Which of the following is a major ocean on Earth?

Physics

2 answers:

drek231 [11]3 years ago

7 0

The answer is d, the Indian Ocean

yarga [219]3 years ago

3 0

The Indian Ocean.
The rest of your options are not oceans, they are bodies of water. (e.g: lakes,rivers,gulfs)

You might be interested in

3. Two Metra trains approach each other on separate but parallel tracks. Train A has a speed of 90 km/ hr, train B has a speed o

Elden [556K]

Answer:he trains take 57.4 s to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

The relative velocity of the train A with respect to B is given by,

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

Substitute 2.71 km for d and 170 km/h for

Express the time in seconds.

Thus, the trains cross each other in 57.4 s.

Explanation:

8 0

3 years ago

If the graph represents speed, what is measured on the x and y axes? You need to specifically identify which axis is which.

Harrizon [31]

X axis - time
Y - distance

4 0

3 years ago

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

joja [24]

Answer:

Approximately $9.7\; \rm m \cdot s^{-2}$ .

Explanation:

Assuming that there is no other force on this vehicle, the $16000\; \rm N$ force from the road would be the only force on this vehicle. The net force would then be equal to this $16000\; \rm N\!$ force. The size of the net force would be $16000\; \rm N\!\!$ .

Let $m$ denote the mass of this vehicle and let $\Sigma F$ denote the net force on this vehicle.

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, $a$ , would be:

$\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}$ .

For this vehicle, $\Sigma F = 16000\; \rm N$ whereas $m = 1650\; \rm kg$ . The acceleration of this vehicle would be:

$\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}$ .