Which type of radiation has the highest penetrating power?

mike's car can accelrate from 0 mph to 40 mph in 5 sec. what is the acceleration of his car miles per hour per sec?

siniylev [52]

1. 40-0=40

3. 40/5=8

8 ml/s

you find the range of acceleration(step one)

divide by the time(step two)

6 0

3 years ago

1. Determine the energy released per kilogram of fuel used.

Serga [27]

The formula for energy release per kilogram of fuel burned is energy release per kg=6.702*10-13. and 19. J 1 Mev = 1.602 X 10 T

Calculate the energy in joules per kilogram of reactants given MeV per reaction. Energy is the ability or capacity to perform tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
Think of a mole of plutonium-239 (molar mass: 239 grams) as a mole of "reactions."

Energy used in the US per person annually = 3-5 X 1011
Population (number of people) = 3.108The required mass of the fuel is 3.5x1011 x3-1x10 8x 10)/6.703 X1013 kg. the mass required: 1.62 x 1033 kg Mev in Joules 6 is equal to 101.60*I0-
19. J 1 Mev = 1.602 X 10 T, which translates to 1.602*1013/2.39x10-3 energy release per kilogram, or 6.702*10-13.

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8 0

2 years ago

The voltage supplied to a circuit is 17 V and the current running through is 10 A. What is the power generated?

Lelu [443]

Answer:

170 W

Explanation:

Applying

P = VI.................... Equation 1

Where P = Power generated in watt, V = Voltage supplied to the circuit, I = Current running through the circuit.

From the question,

Given: V = 17 V, I = 10 A

Substitute these values into equation 1

P = (17×10)

P = 170 Watt.

Hence the power generated is 170 W.

The right option is A. 170 W

3 0

2 years ago

Do thermohaline currents flow vertically or horizontally A. Vertically B. Horizontally

Shkiper50 [21]

B

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Vujkbukbuvuvubkkbbiu

7 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago