Using the same cost and time estimates, consider the time-effectiveness of each engineer.

Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of

Anastasy [175]

Answer:

d. 268 s

Explanation:

Constant Speed Motion

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

$\displaystyle t=\frac{d}{v}$

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

$\displaystyle t=\frac{5}{67}$

t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

8 0

2 years ago

How to find the gradient of a velocity time graph

Sidana [21]

The area-

The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.

Area of light-blue triangle -
The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: area of triangle = 1⁄2 × base × height so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. Area of dark-blue rectangle The width of the rectangle is 6 seconds and the height is 8 meters per second. So the area is 8 × 6 = 48m. Area under the whole graph The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.This is the total area under the distance-time graph. This area represents the distance covered.

7 0

2 years ago

A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m

lbvjy [14]

The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the mass of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the lab cart

$v_{2}_{i}$ : is the initial velocity of the object = 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the lab cart

$v_{2}_{f}$ : is the final velocity of the object

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object will be the same, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

brainly.com/question/2141713?referrer=searchResults

brainly.com/question/2400186?referrer=searchResults

I hope it helps you!

4 0

2 years ago

A real gas will behave most like an ideal gas under conditions of ________.

KengaRu [80]

Answer: high temperature and low pressure

Explanation:

The Ideal Gas equation is:

$P.V=n.R.T$

Where:

$P$ is the pressure of the gas

$V$ is the volume of the gas

$n$ the number of moles of gas

$R=0.0821\frac{L.atm}{mol.K}$ is the gas constant

$T$ is the absolute temperature of the gas in Kelvin

According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them.

Now, real gases can behave approximately to an ideal gas, under the conditions described above and taking into account the following:

When temperature is high a real gas approximates to ideal gas, because the molecules move quickly, preventing the repulsion or attraction forces to take effect. In addition, at low pressures, the volume of molecules is negligible.

4 0

2 years ago

If we heated the ball up and kept the ring room temperature, would the ball be able to fit through the ring? (1 point) Why or wh

Fed [463]

Answer:

no:

Explanation:

it would grow and no longer be able to fit through the loop due to the hot air expanding.

8 0

2 years ago