____ are streamlike movements of water in the ocean.(what is blank)

Which of the following is an example of a force<br> inertia<br> push <br> pull

Whitepunk [10]

A force can be considered a push or pull

hope this helps :)

8 0

3 years ago

Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their

geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

$m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}$

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

$1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\$

The magnitude of the final velocity of the wreck can be found as:

$v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72$

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

$\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}$

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0

3 years ago

Now, using your mass (in kg), and the figures for g (in the table below), you can calculate your weight on other planets.

Licemer1 [7]

Answer:

1) Weight on Mercury

$F =W=mg=68.11 \times 3.61 m.s^{-2}$

Explanation:

do the same to the rest and use your calculator to find the weight in N.

3 0

3 years ago

The Fitness Gram push-up test is a measure of

Morgarella [4.7K]

Answer:

The answer is B.

Explanation:

I meant B. not C so sorry

6 0

3 years ago

It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J of energy whe

boyakko [2]

Answer:

10.4L of water is expended when 1L of crude oil is burned.

Explanation:

This problem requires us to calculate the volume of water that must be expended to absorb the amount of energy released from the burning of 1.00L of crude oil.

In going from water at 18.5°C to steam at 285°C, the water undergoes 3 stages:

Absorption of heat from 18.5°C to 100°C >> phase change to vapour at 100°C >> heating from 100°C to steam at 285°C

Given Cw = 4186 J/kg°C and Cs = 2020 J/kg°C. The latent heat of of vaporization of water Lv = 2.256 ×10⁶ J/kg

The total heat absorbed in the process per kilogram

H = Cw × (100 – 18.5) + Lv + Cs × (285 – 100)

H = 4186 × 81.5 + 2. 256 ×10⁶ + 2020× 185

= 2.696 ×10⁶ J/kg

The amount of heat in J/L of water needed can be calculated as follows:

Density of water = 1000kg/m³ =

1000 kg/m³ × 1m³/1000L = 1kg/L (Basically conversion of density in kg/m³ to kg/L)

Let the volume of water needed be V litres.

Then the mass of water that must be expended = Density × Volume

= 1kg/L × V L = Vkg

The heat that would be absorbed by the water when 1L of crude oil is burned is V×H

= 2.696×10⁶ × V

This is also equal to 2.80×10⁷ J of energy (given).

So,

2.696×10⁶V = 2.8×10⁷

V = (2.80×10⁷)/(2.696×10⁶) = 10.4L of water.

3 0

3 years ago

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