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2 years ago

Which statement best summarizes the theory of evolution by natural selection?

Physics

2 answers:

prisoha [69]2 years ago

8 0

Answer:

c is the one that makes the most sense

Explanation:

eimsori [14]2 years ago

6 0

I’m pretty sure it’s c

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What potential difference is needed to give a helium nucleus (q=2e) 85.0 kev of kinetic energy?

Vilka [71]

The kinetic energy K given to the helium nucleus is equal to its potential energy, which is
$E=q \Delta V$
where q=2e is the charge of the helium nucleus, and $\Delta V$ is the potential difference applied to it.
Since we know the kinetic energy, we have
$E=K=85~keV=q \Delta V$
and from this we can find the potential difference:
$\Delta V = \frac{K}{q}= \frac{85~keV}{2e}=42.5~kV$

6 0

2 years ago

Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8

Kamila [148]

Answer:

34.6 m/s

Explanation:

From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg

Final mass will be 31.5+25.9=57.4 kg

From formula of momentum

M1v1=m2v2

Making v2 the subject of the formula then

$V2=\frac {M1v1}{m2}$

Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then

$V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s$

4 0

3 years ago

35 miles North 6 feet down 136 MB 85 km SE

pantera1 [17]

The answer is 133 Se

4 0

3 years ago

Read 2 more answers

Need answers quick !! will give brainliest !!

stiv31 [10]

Answer:

use the bowl of water because Earth's magnetic field is relatively weak. Allowing it to float freely on the water, allows the magnetized needle to freely react to Earth's magnetic field, causing it to align North to South. If you watched closely, the same end of the needle should always point to the North

Explanation:

6 0

2 years ago

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

$Z_{R}$ = R

R = resistance

$Z_{L}$ = jωL

ω = voltage source angular frequency, L = inductance

$Z_{C}$ = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

$Z_{R}$ = 217Ω

$Z_{L}$ = j(220)(0.875) = j192.5Ω

$Z_{C}$ = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$ + $Z_{C}$

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

$v_{R}$ = I× $Z_{R}$

$v_{L}$ = I× $Z_{L}$

$v_{C}$ = I× $Z_{C}$

Given values:

I = (0.0569∠1.147+220t)A

$Z_{R}$ = 217Ω = (217∠0)Ω

$Z_{L}$ = j192.5Ω = (192.5∠π/2)Ω

$Z_{C}$ = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

$v_{R}$ = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

$v_{L}$ = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

$v_{C}$ = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

$v_{R}$ = (12.35∠1.147+220t)V

$v_{L}$ = (10.95∠2.718+220t)V

$v_{C}$ = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

$v_{R}$ = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

$v_{L}$ = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

$v_{C}$ = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0

3 years ago