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4 years ago

7

If you speed up from the rest to 12m/s in 3 seconds, what is your acceleration?

1 answer:

azamat4 years ago

7 0

I believe that your acceleration would be 4 m/s

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Which of the following is always true about the particles that make up matter? A. They can only be found in solid substances. B.

Neko [114]

B.They are too small to be seen with only our eyes

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3 years ago

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Nuclear explosions can be ______ of times more powerful than the largest conventional weapon. A. Hundreds B. Millions C. Thousan

yuradex [85]

Answer:

C

Explanation:

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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 $5.468\times10^{-6}\ N$

5 0

3 years ago

Two trains, each having a speed of 33 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h fli

Alex Ar [27]

Answer:

66 km

Explanation:

Given that:

The speed of the two trains = 33 km/h

The speed of the bird = 60 km/h

The distance apart between the two trains = 60 km

From the given information, we are being told that the two trains are going at the same speed. Therefore, they will definitely collide at 30 km

We know that:

speed of the train = distance traveled × time

Making the time t the subject of the formula:

time = speed of the train / distance traveled

time = 30 km / 33 km/h

time = 0.909 / hr

Thus, the bird flying at a given speed of 60 km/h in a time of 0.909 / hr will cover a total distance of :

distance (d) = speed of the bird/ time

distance (d) = $\dfrac{60 \ km/hr}{0.909 \ /hr}$

distance (d) = 66 km

3 0

4 years ago

Please help will make brainliest

kumpel [21]

The Gulf of Mexico
OR
near trenches, in the middle of the ocean, the continental shelf, and lastly, in the United States

3 0

3 years ago

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