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Sergio [31]
3 years ago
14

What should you do if someone on foot approaches your car while you are in it?

Physics
1 answer:
tatyana61 [14]3 years ago
6 0
It depends. If you are driving and the person doesn't look like a serial killer, you should stop.
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A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of th
Ber [7]

Answer:

0.546 ohm / μm

Explanation:

Given that :

N = 1.015 * 10^17

Electron mobility, u = 3900

Hole mobility, h = 1900

Ng = 4.42 x10^22

q = 1.6*10^-19

Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

= 0.01578880889 ohm /cm

Resistivity of germanium :

R = 1 / 2q * sqrt(Ng) * sqrt(u*h)

R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)

R = 1 /0.0001831

R = 5461.4964 ohm /cm

5461.4964 / 10000

0.546 ohm / μm

7 0
3 years ago
How reactive is an atom of Sodium(Na) and why?
lara [203]
Sodium (Na) is very reactive because it does not have a full valence shell.
6 0
3 years ago
What has more kinetic energy 15 kg ball rolling north at 15 m/s or a 15 kg ball rolling backwards at 7m/s
Setler79 [48]

Answer:

15 kg ball

Explanation:

6 0
3 years ago
A mass of 10 g of oxygen fill a weighted piston–cylinder device at 20 kPa and 110°C. The device is now cooled until the tempe
mezya [45]

Answer:

The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

\Delta V=V_{2}-V_{1}

\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})

Put the value into the formula

\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)

\Delta V=14.297\ L

\Delta V=14.3\times10^{-3}\ m^3

Hence, The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

6 0
3 years ago
The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22
alexgriva [62]
1) The law of motion of the projectile is
h(t) = 2+22.5 t-4.9 t^2
To find the velocity, we should compute the derivative of h(t):
v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t
So now we can calculate the speed at t=2 s and t=4 s:
v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s
v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
v(t)=0
So we have
22.5-9.8 t=0
And solving this we find
t=2.30 s

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
2+22.5t -4.9 t^2 = 0
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
t=4.68 s
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36  m/s
with negative sign, because it is directed downward.
8 0
3 years ago
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