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likoan [24]
3 years ago
14

Diffusion in physics​

Physics
1 answer:
fgiga [73]3 years ago
5 0

Answer:

sorry but which class your talking 'bout

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3. An explosion of gas from a Hawaiian volcano blows a bunch of rocks into the sky. Each has 2000J of
anzhelika [568]

Answer: 3.906kg

Explanation:

K.E = 2000J

V = 32m/s

m =?

K.E = 1/2MV2

2000 = 1/2M x (32^2)

2000 = 1/2M x 1024

2000 = 512M

M = 2000/512

M = 3.906kg

4 0
3 years ago
a man is trying to move a piano by himself, but it will not move, what force must they overcome to start the piano moving?
m_a_m_a [10]

Answer:

The friction of the piano and the weight

Explanation:

7 0
3 years ago
A satellite m-500 kg orbits the earth at a distance d 215 km, above the surface of the planet. The radius of the earth is re 6.3
AnnZ [28]

Answer:

7.78 * 10³ m/s

Explanation:

Orbital velocity is given as:

v = √(GM/R)

G = 6.67 * 10^(-11) Nm/kg²

M = 5.98 * 10^(24) kg

R = radius of earth + distance of the satellite from the surface of the earth

R = 2.15 * 10^(5) + 6.38 * 10^(6)

R = 6.595 * 10^(6) m

v = √([6.67 * 10^(-11) * 5.98 * 10^(24)] / 6.595 * 10^(6))

v = √(6.048 * 10^7)

v = 7.78 * 10³ m/s

4 0
3 years ago
The bell rings and a physics student heads to class. They stop to talk to a few friends. They slow
grin007 [14]

Answer:

420m

Explanation:

Given parameters:

Time  = 5minutes

Average speed  = 1.4m/s

Unknown:

Distance covered  = ?

Solution:

Speed is the rate of change of distance with time.

 Mathematically;

              Speed  = \frac{distance}{time}  

    Distance  = speed x time

 

Insert the parameters and solve;

   Convert the time to seconds;

              1 minute = 60s

               5 minute = 5  x 60 = 300s

So,

   Insert the parameters and find the distance;

     Distance  = 300 x 1.4  = 420m

3 0
2 years ago
Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o
adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

F = \sqrt{F_h^2+F_v^2}

F = \sqrt{3.36^2+13.98^2}

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

\theta =tan^{-1}(\dfrac{F_v}{F_h})

\theta =tan^{-1}(\dfrac{13.98}{3.36})

   θ = 76.48°

3 0
3 years ago
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