A 5,400 W motor is used to do work. If the motor is used for 640 s, about how much work could it do?

Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

$l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h$ #The speed toward each other.

$\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h$

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0

2 years ago

A 50.0-kg block is being pulled up a 16.0Â° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic

Drupady [299]

When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N

When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:

Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²

8 0

3 years ago

Passenger side, rear-view mirrors on automobiles are generally ____.

Lesechka [4]

I believe it’s C. Plane .

8 0

3 years ago

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug

telo118 [61]

Answer:

a) $V_{x}=3.72m/s$ , b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

$y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}$

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

$0=y_{0}+\frac{1}{2}a_{y}t^{2}$

so we can solve for t, so we get:

$t=\sqrt{\frac{-(2)y_{0}}{a}}$

so now we can substitute the known values, so we get:

$t=\sqrt{\frac{-(2)(1.42)}{-9.8}}$

which yields:

t=0.538s

So we can use this value to find the velocity in x:

$V_{x}=\frac{x}{t}$

When substituting we get:

$V_{x}=\frac{2m}{0.538s}$

which yields:

$V_{x}=3.72m/s$

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

$\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

We know that $V_{0}$ is zero, so we can simplify the expression:

$\Delta y=\frac{V_{yf}^{2}}{2a}$

So we can solve the equation for $V_{yf}^{2}$ so we get:

$V_{yf}=\sqrt{2\Delta y a}$

and when substituting the known values we get:

$V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}$

which yields:

$V_{yf}=-5.28m/s$

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

$tan \theta =\frac{V_{y}}{V_{x}}$

when solving for theta we get:

$\theta = tan^{-1}(\frac{V_{y}}{V_{x}})$

We can substitute so we get:

$\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})$

which yields:

$\theta = -54.83^{o}$

7 0

3 years ago

. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa radius 2cm. a. What is the speed of the water on the

lana66690 [7]

Answer:

Outlet Velocity = 192 m/s

Outlet Pressure = 510 kPa

Explanation:

Givens:

Inlet Velocity, V₁ = 12 m/s

Inlet Pressure, P₁ = 600 kPa = 600,000 Pa

Inlet Radius r₁ = 0.5 cm

Outlet Velocity, V₂ = not given (we are asked to find this)

Outlet Pressure, P₂ = not given (we are asked to find this)

Outlet Radius, r₂ = 0.5 cm

From these, we can find the following:

Inlet Area, A₁ = π (r₁)² = π(2)² = 4π cm²

Outlet Area, A₂ = π (r₂)² = π(0.5)² = 0.25π cm²

Assuming that water is incompressible, we can reason that within the same given time, the amount of volume of water entering the inlet must equal the volume of water exiting the outlet. Hence by the continuity equation (i.e. conservation of mass)

Inlet Volume flow rate = Outlet Volume flow rate

(recall that Volume flow rate in a pipe is given by Velocity x Cross Section Area), Hence the equation becomes

V₁ x A₁ = V₂ x A₂ (substituting the values that we know from above)

12 x 4π = V₂ x 0.25π (we don't have to change all to SI units because the conversion factors on the left will cancel out the conversion factors on the right).

V₂ = (12 x 4π) / (0.25π)

V₂ = 192 m/s (Answer)

For Part B, if we assume a closed ideal system (control volume method), we can simply apply the energy equation (i.e Bernoulli's equation)

P₁ + (1/2)ρV₁ + ρgh₁ = P₂ + (1/2)ρV₂ + ρgh₂

Because the pipe is horizontal, there is no difference between h₁ and h₂, hence we can neglet this term:

P₁ + (1/2)ρV₁ = P₂ + (1/2)ρV₂ (rearranging)

P₂ = P₁ + (1/2)ρV₁ - (1/2)ρV₂

= P₁ + (1/2)ρ (V₁-V₂)

Assuming that the density of water is approx, ρ = 1000 kg/m³

P₂ = 600,000 + (1/2)(1000) (12-192)

= 600,000 + ( -90,000)

= 510,000 Pa

= 510 kPa (Answer)

8 0

3 years ago