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kirill [66]
3 years ago
13

What is an astronomical unit (AU)?

Physics
1 answer:
avanturin [10]3 years ago
4 0

The AU ... Astronomical Unit ... used to be defined as the average distance between the Sun and Earth during the year.

Now it's defined as 149,597,870,700 meters exactly.

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A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
First we need to find the net charge flowing at a certain point of the wire in one second, \Delta t=1.0 s. Using I=0.92 A and re-arranging the previous equation, we find
Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
3 0
3 years ago
Explain how the properties of alpha, beta and gamma radiation affect the level of hazard at different distances.
Mumz [18]

Answer:

Radiation can be absorbed by substances in its path. For example, alpha radiation travels only a few centimetres in air, beta radiation travels tens of centimetres in air, and gamma radiation travels very large distances. ... The thicker the substance, the more the radiation is absorbed

8 0
3 years ago
A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc
Nikitich [7]

Answer:

x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

Explanation:

From the law of conservation of energy

Energy lost  by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x

x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}

x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

The required distance from A to B is x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

5 0
3 years ago
The key idea of the transformation called translation is the gliding or sliding of every point in the plane the same direction.
Mrrafil [7]
That statement is true.

There are several definitions about transformation called translation, but the key idea is the gliding or sliding of every point in the plane the same direction.

Hope this helps
6 0
3 years ago
Read 2 more answers
In the sum A→+B→=C→, vector A→ has a magnitude of 13.6 m and is angled 40.2° counterclockwise from the +x direction, and vector
il63 [147K]

Answer:

|B|=27.00425726m

\alpha =210.3781372°

Explanation:

Let's use the component method of vector addition:

A_x=13.6cos(40.2)=10.38762599\\A_y=13.6sin(40.2)=8.778224553\\Cx=13.8cos(20.7+180)=-12.90912763\\Cy=13.8sin(20.7+180)=-4.877952844

Now, we know:

C_x=A_x+B_x\\\\C_y=A_y+b_y

So:

B_x=C_x-A_x=-23.29675362\\B_y=C_y-A_y=-13.6561774

Now lets calculate the magnitude of the vector B:

|B|=\sqrt{(B_x)^{2} +(B_y)^{2}  }=27.00425726m

Finally its angle is given by:

\alpha =(arctan(\frac{B_y}{B_x}))+180=30.37813438+180=210.3781344°

Keep in mind that I added 180 to the angles of C and B to find the real angles measured from the + x axis counter-clock wise.

8 0
3 years ago
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