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3 years ago

Find h , the maximum height attained by the projectile. express the maximum height in terms of v 0 , Î¸ , and g .

Physics

1 answer:

lutik1710 [3]3 years ago

3 0

solution:

the utmost height would be comprehensive while y'(t)=0. (on the suitable of the trajectory, the y speed is 0.) $y(t)=(88sin20)t-16t^2 y'(t)=88sin20-32t So y'(t)=0 while t=(88sin20)/32$

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A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse

Elanso [62]

Answer:

$v_{f} = 0.51 \frac{m}{s}$

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg : crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Crate weight (W)

W= m*g

W= 90kg*9.8 m/s²

W= 882 N

Friction force : Ff

Ff= μk*N Formula (2)

μk: coefficient of kinetic friction

N : Normal force (N)

Problem development

We apply the formula (1)

∑Fy = m*ay , ay=0

N-W = 0

N = W

N = 882 N

We replace the data in the formula (2)

Ff= μk*N = 0.351* 882 N

Ff= 309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax , ax=0

282 N - 309.58 N = 90*a

a= (282 N - 309.58 N ) / (90)

a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

$v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}$

$v_{f} = 0.51 \frac{m}{s}$

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3 years ago

A piano tuner is tuning a piano when he discovers that the G above middle C is vibrating with a higher frequency than his G tuni

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Answer:

Tuning a piano is the process of adjusting the tension of its strings, thereby altering their pitch, or frequency of vibration, by slightly turning the tuning pins to which they're attached, so that each string sounds pleasingly in harmony with every other string.

<h2>Hopefully u will satisfy with my answer of ur question..!!</h2>

<h2>Have a nice day ahead dear..!!</h2>

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3 years ago

A softball with a mass of 0.11 kg goes moves at a speed of 12 m/s, then the ball is hit by a bat and rebounds in the opposite di

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=1244 and that's from my math book

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3 years ago

Jonathan is studying solutions in his chemistry class. The teacher instructs the students to measure 2 tablespoons of salt into

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8 tablespoons of water.
2 x 4 = 8

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2 years ago

efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

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3 years ago