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iren [92.7K]
3 years ago
9

( Apply Concepts ) Nome , Alaska, lies at 64°N latitude. San Diego , California , lies at 32°N latitude. Which city receives mor

e sunlight?
Physics
1 answer:
tatyana61 [14]3 years ago
6 0
If your asking about the total amount of time the sun is visible, it will be equal over the course of a year. In Nome, Alaska, it will be daylight almost 24 hours a day in the summer and be dark almost 24 hours a day in the winter. In San Diego they are much closer to the equator and will receive sunlight roughly the same amount each day all year long. In the end these times will add up to the same amount.

However if you are asking about intensity of sun light received, San Diego will get considerably more since its near the equator and its getting almost direct sunlight. In Nome, the sunlight is hitting the earth at an angle so fewer rays of light are hitting the ground per square foot.
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An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary
Talja [164]

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

F_{g} - F_{d} - T = 0

7000 - 1800 - T = 0

T = 5200 N

T = 5.2 x 10³ N

Part B)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

T  - F_{g} - F_{d} = 0

T - 7000 - 1800  = 0

T = 8800 N

T = 8.8 x 10³ N

4 0
3 years ago
A cubic sample of a new kind of artificial tissue is subject to an increase in pressure of 160 kPa which results in a reduction
grin007 [14]

Answer:

0.82 MPa

Explanation:

the change in pressure 'σ'=160kPa

K= σ/∈Ф_v => σ/3∈Ф_L

K= 160/(3 x 0.065)

K=820 kPA=0.82 MPa

Thus,the bulk modulus of the tissue 'K' is 0.82 MPa

3 0
3 years ago
Help please! Will give brainly! and thanks.
kipiarov [429]
<h2>Answer:</h2>

All the energy sources are correctly matched with their category.

Explanation:

Renewable energy sources:

These are energy sources which can be replenished as they don't in involves the irreversible phase change.

These resources can never be ended as they can be used again and again.

Wind, geothermal, biomass, bio gas are example of renewable energy sources.

Non renewable energy sources:

These are the energy source which can never be replenished after one time use. They undergo the chemical irreversible change.

These sources are lacking with the passage of time because they can never be reused.

Oil, gas, coal and natural gas are examples.

7 0
3 years ago
Suppose a car manufacturer tested its cars for front-en4 collisions by hauling them up on a crane and dropping then; from a cert
Brrunno [24]

Answer:

a

Generally from third equation of motion we have that

v^2 =  u^2 + 2a[s_i - s_f]

Here v is the final speed of the car

u is the initial speed of the car which is zero

s_i is the initial position of the car which is certain height H

s_i is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

v^2 = 0 + 2g[H - 0]

=> v  =  \sqrt{ 2 g H}

b

H  =  9.86 \  m

Explanation:

Generally from third equation of motion we have that

v^2 =  u^2 + 2a[s_i - s_f]

Here v is the final speed of the car

u is the initial speed of the car which is zero

s_i is the initial position of the car which is certain height H

s_i is the final position of the car which is zero meters (i.e the ground)

a is the acceleration due to gravity which is g

So

v^2 = 0 + 2g[H - 0]

=> v  =  \sqrt{ 2 g H}

When v  = 50 \  km/h = \frac{50 *1000}{3600} = 13.9 \  m/s we have that

13.9  =  \sqrt{ 2 g H}

=> H  =  \frac{13.9^2}{2 *  9.8}

=> H  =  9.86 \  m

6 0
3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
3 years ago
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