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Deffense [45]
3 years ago
9

Pleas ehelp about convergent boundaries and don't spam I really need help on a few question PLLLLLLZZZ

Physics
2 answers:
Lapatulllka [165]3 years ago
8 0
I Don’t know I’m so sorry
ratelena [41]3 years ago
4 0
The top paragraph explains it all. The plates are moving because of the directing the force is creating, a check next to both blocks I’m not good in explaining sorry.
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If a projectile hits a stationary target, and the projectile continues to travel in the same direction, the mass of the projecti
nika2105 [10]

The correct arrangement of the question is;

If a projectile hits a stationary target, and the projectile continues to travel in the same direction,

A) the mass of the projectile is less than the mass of the target.

B) the mass of the projectile is equal to the mass of the target.

C) the mass of the projectile is greater than the mass of the target.

D) nothing can be said about the masses of the projectile and target without further information.

E) this is an unphysical situation and will not actually happen.

Answer:

Option C: The mass of the projectile is greater than the mass of the target.

Explanation:

We want to find what will happen when a projectile continues in motion after it hits a target.

Now, for the projectile to keep moving in that direction after it hits the target, it means it had a force bigger than the force of the target to overpower it and force it to move with it.

Now, from law of inertia, Force = ma.

But in this case acceleration is 0 because the speed of the projectile is constant.

Thus, the force depends on the mass. So for a higher force, the mass of the projectile has to be more than that of the stationary object.

Thus, option C is correct

8 0
3 years ago
What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wave
Firlakuza [10]

Answer:

thinnest soap film is  206.76 nm

Explanation:

Given data

wavelength = 550 nm

index of refraction n  = 1.33

to find out

What is the thinnest soap film

solution

we have wavelength  λ = 550 nm

that is  λ = 550 × 10^{-9} m

and n = 1.3

we will find the thickness of soap film as given by formula that is

thickness = λ/2n

thickness = 550 × 10^{-9}  / 2(1.33)

thickness = 206.76 × 10^{-9}  m

thinnest soap film is  206.76 nm

7 0
3 years ago
Describe two reasons why scientists are looking for alternatives to fossil fuels
Evgesh-ka [11]
1) It is because they produce a lot of pollution in the atmosphere
2) They are limited in number and can't be recycled

Hope this helps!
3 0
3 years ago
A certain material has a mass of 565 g while occupying 50 cm3 of space. What is this material?
EastWind [94]

<u>Answer:</u>

Lead

<u>Explanation:</u>

To get the density of the material, the formula would be:

mass divided by volume which is given by d = \frac { m } { v }.

Here in this problem, we are given a mass of 565 g which occupies a volume of 50 cm^3.

So plugging the data in the above formula to find the density:

Density = \frac { 565 } { 50 } = 11.3

From the table, we can see that the material is Lead which has a density of 11.3c/cm^3.

8 0
3 years ago
Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
3 years ago
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