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andrew-mc [135]

4 years ago

5

Suppose the polar bear was running on land instead of swimming. If the polar bear runs at a speed of about 8.3 m/s, how far will

it travel in 10.0 hours?

1 answer:

olga_2 [115]4 years ago

8 0

Answer:

298800 m

Explanation:

$v = \frac{d}{t}$

V=speed

d=distance

t=time

$d = vt$

Change hours to seconds

It will be 36000s

8.3m/s * 36000s

=298800m

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One small speaker is placed 3m to the east of a second speaker, and a listener stands 4m directly south of one of the speakers.

alexandr402 [8]

Answer:

The value is $\lambda = 2 \ m$

Explanation:

From the question we are told that

The distance of the speaker from the second speaker to the east is $d = 3 \ m$

The distance of the speaker from the listener to the south is $a = 4 \ m$

Generally given that if the speaker move in any direction, their sound become louder , it then mean that the position of the listener of minimum sound (i.e a position of minima ) ,

Generally the path difference of the sound produce by both speaker at a position of minima is mathematically represented as

$y = \frac{\lambda}{2}$

Generally considering the orientation of the speakers and applying Pythagoras theorem we see that distance from the second speaker to the listener is mathematically represented as

$b = \sqrt{d^ 2 + a^2 }$

=> $b = \sqrt{3^ 2 + 4^2 }$

=> $b = 5$

Generally the path difference between the two speaker with respect to the listener is

$y = b - a$

=> $y = 5 - 4$

=> $y = 1$

So

$1 = \frac{\lambda}{2}$

=> $\lambda = 2 \ m$

4 0

3 years ago

I need help ASAP please and thank u very much if u can answer i’m giving a brainliest to btw

NemiM [27]

Answer:

A I Think

Explanation:

8 0

3 years ago

How do you balance this equation Mg+ O2= MgO

Anuta_ua [19.1K]

Answer:

2 Mg(s) + O2(g) → 2 MgO(s)

As magnesium (Mg) has a valency of +2, and oxygen (O) has a valency of -2, the ratio would be 1:1, and magnesium oxide would be represented as MgO. It is insoluble in water, hence it has the subscript as a solid, represented by (s). In order to use up the diatomic oxygen (O2), there needs to be two moles of magnesium (2Mg) on the reactant side. This would produce 2 moles of MgO on the product side.

Explanation:

8 0

4 years ago

What impulse must be applied to a 25.0-kg cart to cause a velocity change<br> of 12.0 m/s?

klio [65]

Answer:

Impulse of force = 300Ns

Explanation:

Given the following data;

Mass = 25kg

Change in velocity = 12m/s

To find the impulse;

Impulse is given by the formula;

$Impulse \; of \; force = mass * change \; in \; velocity$

Substituting into the equation, we have;

$Impulse \; of \; force = 25 * 12$

Impulse of force = 300Ns

8 0

3 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago