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3 years ago

A projectile is launched at an angle into the air at velocity v and angle θ. Determine its vertical acceleration.

2 answers:

8 0

Hi there!

Since you are just trying to find the vertical acceleration and not the horizontal acceleration, angle and velocity do not come into play in this problem. The downward pull of gravity has a vertical acceleration of 9.8 m/s/s, which would be your answer.

-Your friend in physics,
ASIAX Frequent Answerer

drek231 [11]3 years ago

5 0

When you're dealing with projectile problems, you have two dimensions: an x (horizontal) dimension and a y (vertical) dimension. The only force that will be acting upon a projectile in the air is the force of gravity, which points down in the y/vertical dimension.

Gravity causes a downward vertical acceleration of $-9.8 \:m/ s^{2}$ .
The velocity and angle do not affect the vertical acceleration.

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You tie a cord to a pail of water, and you swing the pail in a vertical circle of radius 0.600 m. what minimum speed must you gi

omeli [17]

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6 0

3 years ago

Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a specific enthalpy of 2431.6 kJ/kg and exi

PIT_PIT [208]

Answer:

Explanation:

Given:

Steam Mass rate, ms = 1.5 kg/min

= 1.5 kg/min × 1 min/60 sec

= 0.025 kg/s

Air Mass rate, ma = 100 kg/min

= 100 kg/min × 1 min/60 sec

= 1.67 kg/s

Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.

xf, quality = 0.9.

Tsat = 89.9°C

hf = 376.57 kJ/kg

hfg = 2283.38 kJ/kg

Using the equation for specific enthalpy,

hi = hf + (hfg × xf)

= 376.57 + (2283.38 × 0.9)

= 2431.552 kJ/kg

The specific enthalpy of the outlet, h2 = hf

= 376.57 kJ/kg

Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy

= ms × (hi - h2)

= 0.025 × (2431.552 - 376.57)

= 0.025 × 2055.042

= 51.37455 kW

= 51.38 kW.

5 0

3 years ago

You can increase the vapor pressure of a liquid by:

irga5000 [103]

Answer: Option (c) is the correct answer.

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

When we increase the temperature of a liquid substance then there will occur an increase kinetic energy of the molecules. As a result, they will move readily from one place to another.

Hence, liquid state of a substance will change into vapor state of the substance. This means that an increase in temperature will lead to an increase in vapor pressure of the substance.

Thus, we can conclude that you can increase the vapor pressure of a liquid by increasing temperature.

4 0

3 years ago

PLEASE HELP!!! WOULD REALLY APPRECIATE!

Eva8 [605]

Answer:

a. slope=rise/run

rise=0.02

run=-2

determined using the point (3,0.08) and (1,0.1) on the graph

slope=0.02/-2

= -0.01 or -1/100

b.area= area of trapizoid+ rectangle

((0.07+0.11)÷2)×4+1×0.07

0.36+0.07

=0.43$

c. the area represent the total cost after 5 hours

PLEASE MARK BRAINLIEST

3 0

2 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago