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4 years ago

Purse-seine fishing is known to be the safest fishing method for protecting dolphins.

Physics

2 answers:

Ede4ka [16]4 years ago

5 0

The answer is most definitely false

Citrus2011 [14]4 years ago

4 0

<span>Purse-seine fishing is known to be the safest fishing method for protecting dolphins.

This is false. This type of fishing </span><span>method has been most blamed for the unintentional death of dolphins.</span>

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A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10

sashaice [31]

Answer:

Time interval;Δt ≈ 37 seconds

Explanation:

We are given;

Angular deceleration;α = -1.6 rad/s²

Initial angular velocity;ω_i = 59 rad/s

Final angular velocity;ω_f = 0 rad/s

Now, the formula to calculate the acceleration would be gotten from;

α = Change in angular velocity/time interval

Thus; α = Δω/Δt = (ω_f - ω_i)/Δt

So, α = (ω_f - ω_i)/Δt

Making Δt the subject, we have;

Δt = (ω_f - ω_i)/α

Plugging in the relevant values to obtain;

Δt = (0 - 59)/(-1.6)

Δt = -59/-1.6

Δt = 36.875 seconds ≈ 37 seconds

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4 years ago

How much more matter is in a golf ball than a ping pong?

melisa1 [442]

Explanation:

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3 years ago

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the average intensity

umka21 [38]

Answer: $29.85\ W/m^2$

Explanation:

Given

Power $P=60\ W$

Distance from the light source $r=0.4\ m$

Intensity is given by

$I=\dfrac{P}{4\pi r^2}$

Inserting values

$\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2$

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3 years ago

Read 2 more answers

The dosage of technetium-99m for a lung scan is 20. μCi /kg of body mass. How many millicuries of technetium-99m should be given

expeople1 [14]

Explanation:

Below is an attachment containing the solution.

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3 years ago

2. Two projectiles thrown from the same point at angles 0,=60° and 02=30° with the horizontal,

Liono4ka [1.6K]

Answer:

The answer is below

Explanation:

The maximum height (h) of a projectile with an initial velocity of u, acceleration due to gravity g and at an angle θ with the horizontal is given as:

$h=\frac{u^2sin^2\theta}{2g}$

Given that the two projectile has the same height.

$For\ the\ first\ projectile\ with\ an\ angle\ \60^0 \ with\ the\ horizontal\ and initial\ velocity\ u_1:\\\\h=\frac{u_1^2sin^260}{2g} =\frac{0.75u_1^2}{2g} \\\\For\ the\ second\ projectile\ with\ an\ angle\ \30 \ with\ the\ horizontal\ and initial\ velocity\ u_2:\\\\h=\frac{u_2^2sin^230}{2g} =\frac{0.25u_2^2}{2g}\\\\\frac{0.25u_2^2}{2g}=\frac{0.75u_1^2}{2g}\\\\0.25u_2^2=0.75u_1^2\\\\\frac{u_1^2}{u_2^2} =\frac{0.25}{0.75} \\\\\frac{u_1^2}{u_2^2}=\frac{1}{3} \\$ $\\\frac{u_1}{u_2}=\sqrt\frac{1}{3}$

8 0

3 years ago