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ICE Princess25 [194]
3 years ago
9

The map below shows major ocean currents in the North Atlantic and North Pacific Oceans. In general, currents flowing toward the

Physics
2 answers:
Lina20 [59]3 years ago
4 0

Answer:

d

Explanation:

9966 [12]3 years ago
4 0

Answer:

The United States West Coast. Hope this helps :)

Explanation:

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How many planets are made up of gas
Vadim26 [7]

Answer:

The four gas giants in our solar system are Neptune, Uranus, Saturn, and Jupiter. These are also called the Jovian planets. "Jovian planet" refers to the Roman god Jupiter and was intended to indicate that all of these planets were similar to Jupiter.

Explanation:

i hope this helps

7 0
2 years ago
What is the energy in joules of a photon with a frequency of 3.16e 12 s-1?
erica [24]
We have: Energy(E) = Planck's constant(h) × Frequency(∨)
Here, Planck's constant(h) = 6.626 × 10⁻³⁴ J/s
Frequency (∨) = 3.16 × 10¹² /s

Substitute the values into the expression:
E = (6.626 × 10⁻³⁴)(3.16 × 10¹²) J
E = 2.093 × 10⁻²¹ Joules

In short, Your Final answer would be 2.093 × 10⁻²¹ J

Hope this helps!
5 0
3 years ago
Read 2 more answers
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
Which of the following materials could have been found in the giant cloud that formed the solar system?
iren2701 [21]
Liquid and solid water were not in the giant gas cloudr
6 0
3 years ago
Read 2 more answers
A motorboat accelerates uniformly from a velocity of 6.5m/s to the west to a velocity of 1.5m/s to the west. if its accelerate w
grigory [225]
A motorboat accelerates uniformly from a velocity of 6.5m/s to the west to a velocity of 1.5m/s to the west. if its accelerate was 2.7m/s2 to the east , how far did it travel during the accelration? Give your answer in units of kilometers per hour/sec. To find the acceleration of the car we have to <span>
1.   First determine the suitable formula for this word problem.
Which is a.   A=vf-vi/t</span> which will be  
Given are: Vi= 6.5 m/s Vf= 1.5 m/s a= 2.7 m/sec2 t=1.85s 
 Solution: <span>
x = v0t + ½at2</span>
<span>x = <span>16.645375 m </span></span>    



7 0
3 years ago
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