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2 years ago

Describe electrons. Location: Charge: Mass

Physics

2 answers:

Vanyuwa [196]2 years ago

8 0

Answer:

location:electron cloud

charge:negitive

mass:one-two thousands amu

Explanation:

irga5000 [103]2 years ago

5 0

Protons, neutrons and electrons. Both protons and neutrons have a mass of 1 amu and are found in the nucleus. However, protons have a charge of +1, and neutrons are uncharged. Electrons have a mass of approximately 0 amu, orbit the the nucleus, and have a charge of -1.

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The air around a pool and the water in the pool receive equal amounts of energy from the sun. Why does the air experience a grea

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1 year ago

Cassidy Carver's family child care center has an activities table with four chairs an easel that can accommodate a child on each

Akimi4 [234]

Answer:4

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2 years ago

--->Two aircraft P and Q are flying at the same speed. 300 m/s, The direction along which P is flying is at right angles to t

REY [17]

Answer:

The magnitude of the velocity of the aircraft P relative to aircraft Q is zero

Explanation:

The velocity of the two aircraft, P & Q, v = 300 m/s

The angle of the direction between them, Ф = 90°

The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula

Substituting the values in the above equation

v = 300 x cos 90°

= 300 x 0

= 0

Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.

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2 years ago

Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for

g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0

2 years ago

Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th

Anton [14]

Answer:

Option B, 93 cm

Explanation:

An diagram of the seed's motion is attached to this solution.

This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)

(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2

R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.

QED!

6 0

2 years ago

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