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2 years ago

The particles in the atom that account for most of the mass of the atom are protons and

Physics

1 answer:

valkas [14]2 years ago

8 0

Answer:

Neutrons

Explanation:

There are three particles that make up an atom, proton, neutron and electron. The mass of protons and neutrons is considered to be equal though in reality neutrons are heavier than protons; the mass of electrons is very less and ignored in the calculation of atomic mass.

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2. A truck speeds up from a velocity of 6 m/s to 14 m/s in 4 seconds. What is the trucks acceleration? SHOW YOUR WORK

zavuch27 [327]

As it is given here

initial speed is

$v_i = 6 m/s$

After 4 seconds the final speed is

$v_f = 14 m/s$

so here we can use the formula of acceleration using kinematics

$a = \frac{v_f - v_i}{t}$

$a = \frac{14 - 6}{4}$

$a = 2 m/s^2$

so here it will accelerate at 2 m/s^2 rate.

As it is given here

initial it starts from rest

$v_i = 0 m/s$

After 2.5 seconds the final speed is

$v_f = 15 m/s$

so here we can use the formula of acceleration using kinematics

$a = \frac{v_f - v_i}{t}$

$a = \frac{15 - 0}{2.5}$

$a = 6 m/s^2$

so here it will accelerate at 6 m/s^2 rate.

i think question is not correct as in first line it is saying about a bag of trash and then in next line it is asking for the position of Jumper and bridge.

7 0

2 years ago

the amount of water returning to the earth through precipitation is blank the amount of water leaving the earth through evaporat

nalin [4]

Answer:

The amount of water entering the earth through precipitation is equal to the amount of water leaving earth through transpiration.

Explanation:

Rates of precipitation and evaporation vary widely according to regions and seasons. But in a global scale the rates are equal. Thus the total amount of earth’s water maintains its constancy even though there is a continuous change in forms of water.

Evaporation and transpiration are the forms in which Water leaves the earth and it returns to the earth in various forms of precipitation like rain, snow, dew, fog etc. This water then reaches ocean and land. The water that reaches the land flows as surface run off into rivers and water bodies or seep into the ground replenishing the ground water table.

5 0

2 years ago

A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continue

enot [183]

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

7 0

3 years ago

two electrons are an angstrom (1x10^-10m) apart. What electrostatic force do they exert on one another?

Irina-Kira [14]

Answer:

2.30 × 10⁻⁸ N if the two electrons are in a vacuum.

Explanation:

The Coulomb's Law gives the size of the electrostatic force $F$ between two charged objects:

$\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}}$ ,

where

$k$ is coulomb's constant. $k = 8.99\times 10^{8}\;\text{N}\cdot\text{m}^{2}\cdot\text{C}^{-2}$ in vacuum.
$q_1$ and $q_2$ are the signed charge of the objects.
$r$ is the distance between the two objects.

For the two electrons:

$q_1 = q_2 = 1.60\times 10^{-19}\;\text{C}$ .
$r = 1\times 10^{-10}\;\text{m}$ .

$\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}} = -\frac{8.99\times 10^{8}\times (1.60\times 10^{-19})^{2}}{(1\times 10^{-10})^{2}} = 2.30\times 10^{-8}\;\text{N}$ .

The sign of $F$ is negative. In other words, the two electrons repel each other since the signs of their charges are the same.

8 0

2 years ago

The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1

xxMikexx [17]

The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

8 0

2 years ago