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2 years ago

How do the chemical properties of the halogens compare to those of the noble gases?

1 answer:

5 0

The inert gases do not ordinarily for compound. the halogens are extremely reactive whereas the inert gases tends to be unreactive....hope helps

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Calculate ine gravitational potential energy of the ball using pe=m×g×h.(use g=9.8 n/kg)

Neko [114]

Answer:

58.8J

Explanation:

Given parameters;

Mass of ball = 4kg

Height above the floor = 1.5m

g = 9.8n/kg

Unknown:

Potential energy = ?

Solution:

The potential energy of a body is the energy due to the position of the body.

It is mathematically expressed as:

Potential energy = mass x acceleration due to gravity x height

Potential energy = 4 x 9.8 x 1.5 = 58.8J

5 0

3 years ago

If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the mini

kondaur [170]

<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration. The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N. a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r. The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec. So a_N = 114 m/sec^2. g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>

3 0

2 years ago

If a proton were released from rest at the sphere's surface, what would be its speed far from the sphere?

balu736 [363]

Let the sphere is having charge Q and radius R

Now if the proton is released from rest

By energy conservation we can say

$U = K$

$\frac{kQe}{R} = \frac{1}{2}mv^2$

$\frac{2kQe}{mR} = v^2$

now take square root of both sides

$v =\sqrt{\frac{2kQe}{mR}}$

so the proton will move by above speed and

here Q = charge on the sphere

R = radius of sphere

$k = 9 * 10^9$

5 0

2 years ago

Read 2 more answers

The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa

Nostrana [21]

Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>

A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
We have the equation of time period of a SHM as,

$T=2\pi \sqrt{\frac{m}{k} }$