 10 mph.
Doug runs 2 miles an hour faster than Matt, so let Matt’s speed equal x miles per hour. Then Doug’s speed equals x + 2 miles per hour. Each lap is one-quarter of a mile, so Doug runs 1.5 miles in the time it takes Matt to run 1.25 miles.
Rate of Matt is x
Rate of Dough is (x + 2)
Time taken by Matt is 1.25/x
Time taken by Dough is 1.25/(x + 2)
Distance covered by Matt is 1.25
Distance covered by Dough is 1.5
Dough and Matt took the same amount of time from the time Doug started, so make an equation by setting the two times in the chart equal to each other, and then solve for x:
= 
1.5x = 1.25(x + 2)
1.5x = 1.25x + 2.5
0.25x = 2.5
x = 10
So Matt ran at 10 miles per hour.
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Answer:
-0.7 m/s
Explanation:
Initial velocity (u)= 2.5 m/s
Acceleleration (a)= -0.8 m/s^2
Time taken (t) = 4 seconds
Hence,
v=u+at [1st Equation of motion]
v=2.5+-0.8*4
v=2.5-3.2
v=-0.7 m/s
Note that the negative sign indicates that the ball has changed direction and rolls downwards with gravity
When a gas is heated to become a plasma, the atoms (or the molecules) of the gas become ionised. In the ionisation, the atoms loose electrons from the exterior energy levels and thus heating to achieve a plasma will create free electrons and ionized atoms (or ionized molecules).
A plasma can not contain neutrons, because neutrons together with protons make the nuclei of the atoms. To free the neutrons from the atomn nuclei there would be necessary HUGE temperatures.
Also a plasma does not contain neutral elements (atoms) or (neutral) molecules, but Ionized atoms and/or molecules and free electrons.
Thus the good answer is d)
Because Helium fusion in a shell outside the core generates enough thermal pressure to push the upper layers outward.
Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s
Explanation:
Magnetic force(B) = 4.60×10^-3 T
Electric force(E) = 1.64×10^4 V/m
Both forces having equal magnitude ;
Magnetic force = electric force
qvB = qE
vB = E
v = (1.64×10^4) ÷ (4.60×10^-3)
v = 3.57×10^6 m/s
2.) Assume no electric field
qvB = ma
Where a = v^2 ÷ r
R = radius
a = acceleration
v = velocity
qvB = m(v^2 ÷ R)
R = (m×v) ÷ (|q|×B)
q=1.6×10^-19C
m = 9.11×10^-31kg
R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)
R = 32.5227×10^-25 ÷ 7.36×10^-22
R = 4.42×10^-3m
3.) period(T)
T = (2*pi*R) ÷ v
T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)
T = (27.775×10^-3) ÷(3.57×10^6)
T = 7.78×10^-9 s
