The only ocean surface current that is not driven by winds is the

Physics

1 answer:

Setler79 [48]3 years ago

4 0

Answer:

The only ocean surface current that is not driven by winds is “global conveyor belt.

This current is caused by changes in temperatures, salinity and density. The densest waters have to descend to the bottom of the sea moving the less dense towards the surface. The density of water depends on the degree of salinity and temperature. Colder waters are denser and go to the seabed, displacing the other hotter waters to the surface.

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A book is sitting on a table. Jackie is pushing the book with a force of 9 N to the right. If the force of friction is 4 N to th

Free_Kalibri [48]

Answer:

5 N right

Explanation:

Fx = 9N-4N

Fx = 5N

Since we can define the x and y axis. We have x to the right as positive.

8 0

3 years ago

A 34n force is applied to a 213kg mass how much does the mass accelerate

motikmotik

We Know, F = m*a
Here, F = 34 N
m = 213 Kg

Substitute their values in the equation,
34 = 213 * a
a = 34/213
a = 0.159 m/s²

So, your final answer & the acceleration of the object would be 0.159 m/s²

Hope this helps!

3 0

3 years ago

It takes you 5 min to walk with an average velocity of .75 m/s to the north from the parking lot to the entrance of the amusemen

skad [1K]

A displacement is a vector quantity that takes into account the shortest distance from the starting point to the endpoint.

The given above gave a time interval in minutes which needs to be converted to seconds. Given that each minute is 60 seconds, 5 minutes equal 300 seconds. To determine the distance, multiply time with speed. The product is 225 m.

Thus, the displacement is 225 m.

3 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

Time taken to reach 1180 m is 11.29 seconds

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$