Which of the following use the same units of measurement?

Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the wa

Katarina [22]

Answer:

A 93%

Explanation:

$P_1=P_2$ = Pressure will be equal at inlet and outlet

$\rho$ = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

$v_1$ = Velocity at inlet = 1.2 m/s

$v_2$ = Velocity at outlet

$r_1$ = Radius of inlet = $\dfrac{1.5}{2}=0.75\ cm$

$r_2$ = Radius of outlet

From Bernoulli's relation

$P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s$

From continuity equation

$A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m$

The fraction would be

$\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%$

The fraction is 93.0304%

8 0

1 year ago

Chris and Jamie are carrying Wayne on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 100 N. Wayne weigh

PIT_PIT [208]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The value is $F_j = 550\ N$

Explanation:

From the question we are told that

The length of the stretcher is $d = 2.0 \ m$

The weight of the stretcher is $W = 100 \ N$

The weight for Wayne is $W_w = 800 \ N$

The distance of center of gravity for Wayne from Chris is $c_w = 75 cm = 0.75 \ m$

Generally taking moment about the first end where Chris is

$F_j * d$ => upward moment

Here $F_j$ is the force applied by Jamie

Generally taking moment about the second end where Jamie is

$W * ( \frac{d}{2} ) + W_w * (d - c_w)$ => downward moment

Generally at equilibrium , the upward moment is equal to the downward moment

$F_j * d = W * ( \frac{d}{2} ) + W_w * (d - c_w)$

=> $F_j * 2 = 100 * ( \frac{ 2}{2} ) + 800 * (2 - 0.75)$

=> $F_j = 550\ N$

3 0

2 years ago

An ambulance moving at 42 m/s sounds its siren whose frequency is 450 hz. a car is moving in the same direction as the ambulance

Korvikt [17]

(a) Since the ambulance and the car are moving one relative to each other, we have to use the general formula of the Doppler effect, which gives us the shift of the frequency of the siren as heard by an observer in the car:
$f'=( \frac{v+v_o}{v+v_s} )f$
where
f' is the apparent frequency as heard by the observer in the car
v is the velocity of the wave
$v_o$ is the velocity of the observer (positive if it is moving towards the source, negative if it is moving away)
$v_s$ is the velocity of the source (positive if the source is moving away from the observer, negative if is is moving towards it)
f is the real frequency of the sound

In the first part of the problem:
$v=343 m/s$ (speed of the sound wave)
$v_o =-25 m/s$ (the car is moving away from the ambulance)
$v_s = -42 m/s$ (the ambulance is moving towards the car)
$f=450 Hz$ (original frequency of the sound)

If we plug the numbers into the formula, we find
$f'=( \frac{343 m/s-25 m/s}{343 m/s-42 m/s} )(450 Hz)=475 Hz$

b) This time, the ambulance passes the car, so the ambulance is now moving away from the car; this means that $v_s$ must be positive:
$v_s=+42 m/s$
Moreover, the car is now moving towards the ambulance, so we should reverse also the sign of $v_o$ :
$v_o=+25 m/s$
All the other data do not change, so if we use the same formula as before, we find
$f'=( \frac{343 m/s+25 m/s}{343 m/s+42 m/s} )(450 Hz)=430 Hz$

8 0

1 year ago

What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s? The dia

ycow [4]

The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s

The rotational moment of inertia of the puck is
I = (mr²)/2
= 0.5*(0.15 kg)*(0.038 m)²
= 1.083 x 10⁻⁴ kg-m²

The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
= 0.5*(0.15 kg)*(0.5 m/s)²
= 0.0187 j

The rotational KE is
KE₂ = (1/2)*I*ω²
= 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
= 0.0038 J

The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J

Answer: 0.0226 J

4 0

1 year ago

Sea breezes that occur near the shore are attributed to a difference between land and water with respect to what property?

ddd [48]

Answer:

a. mass density

Explanation:

<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>

When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.

<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

7 0

2 years ago