Scientific theories have strod the test of time and will not change

Water falls without splashing at a rate of 0.370 l/s from a height of 2.90 m into a 0.690-kg bucket on a scale. if the bucket is

raketka [301]

<span>R = rate of flow = 0.370 L/s H = height = 2.9 m T= time = 3.9 s V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2 G value = 9.8 m/s2 Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N Wa = weight of accumulated water after 3.9 s Fi = force of impact of water on the bucket S = reading on the scale = Wa + Wb + Fi mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg Therefore, Wa = 1.443 x 9.8 = 14.1414 N Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt) = 0.37 x 7.539 = 2.78943 N S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>

8 0

3 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago

A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.

alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

$F_g = F_n + F_a$

given that

$F_g = mg$

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

$F_n = 150 N$

now from above equation

$30*10 = 150 + F_a$

$F_a = 300 - 150 = 150 N$

So force applied by the person must be 150 N

7 0

3 years ago

A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb

densk [106]

Voltage=V=120V
Power=P=60W

Current=I

$\\ \rm\rightarrowtail I=\dfrac{P}{V}=\dfrac{60}{120}=\dfrac{1}{2}=0.5A$

8 0

2 years ago

What happens to the object if the line crosses the x-axis from the positive portion of a velocity versus time diagram?

stealth61 [152]

Answer:

A velocity time graph shows the change of velocity of an object with respect ot time. If the slope of the graph is increasing in the postive region, it means that the velocity is changing, if the slope is decreasing, it means the the velocity is decreasing, but the object is moving in the same direction (positve direction).

If this slope intersects the graph at x-axis, it means that the body has 0 velocity and has become still. After that, if the line enters in the negative region, it means that its velocity is started to increases again, but the body is movinging in the opposite direction (negative direction)

7 0

3 years ago