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3 years ago

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

Physics

1 answer:

grandymaker [24]3 years ago

7 0

Answer:

the red at the bottom should not count, but the red at the top is the least dense because it floats upon the other liquids

Explanation:

hope this helps

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A major league baseball pitcher throws a pitch that follows these parametric equations: x(t) = 142t y(t) = –16t2 + 5t + 5. The t

azamat

Answer:

(a) $x'(t)= 142$

(b) 142

(d) 96.8 mph

(e) 0.426 s

(f) 0.061 rad

Explanation:

Velocity is a time-derivative of position.

(a) $x(t) = 142t$

$x'(t)= 142$

(b) Since $x'(t)= 142$ is independent of $t$ , it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.

$y'(t)= -32t+5$

(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to $x(t)$ .

$x(t)= 142t = 60.5$

$t=\dfrac{60.5}{142}= 0.426$ .

In this time, the vertical velocity, $y'(t)$ is

$y(t)= -32\times0.426+5 = -8.632$

The speed of the ball at thus point is $s=\sqrt{142^2+(-8.632)^2}=142$ ft/s

To convert this to mph, we multiply the factor 3600/5280

$s=142\times\dfrac{3600}{5280}=96.8 \text{ mph}$

(e) The time has been determined from (d) above.

$t= 0.426$

(f) This angle is given by

$\theta=\tan^{-1}\dfrac{y'(t)}{x'(t)}$

$\theta=\tan^{-1}\dfrac{-8.632}{142}=\tan^{-1}-0.0607=3.47$ (Note here we are considering the acute angle so we ignore the negative sign)

In radians, this is

$\theta=3.47\times\dfrac{\pi}{180}=0.061 \text{ rad}$

6 0

4 years ago

Length of the year is equivalent to the time it takes for one

Black_prince [1.1K]

<span>The length of the year is equivalent to the time it takes for the earth to go around the sun one time.</span>

4 0

3 years ago

Read 2 more answers

A vertical piston-cylinder device initially contains 0.1 m^3 of air at 400 K and 100 kPa. At this initial condition, the piston

jenyasd209 [6]

Answer:

$Q=-38.15kJ$

Explanation:

From the question we are told that

Piston-cylinder initial Volume of air $v_1=0.1 m^3$

Piston-cylinder initial temperature $T_1=400k$

Piston-cylinder initial pressure $P_1= 100kpa$

Supply line temperature $T_s=400k$

Supply line pressure $P_s= 500kpa$

Valve final pressure $P_v=500kpa$

Piston movement pressure $P_m=200kpa$

Piston-cylinder final Volume of air $v_2=0.2 m^3$

Piston-cylinder final temperature $T_2=440k$

Piston-cylinder final pressure $P_2= 500kpa$

Generally the equation for conservation of mass is mathematically given by

$Q=m_2 \mu_2-m_1 \mu_1 +W-(m_2-m_1)h$

where

Initial moment

$m_1=\frac{p_1 V_1}{RT_1}$

$m_1=\frac{100*0.1}{0.287*400}$

$m_1=8.7*10^-^2kg$

Final moment

$m_2=\frac{p_2 V_2}{RT_2}$

$m_1=\frac{500*0.3}{0.287*440}$

$m_1=79*10^{-2}kg$

Work done by Piston movement pressure

$W=Pd$

$W=P(v_2-v_1)$

$W=200(0.2-0.1))$

$W=20000J$

Heat function

$h=cT_1$

$h=1.005(400)$

$h=402$

Therefore

$Q=(0.792*0.718(440)-0.0871*0.718(400)+20-(0.792-0.087)*402))$

$Q=-38.15kJ$

It is given mathematically that the system lost or dissipated Heat of

$Q=-38.15kJ$

7 0

3 years ago

Transfer or movement of thermal energy is called

Aneli [31]

Answer: Thermal energy transfer involves the transfer of internal energy. The three types of thermal energy transfer are conduction, convection and radiation. Conduction involves direct contact of atoms, convection involves the movement of warm particles and radiation involves the movement of electromagnetic waves.

Explanation:

5 0

3 years ago

A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei

Alenkinab [10]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

$F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\$

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

$F = (150\ N)Sin\ 20^o$

8 0

3 years ago