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Rufina [12.5K]
3 years ago
5

Hi, anyone has any ideas for making your own planet? Be creative : )

Physics
2 answers:
xenn [34]3 years ago
6 0

Answer:

I would make all the people depressed but that would be normal and all the happy people were considerd sped.

Explanation:

miv72 [106K]3 years ago
4 0
Some ideas could be different atmospheres, surfaces (hard, rocky, water), colour, distance from sun. Hope this helps a little bit
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A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude
irina1246 [14]

Answer with Step-by -step explanation:

We are given that

b.\mid A\mid=46 m

\theta=20^{\circ} below the positive x-axis

Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis=x=360-20=340^{\circ}

x-component of vector A=A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24

y-Component of vector A=A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64

Magnitude of vector B=86 m

The vector B makes angle with positive x- axis=x'=42^{\circ}

x-component of vector B=B_x=86cos42=63.64

y-Component of vector B=B_y=86sin42=57.62

Vector A=A_xi+A_yj=43.24i-15.64j

Vector B=B_xi+B_yj=63.64i+57.62j

Vector C=A+B

Substitute the values

C=43.24i-15.64j+63.64i+57.62j

C=106.88i+41.98j

c.Direction=\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}

The direction of the vector C=21.5 degree

6 0
3 years ago
PLEASE HELP!! At what temperature will silver have a resistivity that is four times the resistivity of tungsten at room temperat
Svetach [21]
Consider 20 deg.C. as room temperature.

From tables,
Silver has a resistivity of  1.6*10^-8 ohm-m at 20 deg.C, and it increases by 0.0038 ohm-m per deg.K increase.
Therefore if the temperature rise above 20 deg.C is T, then silver will have resistivity of
1.6*10^-8(1 + 0.0038T) ohm-m

At room temperature, the resistivity of tungsten (from tables) is 5.6*10^-8.

The resistivity of silver will be 4 times that of tungsten (at room temperature) when
1.6*10^-8(1 + 0.0038T) = 4*5.6*10^-8
1 + 0.0038T = 14
T = 13/.0038 = 3421 deg.K approx

Answer: 20 + 3421 = 3441 °C
4 0
4 years ago
Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri
Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part kx \pm \omega t, the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but <em>here is an explanation</em> of this:

For both cases, + and -, after a certain time \delta t (\delta t >0), the displacement <em>y</em> of the wave will be determined by the kx\pm\omega (t+\delta t) term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply \pm \omega (t+\delta t).

To know which side, right or left of the origin, would go through the origin after a time \delta t (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation \delta x (we would be looking where in space is what we would have in the future in time). The term would be then k(x+\delta x)\pm\omega t, which at the origin is k \delta x \pm \omega t. This would mean that, when the original equation has kx+\omega t, we must have that \delta x>0 for k\delta x+\omega t to be equal to kx+\omega\delta t, and when the original equation has kx-\omega t, we must have that \delta x for k\delta x-\omega t to be equal to kx-\omega \delta t

<em>Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .</em>

<em />

In conclusion, when kx+\omega t, the part of the wave on the positive side (\delta x>0) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0
3 years ago
ILL GIVE BRAINLYEST
lawyer [7]

Answer:

3rd picture straight line going up right

Explanation:

3rd picture

5 0
4 years ago
Using a crowbar, a person can remove a nail by exerting little force, whereas pulling directly on the nail requires a large forc
algol [13]
Here we deal with a lever law. It states that product of force and distance from a fixed point on a lever is equal on both sides.

F₁*d₁ = F₂*d₂

By analysing this formula we can see that applying small force on a great length equals great force on a small length.
To remove nail we need to apply certain force. If we use F₁ for this required force we can see that on other side we need to apply certain force. If we have greater arm length we need smaller force. In a crowbar arm length along which we apply force is greater than length of our arm. This leads to a conclusion that we need smaller force when using crowbar. Depending on the length of a nail it is possible that we need to apply force that is greater than force required to remove nail.
7 0
3 years ago
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