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1 year ago

5

What determines whether the equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler

or warmer water?

1 answer:

larisa [96]1 year ago

6 0

Answer:

Explanation:

It is determined by the amount of cold or hot water in the mixture to know if it will be closer to been hot or cold.

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A man stands on his balcony, 130 feet above the ground. He looks at the ground, with his sight line forming an angle of 70° with

jenyasd209 [6]

Answer:

d = 380 feet

Explanation:

Height of man = perpendicular= 130 feet

Angle of depression = ∅ = 70 °

distance to bus stop from man = hypotenuse = d = 130 sec∅

As sec ∅ = 1 / cos∅

so d = 130 sec∅ or d = 130 / cos∅

d = 130 / cos(70°)

d = 380 feet

8 0

2 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

2 years ago

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

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

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

1 year ago

When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe

ch4aika [34]

Answer:

4.41 W

Explanation:

P = IV, V = IR

P = V² / R

Given that P = 0.0625 when V = 1.50:

0.0625 = (1.50)² / R

R = 36

So the resistor is 36Ω.

When the voltage is 12.6, the power consumption is:

P = (12.6)² / 36

P = 4.41

So the power consumption is 4.41 W.

5 0

2 years ago

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t

ipn [44]

Answer:

ball clears the net

Explanation:

$v_{o}$ = initial speed of launch of the ball = 20 ms^{-1}

$\theta$ = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

$v_{ox}$ = initial velocity = $v_{o} Cos\theta = 20 Cos5 = 19.92 ms^{-1}$

$t$ = time of travel

$X$ = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

$X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}$ Eq-1

Consider the motion of the ball along the vertical direction

$v_{oy}$ = initial velocity = $v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}$

$t$ = time of travel

$Y_{o}$ = Initial position of the ball at the time of launch = 2 m

$Y$ = Final position of the ball at time "t"

$a_{y}$ = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

$Y = Y_{o} + v_{oy} t + (0.5) a_{y} t^{2}\\Y = 2 + (20 Sin5) (\frac{7}{20 Cos5}) + (0.5) (- 9.8) (\frac{7}{20 Cos5})^{2}\\Y = 2.00758 m\\$

Since Y > 1 m

hence the ball clears the net

7 0

2 years ago