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3 years ago

Why is accuracy important in dodgeball?

1 answer:

8 0

Accuracy and power are the only things that matter in dodge ball. If you are not accurate, then your target does not get hit.

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Determine the surface gravity of Earth. Its mass is 6.0 x 1024 kg and its radius is 6.4 x 106 m.

dexar [7]

Answer:

8.6m/s2

Explanation:

cause im a boss

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3 years ago

Is the color spectrum simply a small segment of the electromagnetic spectrum?

ch4aika [34]

Answer:

yup, u r correct

Explanation:

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3 years ago

Please complete it if you know the answer. "The active region of a transistor is for.........

zubka84 [21]

Answer:

the active region is bound by cutoff region and saturation or power dissipation region.

Explanation:

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3 years ago

The security alarm on a parked car goes off and produces a frequency of 960 Hz. The speed of sound is 343 m/s. As you drive towa

Greeley [361]

Answer: $13.4\ m/s$

Explanation:

Given

The frequency of the source is $f_o=960\ Hz$

Change in frequency is $75\ Hz$

Speed of sound $c=343\ m/s$

Suppose $v$ is the velocity of the observer

Doppler frequency is given by

$f'=f_o\left(\dfrac{c\pm v_o}{c\pm v_s}\right)$

Here, the source is at rest

While approaching source, frequency is

$f_1=f_o\left(\dfrac{c+v}{c}\right)\quad \ldots(i)$

While leaving, frequency is

$f_2=f_o\left(\dfrac{c-v}{c}\right)\quad \ldots(ii)$

The difference in the frequency is

$\Rightarrow f_1-f_2=75\\\\\Rightarrow f_o\left(\dfrac{c+v}{c}\right)-f_o\left(\dfrac{c-v}{c}\right)=75\\\\\Rightarrow f_o\left(\dfrac{2v}{c}\right)=75\\\\\Rightarrow v=\dfrac{75\times 343}{2\times 960}\\\\\Rightarrow v=13.39\approx 13.4\ m/s$

7 0

3 years ago

A nuclear fission power plant has an actual efficiency of 32%. If 0.18 MW of power are produced by the nuclear fission, how much

7nadin3 [17]

Answer:

P₀ = 5.76 x 10⁻² MW

Explanation:

given,

efficiency of the power plant = 32%

Power produced by the nuclear fission = 0.18 MW

the power plant output = ?

using formula of efficiency

$\eta = \dfrac{P_0}{P}$

where P is the power produced in the power plant

P₀ is the power output of the power plant

$\eta = \dfrac{P_0}{P}$

$0.32 = \dfrac{P_0}{0.18}$

P₀ = 0.18 x 0.32

P₀ = 0.0576 MW

P₀ = 5.76 x 10⁻² MW

Power plant output is equal to P₀ = 5.76 x 10⁻² MW

8 0

4 years ago