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2 years ago

A train slows from 60m/s to 20m/s in 50s

Physics

2 answers:

Dovator [93]2 years ago

7 0

Answer:

a = -4/5 m/s^2

Explanation:

Acceleration = change in velocity / time

change in velocity = final velocity - initial velocity

a = (20 m/s - 60 m/s) / 50 s

a = -40 m/s / 50 s

a = -4/5 m/s^2

hope this helps! <3

Alexeev081 [22]2 years ago

4 0

Answer:

- 4/5 m/s^2

Explanation:

It is going from 60 to 20 m/s so ACCELERATION is negative

acceleration = change in velocity / time

= -40 m/s / 50 s = - 4/5 m/s^2

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Answer:

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3 years ago

What is the net force exerted by these two charges on a third charge q3 = 49.0nC placed between q1 and q2 at x3 = -1.085m ?Your

densk [106]

Answer:

The net force exerted by these two charges on a third charge is $5.468\times10^{-6}\ N$

Explanation:

Given that,

Third charge $q_{3}=49.0\ nC$

Distance $x_{3}=-1.085\ m$

Suppose The magnitude of the force F between two particles with charges Q and Q' separated by a distance d. Consider two point charges located on the x axis one charge, q₁ = -12.5 nC , is located at x₁ = -1.650 m, the second charge, q₂ = 31.5 nC , is at the origin.

We need to calculate the total force will be the vector sum of two forces

Using Coulomb's law,

$F_{13}=\dfrac{kq_{1}q_{3}}{(x_{1}-x_{3})^2}$

Put the value into the formula

$F_{13}=\dfrac{9\times10^{9}\times(-12.5\times10^{-9})\times49\times10^{-9}}{(-1.650-(-1.085))^2}$

$F_{13}=-17268.3\times10^{-9}\ N$

We need to calculate the force will be to the negative charge with opposite charges

Using Coulomb's law,

$F_{23}=\dfrac{kq_{2}q_{3}}{(x_{2}-x_{3})^2}$

Put the value into the formula

$F_{23}=\dfrac{9\times10^{9}\times(31.5\times10^{-9})\times49\times10^{-9}}{(-1.085)^2}$

$F_{23}=11800.2\times10^{-9}\ N$

The force also will be to the negative side, charges with same charge sign

We need to calculate the net force exerted by these two charges on a third charge

Using formula of net force

$F_{net}=F_{13}+F_{23}$

$F_{net}=-17268.3\times10^{-9}+11800.2\times10^{-9}$

$F_{net}=-0.0000054681\ N$

$F_{net}=-5.468\times10^{-6}\ N$

Negative sign shows the negative direction.

Hence, The net force exerted by these two charges on a third charge is $5.468\times10^{-6}\ N$

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3 years ago

Read the passage about the pygmy shrew.

cestrela7 [59]

Answer:

A very high metabolism and a very small size.

Explanation:

The pygmy shrew is a very small mammal, that forages day and night. The metabolism of the Pygmy shrew is so high that it must eat at least every 30 minutes or it might die. The best explanation for what happens to the food's mass and energy is that most of the food mass is rapidly used fro building up of the shrew due to its very high metabolism, and a bigger portion of the food is lost from the surface of the body of the shrew, due to its very small size. The combination of these two factors; a very high metabolism (rapidly uses up food material, and generates a large amount of heat in a very short time) and the very small size (makes heat loss due to surface area to volume ratio high) explains what happens to the food mass and energy.

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3 years ago

In which thermodynamic process work done maximum

gulaghasi [49]

Isobaric process has max work done

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4 years ago

An infinite long straight wire is uniformly charged, the charge density is a. Use Coulomb's law to calculate the electric field

bixtya [17]

Answer:

$\vec{E} = \frac{a}{2\pi \epsilon_0 R}\^R$

Explanation:

Since the wire is infinitely long, we will use Gauss' Law:

$\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

We will draw an imaginary cylindrical surface with height h around the wire. The electric flux through the imaginary surface will be equal to the net charge inside the surface.

In that case, the net charge inside the imaginary surface will be the portion of wire with height h. Then the charge of that portion will be equal to

$Q_{enc} = ah$

The left-hand side of the Gauss' Law is the flux through the imaginary surface. Since we choose our surface as a cylinder, of which we know the area, we do not have to take the surface integral.

$\int\vec{E}d\vec{a} = E2\pi R h$

where R is the radius of the imaginary cylinder.

Finally, Gauss' Law gives

$E2\pi Rh = \frac{ah}{\epsilon_0}\\E = \frac{a}{2\pi \epsilon_0 R}$

The vector expression is

$\vec{E} = \frac{a}{2\pi \epsilon_0 R}\^R$

As you can see, the electric field is independent from the height h, since that is merely an imaginary cylinder to apply Gauss' Law. In the end, what matters is the charge density of the wire and the distance from the wire.

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