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3 years ago

Neutrons, protons, electrons are known as subatomic particles explain why that is?

2 answers:

8 0

Together, they constitute an atom, so they are called sub-atomic particles, where neutrons & protons reside under the nucleus whereas electrons revolve outside the nucleus

Hope this helps!

Alekssandra [29.7K]3 years ago

5 0

<span>Particles that are smaller than the atom are called subatomic particles. The three main subatomic particles that form an atom are protons, neutrons, and electrons. The center of the atom is called the nucleus.</span>

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A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot

Leokris [45]

Answer:

a. 32.67 rad/s² b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0

3 years ago

A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,

Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

$V_{r}$ = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0

3 years ago

A train on a straight track goes in the positive direction for 5.9 km, and then backs up for 3.8 km

kenny6666 [7]

Answer:

2.1km

Explanation:

Ill take it as u are talking about the displacement

Since displacement has negatives and positves

5.9 - 3.8 = 2.1km

6 0

3 years ago

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

ratelena [41]

Answer:

a = 1.3 m/s2; fn = 63.1 n :)

8 0

3 years ago

A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. I

natulia [17]

Answer:

x = 25 / μ [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

Y axis

N_B + N_A - W_van - W_load = 0

N_B + N_A = W_van + W_load

X axis

fr = ma

a = fr / m

the total mass is

m = (W_van + W_load) / g

the friction force has the expression

fr = μ N_{total}

fr = μy (W_van + W_load)

we substitute

a = μ (W_van + W_load) $\frac{g}{W_van + W_load}$

a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

v² = v₀² - 2 a x

0 = v₀² -2a x

x = $\frac{v_o^2}{2a}$

x = $\frac{v_o^2}{2 \mu g}$

x = $\frac{40^2}{2 \ 32 \ \mu}$

x = 25 / μ [ ft]

5 0

3 years ago