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Paladinen [302]

3 years ago

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On, Inc., publishing as pe

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enyata [817]3 years ago

8 0

-carbon
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when the mass of an object is increased it would decrease its acceleration in the force is left alone

mojhsa [17]

Answer:

See the explanation below.

Explanation:

This analysis can be easily deduced by means of Newton's second law which tells us that the sum of the forces or the total force on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = total force [N]

m = mass [kg]

a = acceleration [m/s²]

We must clear the acceleration value.

$a=\frac{F}{m}$

We see that the term of the mass is in the denominator, so that if the value of the mass is increased the acceleration decreases, since they are inversely proportional.

4 0

3 years ago

Select all of the answers that apply. which of the following cause volcanic eruptions?

Arada [10]

I think it's A. Pressure and heat built up in the magma chamber.

4 0

3 years ago

Read 2 more answers

Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10

USPshnik [31]

Answer:

D.

Explanation:

In order to know how long after Bill released his rock should Ted throw his if they want the stones to hit the water simultanously, we need to calculate the time needed to hit the water to both rocks independent each other, and just take the difference.

For the rock dropped by Bill, as the only influence on it is gravity (accelerating it downwards with an acceleration equal to g), and v₀ =0, we can use the following kinematic equation:

$y = \frac{1}{2} * g * t^{2}$

where y = height = 40 ft.

As all the parameters are given in SI units, it is advisable to convert this value to m, as follows:

$y = 40 ft*\frac{0.3048m}{1 ft} = 12.2 m$

Now, we can solve for t, as follows:

$t = \sqrt{\frac{2*y}{g}} = \sqrt{\frac{2*12.2m}{9.8m/s2}} = 1.58 s$

For the rock thrown down at 10 m/s, the kinematic equation we just have used becomes:

$y = v0*t +\frac{1}{2} * g * t^{2}$

This leaves us a quadratic equation on t, as follows:

$t = \frac{-10m/s}{9.8m/s2} +/- \sqrt{10m/s^{2} -4*4.9m/s2*(-12.2m)} = -1.02 s +/- 1.88s$

Taking the positive root, we have:

t = -1.02 s + 1.88 s = 0.86 s

So, in order to get that both rocks hit the water at the same time, Ted will need to wait the difference between both times:

Δt = 0.86 s - 1.58s = -0.72 s

5 0

3 years ago

A flea can jump with an initial velocity of 2.2 m/s at an angle of 21° with respect to the

Serga [27]

Answer:

Explanation:

If no one can see it because the lights were out. Did the flea really jump?

What do you want here?

Max height (2.2sin21)²/ 2(9.8) = 3.2 cm

Time of flight 2(2.2sin21)/ (9.8) = 0.16 s

distance of flight (2.2cos21)(0.16) = 33 cm

8 0

2 years ago

A thin rod of length L and total charge Q has the nonuniform linear charge distribution λ(x)=λ0x/L, where x is measured from the

marissa [1.9K]

Answer:

Explanation:

λ(x) = λo x/ L

(a) The total charge is Q.

$Q=\int_{0}^{L}dq$

$Q=\int_{0}^{L}\frac{\lambda _{0}x}{L}dx$

$Q=\frac{\lambda _{0}}{2L}\left ( x^{2} \right )_{0}^{L}$

$Q=\frac{\lambda _{0}}{2L}$

λo = 2Q/L

(b)

Let at a distance x from the origin the charge is dq.

so, dq = (2Q/L) x/ L dx

$dq=\frac{2Qx}{L^{2}}dx$

The potential due to this small charge at a distance d to the left of origin

$dV = \frac{KdQ}{d+x}$

$\int_{0}^{V}dV = \frac{2KQ}{L^{2}}\int_{0}^{L}\frac{xdx}{d+x}$

$V = \frac{2KQ}{L^{2}}\int_{0}^{L}\left ( 1- \frac{d}{d+x}\right )dx$

$V = \frac{2KQ}{L^{2}}\times \left ( x-dln(d+x) \right )\int_{0}^{L}$

$V = \frac{2KQ}{L^{2}}\times \left ( L-dln(d+L)-0+dlnd \right )$

$V = \frac{Q}{4\pi \epsilon _{0}L^{2}}\times \left ( L+d\times ln\left (\frac{d}{d+L} \right )\right )$

4 0

3 years ago