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2 years ago

What is the similarities of a furnace and the sun?

Physics

1 answer:

Svetlanka [38]2 years ago

6 0

Answer:

They are both extremely hot, they both produce a form of light, they both have/use fire(typically)

Explanation:

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If Zack has a mass of 80 kg calculate his weight on Earth. Please show work, thank you!

Alina [70]

784 Newton ( mass x gravity = 80 x 9.8)

4 0

3 years ago

On the Moon, the acceleration due to the effect of gravity is only about 1/6 of that on Earth. An astronaut whose weight on Eart

vazorg [7]

And on the moon, the with is 595/6 N

3 0

2 years ago

2) A 0.4kg ball moves in horizontal circle of radius 3 m at speed of 100m/s. What

sweet-ann [11.9K]

Answer:

F = 1300 N

Explanation:

F = mv²/R = 0.4(100²)/3 = 1333.3333...

6 0

2 years ago

A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.

charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0

2 years ago

The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water

sashaice [31]

Answer: the constant angular velocity of the arms is 86.1883 rad/sec

Explanation:

First we calculate the linear velocity of the single sprinkler;

Area of the nozzle = π/4 × d²

given that d = 8mm = 8 × 10⁻³

Area of the nozzle = π/4 × (8 × 10⁻³)²

A = 5.024 × 10⁻⁵ m²

Now total discharge is dived into 4 jets so discharge for single jet will be;

Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec

So using continuity equation ;

Q_single = A × V_single

V_single = Q_single/A

we substitute

V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)

V_single = 29.8566 m/s

Now resolving the forces as shown in the second image,

Vt = Vcos30°

Vt = 29.8566 × cos30°

Vt = 25.8565 m/s

Finally we calculate the angular velocity;

Vt = rω

ω_single = Vt / r

from the given diagram, radius is 300mm = 0.3m

so we substitute

ω_single = 25.8565 / 0.3

ω_single = 86.1883 rad/sec

Therefore the constant angular velocity of the arms is 86.1883 rad/sec

7 0

2 years ago