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3 years ago

6

A hot-air balloon with a mass of 400 kilograms moves across the sky with 3,200 joules of kinetic energy. The velocity of the bal

loon is meters/second.

1 answer:

kotegsom [21]3 years ago

4 0

Answer:

4 m/s

Explanation:

KE =

Velocity of balloon will be 4 m/s.

!! Hope It Helps !!

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What is the displacement of the GBS cross-country team if they begin at the school, run 10 miles, and finish back at the school?

olga2289 [7]

Answer:

0

Explanation:

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3 years ago

The velocity of a moving body increases from 10m/s to 15 m/s in 5 sec . calculate its acceleration

Gnoma [55]

Answer:

v=u+at

15=10+a x 5

15-10= 5a

5= 5a

a=1 m/sec^2

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3 years ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = $\sqrt{\frac{2qV}{m}}$

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

radius of circular orbit is 2.76 × $10^{-4}$ m

8 0

3 years ago

What’s the power if a student does 2240j of work for 2.8 seconds

Alex787 [66]

Answer:

800 Watts

Explanation:

Power = Work/time

Working in SI units, Power = Watts, Work = Joules, Time = seconds.

Power = 2240J/2.8s = 800 Watts.

7 0

3 years ago

Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.3

fiasKO [112]

Explanation:

A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,

$\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}$

h = 2.3 m

$v=\sqrt{2\times 9.8\times 2.3} \\\\v=6.71\ m/s$

So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.

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3 years ago