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3 years ago

I’m stuck in B and D. which one is it?

Physics

2 answers:

Varvara68 [4.7K]3 years ago

8 0

Answer:

Option b is correct - Rocky, nearly continuous, found on earth surface.

The reason is it is the rocky surface area composed of crust and upper mantle part. Basically, it is the outer shell of earth.

tatuchka [14]3 years ago

6 0

It is b! the lithosphere is the crust and upper mantle

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Football player 1 has a mass of 80 kg and a velocity of 2 m/s east while player 2 has a mass of 70 kg and a velocity of 3 m/s we

sukhopar [10]

The total momentum of the system is equal to 50 Kgm/s.

Given the following data:

Mass 1 = 80 kg

Velocity 1 = 2 m/s east.

Mass 2 = 70 kg

Velocity 2 = 3 m/s west.

To determine the total momentum of the system:

Mathematically, momentum is given by the formula;

$Momentum = mass \times velocity$

For Football player 1:

$Momentum = 80 \times 2$

Momentum 1 = 160 Kgm/s.

For Football player 2:

$Momentum = 70 \times 3$

Momentum 1 = 210 Kgm/s.

Now, we can calculate the total momentum of the system:

$Total\;momentum = momentum \;2 - momentum \;1\\\\Total\;momentum = 210-160$

Total momentum = 50 Kgm/s.

Note: We subtracted because the football players were moving in opposite directions.

6 0

2 years ago

As a space shuttle climbs, _______

n200080 [17]

Rocket fuel will and smoke will emit from the thrusters.

4 0

3 years ago

Read 2 more answers

What is the mechanical advantage of the wheel and axle shown below?

BigorU [14]

The correct answer is A. 32.5

Mechanical advantage is the ratio of force that is input into a machine to the force output.

Mechanical advantage of a wheel and axle is calculated by dividing the radius of the wheel by that of the axle.

MA=R/r where R is the radius of the wheel and r is the radius of the axle.

Substituting for the values in the question gives:

MA=26cm/0.8cm

=32.5

6 0

3 years ago

Read 2 more answers

A gas storage cylinder in an ordinary chemical laboratory measures wide and high. This is the label on it. Contents: gas Pressur

zhenek [66]

Complete Question

A gas storage cylinder in an ordinary chemical laboratory measures 3.5 cm wide and 9.2 cm high with this label on it:

Contents: N2 Gas

Pressure 18.6 atm

If the cylinder is opened and the gas allowed to escape into a large empty plastic bag, what will be the final volume of nitrogen gas, including what’s collected in the plastic bag and what’s left over in the cylinder?

Use 1 atm for the gas final pressure

Answer:

The final volume is 1647.03cm³ or 1.65 litres

Explanation:

Given

Width of cylinder, W = 3.5 cm

Height of Cylinder, H = 9.2 cm

Gas Initial Pressure, P1 = 18.6 atm

Gas Final Pressure, P2 = 1 atm

First, the initial volume of gas has to be calculated to known the quantity of gas before gas outflow

Volume of Gas, V = Volume of a cylinder

V = πr²h

Given the the width of the cylinder is 3.5cm

This means that the diameter of the cylinder is also 3.5cm

Hence; Diameter, D = 3.5cm

Radius, r = ½D

r = ½ * 3.5cm

r = 1.75cm

So, V1 = πr²h becomes

V1 = π * 1.75² * 9.2

V1 = 28.175π cm³

V1 = 88.55 cm³

From Ideal Gas Law;

PV = nRT

Where k = PV (From Boyle's Law)

P1V1 = P2V2 ---- Make V2 the subject of formula

V2 = (P1V1)/P2

Where P1,P2,V1 and V2 represent the initial pressure, final pressure, initial volume and final volume, respectively.

Recall that P1 = 18.6 atm

P2 = 1 atm

V1 = 88.55 cm³

Final volume, V2 = (P1V1)/P2

V2 = (18.6 atm * 88.55 cm³)/1 atm³

V2 = 1647.03cm³

Convert to Litre

V2 = 1647.03/1000

V2 = 1.64703 litres

V2 = 1.65 L

Hence, the final volume is 1647.03cm³ or 1.65litres

8 0

3 years ago

A long, straight wire with 2 A current flowing through it produces magnetic field strength 1 T at its surface. If the wire has a

jeka57 [31]

Answer:

$r = 3.36 \times 10^{-7} m$

Explanation:

As per Ampere's law of magnetic field we know that

line integral of magnetic field along closed ampere's loop is equal to the product of current enclosed and magnetic permeability of medium

So it is given as

$\int B. dl = \mu_0 i_{en}$

here we can say that enclosed current is given as

$i_{en} = \frac{i}{\pi R^2} (\pi r^2)$

now from ampere'e loop law for any point inside the wire we will have

$B.(2\pi r) = \mu_o (\frac{ir^2}{R^2}$

$B = \frac{\mu_0 i r}{2\pi R^2}$

now we know that magnetic field inside the wire is 84% of the field at its surface

so we will have

$0.84 \frac{\mu_o i}{2\pi R} = \frac{\mu_o i r}{2\pi R^2}$

so we have

$r = 0.84 R$

now we know

$\frac{\mu_o i}{2\pi R} = 1$

here i = 2 A

$R = 2\times 10^{-7} m$

so now we have

$r = 3.36 \times 10^{-7} m$

7 0

3 years ago