1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
sineoko [7]
3 years ago
15

The ocean’s surface temperatures vary with time of year and a. latitude. c. seasons. b. longitude. d. moon phases.

Physics
2 answers:
Natali5045456 [20]3 years ago
5 0

answer is A your welcome

nignag [31]3 years ago
4 0

Answer: a. Latitude

Explanation:

The average temperature of ocean's surface is about 17°C. The temperature varies with the seasons and location of the ocean. It means the time of the year -whether it is winters , summers, autumn or spring and location - in which zone it lies.

In winters the temperature dips down. The arctic ocean and Antarctic ocean have extremely low temperatures. On the other ocean lying near equator have higher temperature.

Thus, the ocean’s surface temperatures vary with time of year and<u> a. latitude. </u>

You might be interested in
Plz help ASAP I'll mark as brainliest ​
gogolik [260]

Hi there!

1.

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

2.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

3.

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

4.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

m = 0.8 kg

8 0
3 years ago
suppose you want to determine the surface area of this sugar cube. it has edges that are each 2 cm long. if you cut the cube in
ivolga24 [154]

Answer:

Half: 6 cm^2    Whole: 12 cm^2

Explanation:

First, we know that the edges of the cube are 2 cm long. So there are 6 faces on a cube. We do 2x6=12 cm^2 as our total surface area. Then it asks for each half. So you would divide it by 2 and get 6 cm^2 as your half.

5 0
3 years ago
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
MatroZZZ [7]

Answer:

The wavelength is \lambda_2 =  534 *10^{-9} \ m

Explanation:

From the question we are told that

   The wavelength of the first light is  \lambda _ 1 =  587 \ nm

    The order of the first light that is being considered is  m_1  =  10

     The order of the second light that is being considered is  m_2  =  11

Generally the distance between the fringes for the first light is mathematically represented as

      y_1 =  \frac{ m_1  * \lambda_1 *  D}{d}

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         y_2 =  \frac{ m_2  * \lambda_2 *  D}{d}

Now given that both of the light are passed through the same double slit

       \frac{y_1}{y_2}  =  \frac{\frac{m_1 *  \lambda_1 * D}{d}  }{\frac{m_2 *  \lambda_2 * D}{d}  } = 1

=>    \frac{ m_1 *  \lambda _1  }{ m_2  *  \lambda_2} =  1

=>     \lambda_2 =  \frac{m_1 *  \lambda_1}{m_2}

=>    \lambda_2 =  \frac{10  *   587 *10^{-9}}{11}

=>   \lambda_2 =  534 *10^{-9} \ m

4 0
3 years ago
The velocity of a 48.0 g shell leaving a 2.95 kg rifle is 391. m/s. What is the recoil velocity of the rifle?
Monica [59]

Hi there!

\large\boxed{  -6.36m/s}

Use the equation:

v_{2} = -\frac{m_{1}}{m_{2}} v_{1}

Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:

v2 = ?

m1 = 0.048 kg (converted)

m2 = 2.95

v1 = 391

v_{2} = -\frac{0.048}{2.95} *391

v_{2} = -6.36m/s

8 0
3 years ago
A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a s
Natalka [10]

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

KE_i = KE_f

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

Here,

m = Mass

V = Velocity

Replacing,

KE_f = \frac{1}{2} (12000)(11)^2

KE_f = 72600J

Therefore the  final kinetic energy of the two car system is 72.6kJ

8 0
3 years ago
Other questions:
  • The signal to a speaker causes .5 amps of current. The signal level is increased to 2.5 amps. What is the gain, rounded to the n
    15·1 answer
  • How did rutherfords discovery effect the model of the atom
    11·1 answer
  • A force is directly proportional to what ?
    7·1 answer
  • A technician A says that a voltmeter can be used to monitor the output signal of an optical sensor. Technician B says that optic
    5·1 answer
  • A 0.50 kg object is at rest. A 2.88 N force to the right acts on the object during a time interval of 1.48 s. a) What is the vel
    6·1 answer
  • Which of the following is a chemical property of matter?
    11·2 answers
  • Part B<br> Does the size of the particles change as the substance changes state?
    10·1 answer
  • En las olimpiadas del 2012 del colegio villapalos maria gano la carrera de los 100 m en 10,56 s y la de 200 m en 22,34 s ¿en cua
    5·1 answer
  • Which trait shared by dolphins and bats possibly lead to the evolution of echolocation in these two animal groups? the need to m
    10·1 answer
  • Which of the following are examples of some of the environmental costs of technological development? (Choose all that apply)
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!