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aleksandrvk [35]
3 years ago
15

What is the effect on climate from these factors??

Physics
1 answer:
fenix001 [56]3 years ago
7 0
The effect on the climate is that the climate goes up
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The current theory of the structure of the
Mariana [72]

Answers:

a) 2.82(10)^{21} kg

b) 1410 J

c) 36.62 m/s

Explanation:

<h3>a) Mass of the continent</h3>

Density \rho  is defined as a relation between mass m and volume V:

\rho=\frac{m}{V} (1)

Where:

\rho=2720 kg/m^{3} is the average density of the continent

m is the mass of the continent

V is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}

Finding the mass:

m=\rho V (2)

m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3}) (3)

m=2.82(10)^{21} kg (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy K is given by the following equation:

K=\frac{1}{2}mv^{2} (5)

Where:

m=2.82(10)^{21} kg is the mass of the continent

v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s is the velocity of the continent

K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2} (6)

K=1410 J (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass m=77 kg and the same kinetic energy as that of the continent 1413 J, we can find its velocity by isolating v from (5):

v=\sqrt{\frac{2 K}{m}} (6)

v=\sqrt{\frac{2 (1413 J)}{77 kg}}

Finally:

v=36.62 m/s This is the speed of the jogger

5 0
3 years ago
regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac
solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum p_{i} before the collision must be equal to the final momentum p_{f} after the collision:

p_{i}=p_{f} (1)

Being:

p_{i}=MV_{i}+mU_{i}

p_{f}=(M+m) V

Where:

M=480 g \frac{1 kg}{1000 g}=0.48 kg the mas of the peregrine falcon

V_{i}=45 m/s the initial speed of the falcon

m=240 g \frac{1 kg}{1000 g}=0.24 kg is the mass of the pigeon

U_{i}=0 m/s the initial speed of the pigeon (at rest)

V the final speed of the system falcon-pigeon

Then:

MV_{i}+mU_{i}=(M+m) V (2)

Finding V:

V=\frac{MV_{i}}{M+m} (3)

V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg} (4)

V=30 m/s (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

F=\frac{\Delta p}{\Delta t} (6)

Where:

F is the force exerted on the pigeon

\Delta t=0.015 s is the time

\Delta p is the pigeon's change in momentum

Then:

\Delta p=p_{f}-p_{i}=mV-mU_{i} (7)

\Delta p=mV (8) Since U_{i}=0

Substituting (8) in (6):

F=\frac{mV}{\Delta t} (9)

F=\frac{(0.24 kg)(30 m/s)}{0.015 s} (10)

Finally:

F=480 N

7 0
3 years ago
5. Find the mass of a car that is traveling at a velocity of 35 m/s West.
sattari [20]

Answer:

m = 9795.9 kg

Explanation:

v = 35 m/s

KE = 6,000,000 J

Plug those values into the following equation:

KE = \frac{1}{2} mv^{2}

6,000,000 J = (1/2)(35^2)m

---> m = 9795.9 kg

3 0
3 years ago
A plane flies from base camp to lake a, 200 km away in the direction 20.0° north of east. after dropping off supplies it flies t
Liono4ka [1.6K]

Distance of lake a is 200 km at 20 degree north of east

distance between lake a and b is 230 km at 30 degree west of north

now the distance between base and lake b is given as

d = d_1 + d_2

given that

d_1 = 200 cos20 i + 200 sin20 j

d_1 = 187.94 i + 68.4 j

d_2 = -230 sin30 i + 230 cos30 j

d_2 = -115 i + 199.2 j

now the total distance is

d = (187.94 - 115)i + (199.2 + 68.4)j

d = 72.94 i + 267.6 j

now the magnitude of the distance is given as

d = \sqrt{72.94^2 + 267.6^2}

d = 277.4

also the direction is given as

\theta = tan^{-1}\frac{267.6}{72.94}

\theta = 74.7 degree

<em>so it is 277.4 km at 74.7 degree North of East</em>

8 0
3 years ago
describe the time of day that an early explorer might have planned to enter a harbor and when he might have planned to leave
inessss [21]
He would try to enter as the tide is rising, and leave as the tide is falling. Those things happen at all different times of day during a month.
4 0
3 years ago
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