1. A 2.10 m rope attaches a tire to an overhanging tree limb. A girl swinging on the

How long will it take for a car to travel 200 km if it has an average speed of 55 km/hr?

andreyandreev [35.5K]

Answer: Approximately 3.65 hours

Explanation:

55 km/h x 3.65 hrs = 200.75 Km/h

4 0

3 years ago

Give an example of a force applied to an object that does not change the object's velocity. Why does the object's velocity not c

Xelga [282]

Answer:

A chair at rest on the floor has two forces acting on it its own weight that pulls it downward and the floor pushing upward on the chair, both of these forces are acting on it but the net force is 0, so the chair remains at rest and its velocity stays at 0.

8 0

2 years ago

The freight cars a and b have a mass of 20 mg and 15 mg, respectively. if the cars collide and couple together, what is the velo

igomit [66]

Suppose car A is moving with a velocity Va, and car b with a velocity Vb,

According the principle of conservation of momentum:

Va x Ma + Vb x Mb = (Ma + Mb) V

V = (Va x Ma + Vb x Mb)/(Ma +Mb)

V = speed of cars after coupling

V = (Va x 20 mg + Vb x 15 mg)/(20 mg + 15 mg)

Put in the values of Va and Vb, and get the V

7 0

3 years ago

Read 2 more answers

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

3 0

3 years ago

What happens to light waves at the interface between different media?

nevsk [136]

Answer:

Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.

Explanation:

Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.

Hope this helps.

6 0

3 years ago