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3 years ago

How do protons neutrons and electrons differ

Physics

2 answers:

WITCHER [35]3 years ago

6 0

Answer:

They differ because

Proton means positive charge

Electrons are negatively charged

Neutrons are neutral

Alexus [3.1K]3 years ago

4 0

Answer/Explanation:

They have a relatively small mass compared to Protons and Neutrons. Protons are electrochemically positive in charge and the Neutrons are electrochemically neutral in charge.

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The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F

photoshop1234 [79]

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

T - 25 N = (8 kg) a

where T is the tension in the rope and a is the acceleration.

The hanging mass has a net force of

(6 kg) g - T = (6 kg) a

where g = 9.8 m/s².

Adding these equations together eliminates T, and we can solve for a :

(T - 25 N) + ((6 kg) g - T ) = (14 kg) a

33.8 N = (14 kg) a

a = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

T - 25 N = (8 kg) (2.4 m/s²)

T ≈ 25 N + 19.31 N ≈ 44 N

5 0

3 years ago

A centrifuge in a medical research laboratory rotates at an angular speed of 3,650 rev/min. When switched off, it rotates 46.0 t

erma4kov [3.2K]

Answer:

The constant angular acceleration of the centrifuge = -252.84 rad/s²

Explanation:

We will be using the equations of motion for this calculation.

Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.

First of, we write the given parameters.

w₀ = initial angular velocity = 2πf₀

f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s

w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s

θ = 46 revs = 46 × 2π = 289.14 rad

w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)

α = ?

Just like v² = u² + 2ay

w² = w₀² + 2αθ

0 = 382.38² + [2α × (289.14)]

578.29α = -146,214.4644

α = (-146,214.4644/578.29)

α = - 252.84 rad/s²

Hope this Helps!!!

6 0

4 years ago

Read 2 more answers

A mango is dropped and falls freely from rest. What are its position and velocity after

marissa [1.9K]

Answer:

$1s^{2}$

Explanation:

5 0

3 years ago

Read 2 more answers

The velocity of a car changes from 20 m/s east to 5 m/s east in 5 seconds. What is the acceleration of the car?

Klio2033 [76]

$acceleration = \frac{ v_{2}- v_{1} }{t} = \frac{5-20}{5} =-3m/s^{2}$

3 0

3 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0

3 years ago