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Lunna [17]
2 years ago
15

A dolphin swims due east for 1.90 km, then swims 7.20 km in the direction south of west. What are the magnitude and direction of

the vector that will take the dolphin back to its starting point? (Enter the magnitude in km and the direction in degrees north of east.)
Physics
1 answer:
kykrilka [37]2 years ago
3 0

Answer:

magnitude = 7.446 km, direction = 75.22° north of east

Explanation:

From the questions,

To get the the magnitude of the resultant vector we use Pythagoras theorem

a² = b²+c²

From the diagram,

y² = 1.9²+7.2²

y² = 55.45

y = √(55.45)

y = 7.446 km.

The direction of the dolphin is given as,

θ = tan⁻¹(7.2/1.9)

θ = tan⁻¹(3.7895)

θ = 75.22° north of east

Hence the magnitude of the resultant vector = 7.446 km, and it direction is 75.22° north of east

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For which of the following problems would a scientist most likely use carbon-14?
spayn [35]

Answer:

To calculate the age of a piece of bone

Explanation:

Carbon 14 is an isotope of carbon that is unstable and decays into Nitrogen 14 by emitting an electron. The decay rate of radioactive material is  normally expressed in terms of its "half-life" (the time required by half the radioactive nuclei of a sample to undergo radioactive decay). The nice thing about carbon 14 is that its "half-life" is about 5730 years, which gives a nice reference to measure the age of fossils that are some thousand years old.

Carbon 14 dating is used to determine the age of objects that have been living organisms long ago. They measure how much carbon 14 is left in the object after years of decaying without having exchange with the ambient via respiration, ingestion, absorption, etc. and therefore having renewed the normal amount of carbon 14 that is in the ambient.

A rock is not a living organism, so its age cannot be determined by carbon 14 dating.

3 0
2 years ago
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th
Anton [14]

Answer:

Option B, 93 cm

Explanation:

An diagram of the seed's motion is attached to this solution.

This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)

(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2

R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.

QED!

6 0
1 year ago
Read 2 more answers
A capacitor, C1, consists of two parallel circular plates with radius R and separation of d. A second capacitor, C2, consists of
qaws [65]

Answer:

\frac{E_1}{E_2}= 4

Explanation:

Capacitance C is given by

C= \frac{\epsilon_0A}{d}

A= area of capacitor cross section

d= distance

therefore,

C_1= \frac{\epsilon_0A_1}{d_1}

A_1= πR^2

d_1= d

C_2= \frac{\epsilon_0A_2}{d_2}

A_= π(2R)^2

d_2 = 2d

q= \frac{\epsilon_0A_1}{d_1}V_1

threfore

V_1= \frac{qd_1}{\epsilon_0A_1}

and

V_2= \frac{qd_2}{\epsilon_0A_2}

also we know that E= V/d

⇒\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}

⇒\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}

= A_1/A_2= \frac{4R^2}{R^2}=4

therefore,

\frac{E_1}{E_2}= 4

5 0
1 year ago
A hose has been clamped so that the area at the clamp is only one quarter the area of the rest of the hose. When we ignore the v
Bad White [126]

Answer:

1

Explanation:

A hose has been clamped so that the area at the clamp is only one quarter the area of the rest of the hose. When we ignore the viscosity of water, the ratio of the volume of water delivered per unit time when the clamp is on to the volume of water delivered per unit time without the clamp is 1 as continuity says the same amount of water must flow out

4 0
2 years ago
A fireman clings to a vertical ladder and directs the nozzle of a hose horizontally toward a burning building. The rate of water
Aleks [24]

Answer:

A torque of 102.5375 Nm must be exerted by the fireman

Explanation:

Given:

The rate of water flow = 6.31 kg/s

The speed of nozzle  = 12.5 m/s

Now, from the Newton's second law we have  

The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction

thus we have,

F = 6.31 kg/s x 12.5m/s

or

F = 78.875 N  

Now,

The torque (T) exerted by water force about the fireman's will be

T = (F x d)

or

T = 78.875 N x 1.30 m

T = 102.5375 Nm

hence,

<u>A torque of 102.5375 Nm must be exerted by the fireman</u>

4 0
2 years ago
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