All categories

3 years ago

The pressure exerted at the bottom of a column of liquid is 30 kPa. The height of the

Physics

1 answer:

ella [17]3 years ago

7 0

Answer:

For example, the pressure acting on a dam at the bottom of a reservoir is ... pressure = height of column × density of the liquid × gravitational field ... The density of water is 1,000 kg/m 3.

You might be interested in

When it gets that hot the evergreens can explode?

Ghella [55]

If you mean the tree, evergreen trees can exploded if theres extreme stress on the trunk

3 0

3 years ago

The charges that are free to move in a metallic conducting wire and that are responsible for the flow of electric current are

Anna007 [38]

Answer:Negatively charged particle called Free Electrons

Explanation:

Current is the flow of charged particles called Free electrons. Electrons are free to move from one atom to another and we call them a sea of de-localized electrons. In absence of any externally applied emf, these electrons are randomly moving but with the onset of emf, these electrons flow in a particular direction.

6 0

3 years ago

A 5 kg block is resting on a ramp inclined at 35 degrees above the horizontal. What is the magnitude of the normal force acting

Mamont248 [21]

I'm pretty sure the answer is b 28n hope helps :)

4 0

3 years ago

II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically st

AveGali [126]

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

$a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}$

Since we start counting at 0s and the ball departs from rest, this is just $a=\frac{v}{t}$

So we can write:

$F=ma=\frac{mv}{t}=\frac{gmv}{gt}$

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

$F=\frac{Wv}{gt}=\frac{v}{gt}W$

Substituting our values:

$F=\frac{(73.14m/s)}{(9.81m/s^2)(30\times10^{-3}s)}W=248.5W N$

Where the average force exerted has been written it terms of the tennis ball's weight W.

8 0

3 years ago

One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what

Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

$F_y=ma_y$ (2)

$F_x=ma_x$ (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

$F_y=W+n=0$

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

$F_y=(755)(-9.81)+n=0$

$n=7406.55 N$ (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

$F_x=F+f=ma_x$

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

$1988+f=m(1.36)$

$f=(755)(1.36)-1988=-961.2 N$

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

$|f|=\mu_k n$

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

$\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13$

4 0

4 years ago