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alukav5142 [94]
3 years ago
5

A long distance runner running a 5.0km track is pacing himself by running 4.5km at 9.0km/h and the rest at 12.5km/h. What is his

average speed?
Physics
1 answer:
sweet [91]3 years ago
6 0

Answer:

9.26 km/h

Explanation:

Applying,

V' = D'/t'............... Equation 1

Where V' = Average speed, D' = Total distance, t' = total time.

Given: D' = 5 km

But,

v = d/t............ Equation 2

Where v = speed , d = distance, t = time

t = d/v............ Equation 3

Given: d = 4.5 km, v = 9 km/h, and d = 0.5 km, v = 12.5 km/h

Therefore,

t₁ = 4.5/9 = 0.5 hours

t₂ = 0.5/12.5

t₂ = 0.04 hours

Therefore,

V' = 5/(0.5+0.04)

V' = 5/0.54

V' = 9.26 km/h

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calculate the amount of heat (in btu) needed to raise the temp of 2.5lb of glass from 45°f to 350°f. the specific heat capacity
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So you would use the equation Q=cmΔT, where c is the specific heat, m is the mass, and ΔT is change in temperature. Q, or heat added, would equal (0.187)(2.5)(350-45), which simplifies to 142.5875 btu.
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It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an
Maru [420]

Answer:

q=7.965*10^-^6C

Explanation:

From the question we are told that

Altitude of  d_1 390m

MagnitudeM_1=60.0 N/C

Altitude of d_2=240 m

Magnitude is M_2= 100 N/C

Distance of cube d_c=150 m

Generally the flux \phi is mathematical given as

 \phi=60(150)^2cos180+100(150)^2*cos0

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Generally Quantity of charge  q is mathematically given as

 q=\varepsilon _0 *\phi

 q=8.85*10^-^1^2 *9*10^5

 q=7.965*10^-^6C

5 0
3 years ago
A hot-air balloon is 11.0 m above the ground and rising at a speed of 7.00 m/s. A ball is thrown horizontally from the balloon b
Nesterboy [21]

Answer:

18.6 m/s

Explanation:

h = Initial height of the balloon = 11 m

v_{o} = initial speed of the ball

v_{oy} = initial vertical speed of the ball = 7 m/s

v_{ox} = initial horizontal speed of the ball = 9 m/s

initial speed of the ball is given as

v_{o} = \sqrt{v_{ox}^{2} + v_{oy}^{2}} = \sqrt{9^{2} + 7^{2}} = 11.4 m/s

v_{f} = final speed of the ball as it strikes the ground

m = mass of the ball

Using conservation of energy

Final kinetic energy before striking the ground = Initial potential energy + Initial kinetic energy

(0.5) m v_{f}^{2} = (0.5) m v_{o}^{2} + mgh \\(0.5) v_{f}^{2} = (0.5) v_{o}^{2} + gh\\(0.5) v_{f}^{2} = (0.5) (11.4)^{2} + (9.8)(11)\\(0.5) v_{f}^{2} = 172.78\\v_{f} = 18.6 m/s

7 0
3 years ago
Read 2 more answers
A 11.0 kg test rocket is fired vertically from Cape Canaveral. Its fuel gives it a kinetic energy of 1985 J by the time the rock
nirvana33 [79]

Answer:

h = 18.41 m

Explanation:

Given that,

Mass of a test rocket, m = 11 kg

Its fuel gives it a kinetic energy of 1985 J by the time the rocket engine burns all of the fuel.

According to the law of conservation of energy,

PE = KE = mgh

h is height will the rocket rise

h=\dfrac{E}{mg}\\\\h=\dfrac{1985 }{11\times 9.8}\\\\h=18.41\ m

So, the rocket will rise to a height of 18.41 m.

5 0
3 years ago
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