All categories

3 years ago

Please help! I dont get it!

1 answer:

5 0

I know its late but its <span>B.Temperature plays a greater role in thermohaline circulation than salinity does, since it is the temperature of the ocean interacting with the atmosphere that causes deep sea currents.
</span>

You might be interested in

You're driving down the highway late one night at 22 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim

sleet_krkn [62]

speed of the car is given on the road

$v = 22 m/s$

reaction time is given as

$t = 0.50 s$

now we can find the distance that it will move during this reaction time

$d_1 = 22* 0.50 = 11 m$

now the deceleration of the car is 10 m/s^2

so the distance that it will move before stop is given by

$v_f^2 - v_i^2 = 2 a d$

$0^2 - 22^2 = 2*(-10)*d$

$d = 24.2 m$

so the total distance that it require to stop is given as

$d = 24.2 + 11 = 35.2 m$

while the deer is standing at distance 38 m

so the car will stop 2.8 m before the position of deer

3 0

3 years ago

What are some examples of pressure

aniked [119]

Squeezing a ball is one

3 0

3 years ago

Based on the soil texture triangle, how much slit is presented in silty clay?

morpeh [17]

B) 20-40 percent
I hope this helps

4 0

3 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric f

musickatia [10]

Answer:

Er = 231.76 V/m, 27.23° to the left of E1

Explanation:

To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:

E1:

$E_1_x = 0V/m\\E_1_y=100V/m$

E2:

Keep in mind that the x component of electric field E2 is directed to the left.

$E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m$

∑x: $E_1_x+E_2_x = 0V/m - 106.07V/m = -106.07V/m$

∑y: $E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m$

The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.

$||E_r||= \sqrt{(-106.07V/m)^2+(206.07V/m)^2} = 231.76 V/m\\\\\alpha = arctan(\frac{206.7 V/m}{-106.07 V/m}) = 117.24degrees$

or 27.23° to the left of E1.

8 0

3 years ago