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3 years ago

The spinning of earth on its axis causes the apparent rising and setting of the

Physics

2 answers:

Alenkasestr [34]3 years ago

7 0

The sun

Hope this helps

rodikova [14]3 years ago

4 0

Sun, moon, and some stars

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Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of

Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) <em>High and low pressures in kilopascals</em>:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{high} = 15.999\,kPa$

$p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{low} = 10.666\,kPa$

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) <em>High and low pressures in pounds per square inch</em>:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}$

$p_{high} = 2.320\,psi$

$p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}$

$p_{low} = 1.547\,psi$

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) <em>High and low pressures in meter water column in meters water column</em>:

We can calculate the equivalent water column of a mercury column by the following relation:

$\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}$

$h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg}$ (Eq. 1)

Where:

$\rho_{w}$ , $\rho_{Hg}$ - Densities of water and mercury, measured in kilograms per cubic meter.

$h_{w}$ , $h_{Hg}$ - Heights of water and mercury columns, measured in meters.

If we know that $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , $h_{Hg, high} = 0.120\,m$ and $h_{Hg, low} = 0.080\,m$ , then we get that:

$h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m$

$h_{w, high} = 1.632\,m$

$h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m$

$h_{w, low} = 1.088\,m$

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

4 0

3 years ago

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike

zysi [14]

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

$s=h$ (the vertical displacement is the height of the counter)

$u=0$ (the initial vertical velocity of the cup is zero)

$a=g$ (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

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

where

d is the distance covered

t is the time taken

Here, the distance covered is $d$ , the distance from the base of the counter, while the time taken is the time of flight:

$t=\sqrt{\frac{2h}{g}}$

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

$v=\frac{d}{\sqrt{\frac{2h}{g}}}$

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

$v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}$

The vertical speed instead is given by the suvat equation:

$v_y=u_y + at$

where:

$u_y=0$ is the initial vertical velocity

$a=g$ is the acceleration

$t=\sqrt{\frac{2h}{g}}$ is the time of flight

Substituting,

$v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}$

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

Just before hitting the floor, the velocity of the cup has two components:

$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

$tan \theta = \frac{v_y}{v_x}$

where here $\theta$ is measured as below the horizontal direction.

Substituting the expressions for $v_x,v_y$ , we find:

$tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}$

$\theta=tan^{-1}(\frac{2h}{d})$ (radians)

4 0

3 years ago

A 4 kilogram box is at rest on a frictionless floor. A net force of 12 newtons acts on it. What is the acceleration of the box?

Rashid [163]

Answer:

Explanation:

Step one:

given

Mass m= 4kg

Force F= 12N

Required

Acceleration the relation between force, acceleration, and mass is Newton's first equation of motion, which says a body will continue to be at rest or uniform motion unless acted upon by an external force

F=ma

a=F/m

a=12/4

a=3 m/s^2

4 0

3 years ago

A 10 kg block is attached to a light cord that is wrapped around the pulley of an electric motor, as shown above. If the motor r

Anika [276]

Answer:

156.8 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 10 kg

Height (h) = 8 m

Time (t) = 5 s

Power (P) =?

Next, we shall determine the energy used by the motor to raise the block. This can be obtained as follow:

Mass (m) = 10 kg

Height (h) = 8 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 10 × 9. 8 × 8

E = 784 J

Finally, we shall determine the power output of the motor. This can be obtained as illustrated below:

Time (t) = 5 s

Energy (E) = 784 J

Power (P) =?

P = E/t

P = 784 / 5

P = 156.8 Watts

Therefore, the power output of the motor is 156.8 Watts

7 0

3 years ago

What happens if you move a bar magnet back and forth along the axis of the

nadezda [96]

C) A current is induced in the coiled wire, which lights the light bulb

The moving magnetic field creates electricity which lights the light bulb

Hope it helps!

8 0

2 years ago

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