How much time would it take for the sound of thunder to travel 1500 meters of sound travels at a speed of 330 m/sec?

The name of the Sl unit for force is the ___?<br><br> A. Joule<br> B. Newton<br> C.Watt

tiny-mole [99]

Answer:

the name of the SI unit for force is the newton

8 0

3 years ago

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A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago

A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

3 years ago

An object accelerates from rest and travels 53 m west in 5.2 s. Determine the acceleration

zmey [24]

Answer:

20.4m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Distance = 53m

Time = 5.2s

Unknown:

Acceleration = ?

Solution:

This is a linear motion and we use the right motion equation;

S = ut + $\frac{1}{2}$ at²

S is the distance

u is the initial velocity

a is the acceleration

t is the time

Insert the parameters and solve;

53 = (0x 5.2) + $\frac{1}{2}$ x a x 5.2

53 = 2.6a

a = $\frac{53}{2.6}$ = 20.4m/s²

6 0

3 years ago

If you have a gold brick that is 2.0 cm by 3.0 cm by 4.0 cm and has a density of 19.3 g/ml, what is its mass

34kurt

Answer:

It is

Explanation:

1 Answer. The volume is 37.0 cm3Au .

8 0

3 years ago