An unstable nucleus results from too many or too few A) electrons. B) neutrons. C) protons. D) radons.

An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a

Virty [35]

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = $3.1$ m

Width of the window = $2.1$ m

Area of the window = $(3.1\times 2.1) = 6.51\ m^2$

Difference in air pressure = Inside pressure - Outside pressure

= $(1.0-0.954)$ atm = $0.046$ atm

Conversion of the pressure in its SI unit.

⇒ $1$ atm = $101325$ Pa

⇒ $0.046$ atm = $0.046\times 101325 =4660.95$ Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ $Pressure=\frac{Force }{Area}$

⇒ $Force =Pressure\times Area$

⇒ Plugging the values.

⇒ $Force =4660.95\times 6.51$

⇒ $Force=30342.78$ Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

3 0

4 years ago

The exosphere is the layer of the atmosphere

gayaneshka [121]

The exosphere is the layer of the atmosphere "Where gas molecules can be exchanged between Earth's atmosphere and outer space." Thus, the answer would be C.

8 0

3 years ago

Read 2 more answers

Two metallic rods A and B of different materials have same length. The linear expansivity of A is 12×10–6 oC–1and cubical expans

uysha [10]

Answer:

The length of rod A will be <u>greater than </u>the length of rod B

Explanation:

We, know that the formula for final length in linear thermal expansion of a rod is:

L' = L(1 + ∝ΔT)

where,

L' = Final Length

L = Initial Length

∝ = Co-efficient of linear expansion

ΔT = Change in temperature

Since, the rods here have same original length and the temperature difference is same as well. Therefore, the final length will only depend upon the coefficient of linear expansion.

For Rod A:

∝₁ = 12 x 10⁻⁶ °C⁻¹

For Rod B:

∝₂ = β₂/3

where,

β₂ = Coefficient of volumetric expansion for rod B = 24 x 10⁻⁶ °C⁻¹

Therefore,

∝₂ = 24 x 10⁻⁶ °C⁻¹/3

∝₂ = 8 x 10⁻⁶ °C⁻¹

Since,

∝₁ > ∝₂

Therefore,

L₁ > L₂

So, the length of rod A will be <u>greater than </u>the length of rod B

6 0

3 years ago

An experiment is conducted on a long straight wire of diameter d. A constant current is sent through the wire and the magnetic f

soldi70 [24.7K]

Answer:

D.

Explanation:

To solve the exercise it is necessary to apply the concepts related to the Magnetic Field described by Faraday.

The magnetic field is given by the equation:

$B = \frac{\mu_0 I}{2\pi d}$

Where,

$\mu =$ Permeability constant

d = diameter

I = Current

For the given problem we have a change in the diameter, twice that of the initial experiment, therefore we define that:

$B_1 = \frac{\mu_0 I}{2\pi d}$

$B_2 = \frac{\mu_0 I}{2\pi 2d}$

The ratio of change between the two is given by:

$\frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{2\pi d}}{\frac{\mu_0 I}{2\pi 2d}}$

$\frac{B_2}{B_1} = \frac{d}{2d}$

$\frac{B_2}{B_1} = \frac{1}{2}$

$B_2 = B_1 \frac{1}{2}$

Therefore the correct answer is D.

4 0

3 years ago

A cue ball, moving with 9.0 N·s of momentum strikes the nine-ball at rest. The nine-ball moves off with 2.0 N·s in the original

lisov135 [29]

Answer:

P = 7.28 N.s

Explanation:

given,

initial momentum of cue ball in x- direction,P₁ = 9 N.s

momentum of nine ball in x- direction, P₂ = 2 N.s

momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s

momentum of the cue after collision = ?

using conservation of momentum

in x- direction

P₁ + p = x + P₂

p is the initial momentum of the nine balls which is equal to zero.

9 + 0 = x + 2

x = 7 N.s

momentum in x-direction.

equating along y-direction

P'₁ + p = y + P'₂

0 + 0 = y + 2

y = -2 N.s

the momentum of the cue ball after collision is equal to resultant of the momentum .

$P = \sqrt{x^2+y^2}$

$P = \sqrt{7^2+(-2)^2}$

P = 7.28 N.s

the momentum of the cue ball after collision is equal to P = 7.28 N.s

7 0

3 years ago