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3 years ago

In a plane mirror, the image is the same size as the object. O A. True B. False

Physics

2 answers:

Sindrei [870]3 years ago

6 0

It is true * if this helped you in anyway please be sure to leave a thanks:)

lisov135 [29]3 years ago

4 0

Answer:

A. true

Explanation:

a plane mirror is the one having angle of incidence equal to angle reflection.

the image appears of the same size as the object.

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What color dose red cabbage juice turn when it is mixed with lemon juice

miss Akunina [59]

I think it’s green but idk

7 0

3 years ago

A 75 g, 30 cm long rod hangs vertically on a friction less, horizontal axel passing through the center. A 10 g ball of clay trav

kotegsom [21]

Angular momentum is conserved, just before the clay hits and just after;
mv(L/2) = Iw 

I is the combined moment of inertia of the rod, (1/12)ML^2 , and the clay at the tip, m(L/2)^2 ; 
I = [(1/12)ML^2 + m(L/2)^2] 

Immediately after the collision the kinetic energy of rod + clay swings the rod up so the clay rises to a height "h" above its lowest point, giving it potential energy, mgh. From energy conservation in this phase of the problem; 
(1/2)Iw^2 = mgh 

Use the "w" found in the conservation of momentum above; and solve for "h" 
h = mv^2L^2/8gI 

Next, get the angle by noting it is related to "h" as; 
h = (L/2) - (L/2)Cos() 

So finally 
Cos() = 1- 2h/L = 1 - mv^2L/4gI 

m=mass of clay 
M=mass of rod 
L=length of rod 
v=velocity of clay

7 0

3 years ago

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500°C, and 80 m/s and the exit c

FrozenT [24]

Answer:

a) $\Delta \dot K = 24.570\,kW$ , b) $\dot W_{out} = 12729.15\,kW$ , c) $A_{in} = 0.0136\,m^{2}$

Explanation:

A turbine is a device which works usually in steady state and assumption of being adiabatic means no heat interactions between steam through turbine and surroudings and produce mechanical work from fluid energy. Changes in gravitational energy can be neglected. This system can be modelled after the First Law of Thermodynamics:

$-\dot W_{out} + \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2}) = 0$

a) Change in kinetic energy

$\Delta \dot K = \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2} - v_{out}^{2})$

$\Delta \dot K = \frac{1}{2} \cdot \left(12.6\,\frac{kg}{s} \right) \cdot \left[\left(80\,\frac{m}{s} \right)^{2}-\left(50\,\frac{m}{s} \right)^{2}\right]$

$\Delta \dot K = 24570\,W$

$\Delta \dot K = 24.570\,kW$

b) Power output

$\dot W_{out} = \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2})$

$\dot W_{out} = \left(12.6\,\frac{kg}{s}\right)\cdot \left(3446\,\frac{kJ}{kg} - 2437.7\,\frac{kJ}{kg} \right) + 24.570\,kW$

$\dot W_{out} = 12729.15\,kW$

c) Turbine inlet area

Turbine inlet area can be found by using the following expressions:

$\dot V_{in} = \dot m \cdot \nu_{in}$

$\dot V_{in} = \left(12.6\,\frac{kg}{s}\right) \cdot \left(0.086442\,\frac{m^{3}}{kg} \right)$

$\dot V_{in} = 1.089\,\frac{m^{3}}{s}$

$A_{in} = \frac{\dot V_{in}}{v_{in}}$

$A_{in} = \frac{1.089\,\frac{m^{3}}{s} }{80\,\frac{m}{s} }$

$A_{in} = 0.0136\,m^{2}$

8 0

3 years ago

Particle q1 has a positive 6 µc charge. particle q2 has a positive 2 µc charge. they are located 0.1 meters apart. recall that k

Dmitrij [34]

(a) Force between the two charges

The electrostatic force between the two charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, r their separation.

In this problem:

$q_1 =6 \mu C=6 \cdot 10^{-6}C$

$q_2=2 \mu C=2 \cdot 10^{-6}C$

$r=0.1 m$

Substituting into the equation, we find

$F=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(6 \cdot 10^{-6}C)(2 \cdot 10^{-6}C)}{(0.1 m)^2}=10.8 N$

(b) direction of particle q2

Particle q2 wants to move in the direction of the force acting on it. The direction of the force depends on the relative sign of the two charges: like charges attract each other, opposite charges repel each other. In this case, the two charges are both positive, so they repel each other and q2 tends to move away from particle q1.

7 0

3 years ago

Read 2 more answers

A pelican flying along a horizontal path drops a fish from a height of 4.7 m. The fish travels 9.3 m horizontally before it hits

slavikrds [6]

Answer:

(A) 9.5 m/s

(B) 5.225 m

Explanation:

vertical height (h) = 4.7 m

horizontal distance (d) = 9.3 m

acceleration due to gravity (g) = 9.8 m/s^{2}

initial speed of the fish (u) = 0 m/s

(A) what is the pelicans initial speed ?

lets first calculate the time it took the fish to fall

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

since u = 0

s = $(\frac{1}{2}) at^{2}$

t = $\sqrt{\frac{2s}{a} }$

where a = acceleration due to gravity and s = vertical height

t = $\sqrt{\frac{2 x 4.7 }{9.8} }$ = 0.98 s

pelicans initial speed = speed of the fish

speed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s

initial speed of the pelican = 9.5 m/s

(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?

vertical height = 1.5 m

pelican's speed = 9.5 m/s

lets also calculate the time it will take the fish to fall

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

since u = 0

s = $(\frac{1}{2}) at^{2}$

t = $\sqrt{\frac{2s}{a} }$

where a = acceleration due to gravity and s = vertical height

t = $\sqrt{\frac{2 x 1.5 }{9.8} }$ = 0.55 s

distance traveled by the fish = speed x time = 9.5 x 0.55 = 5.225 m

8 0

2 years ago