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An ordinary drinking glass is filled to the brim with water (268.4 mL) at 2.0 ° C and placed on the sunny pool deck for a swimmer to enjoy. If the temperature of the water rises to 32.0 ° C before the swimmer reaches for the glass, how much water will have spilled over the top of the glass? Assume the glass does not expand.

Answers

Answer 1

Answer:

Answer:

$\Delta V=1.667*10^(-3)$

Explanation:

Given the initial temperature T_i=2° C

final temperature T_f= 32° C

The original volume of water Vo=268.8 mL= 0.2688 L

we need to calculate the change in the volume

As we know that volume expansion is given by

$(\Delta V)/(V_0)= \beta\Delta T$

ΔV= change in Volume

β= expansion coefficient = $207*10^(-6) K^(-1)$

therefore,

$\Delta V= \beta\Delta T V_0$

plugging values we get

$\Delta V=207*10^(-6) K^(-1) (32-2)*0.2688$

$\Delta V=1.667*10^(-3)$

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Answers

Answer:

a=-2.77 m/s^2

Explanation:

Assuming constant acceleration,

v=at + v_0

where v_0 is the initial velocity.

At rest, v=0, so

0=at+v_0

So solving the equation for a:

a=(-v_0)/t

Inserting the numbers yields

a=-2.77 m/s^2

The switch in the circuit has been in the left position for a long time. At t=0 it moves to the right position and stays there.a. Write the expression for the capacitor voltage v(t), fort≥0. b. Write the expression for the current through the 2.4kΩ resistor, i(t), fort≥0+.

Answers

Answer:

Pls refer to attached file

Explanation:

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400g over a distance of 0.52 cm as it rapidly bends its thorax, making the "click" that gives it its name. part a assuming the beetle jumps straight up, at what speed does it leave the ground? part b how much time is required for the beetle to reach this speed? part c ignoring air resistance, how high would it go?

Answers

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

$v^2=u^2+2as$

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\n = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\n =40.768 (m/s)^2\n v=6.385 m/s$

The launching speed of the beetle is 6.4 m/s.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\n 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\n t = (6.385 m/s)/(3920 m/s^2) = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is 1.62 ms.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\n (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\n s=((6.385 m/s)^2)/(2(9.8m/s^2)) =2.08 m$

The beetle can jump to a height of 2.1 m

We have that for the Question the Speed,Time and Height are

u=6.38m/s
T=13sec
h=2m

From the question we are told

Certain insects can achieve seemingly impossible accelerations while jumping.

the click beetle accelerates at an astonishing 400g over a distance of 0.52 cm as it rapidly bends its thorax, making the "click" that gives it its name.

Speed,Time and Height

Generally the equation for the average velocity is mathematically given as

$v^2-u^2=-2ah\n\nTherefore\n\nu=√(2*400*9.8*0.0052)\n\n$

u=6.38m/s

Generally the equation for the Time of flight is mathematically given as

$T=(2u)/(g)\n\nTherefore\n\nT=(2(6.38))/(9.8)$

T=13sec

Generally the equation for the air resistance is mathematically given as

$v^2-u^2=2gh\n\nTherefore\n\nh=(6.38^2)/(2*9.8)\n\$

h=2m

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A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like protons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create x-rays by bombarding a hard metal target with the beam. Consider a beam of protons at 1.90 keV and a current of 4.95 mA produced by the generator. (a) What is the speed of the protons (in m/s)?

Answers

Answer:

603383.67253 m/s

Explanation:

m = Mass of proton = $1.67* 10^(-27)\ kg$

K = Kinetic energy = 1.9 keV

$1\ ev=1.6* 10^(-19)\ J$

Velocity of proton is given by

$v=\sqrt{(2K)/(m)}\n\Rightarrow v=\sqrt{(2* 1.9* 10^3* 1.6* 10^(-19))/(1.67* 10^(-27))}\n\Rightarrow v=603383.67253\ m/s$

The speed of the protons is 603383.67253 m/s

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine provides anet thrust of 0.795 N perpendicular tothe tethering wire.(a) Find the torque the net thrust producesabout the center of the circle.
N·m

(b) Find the angular acceleration of the airplane when it is inlevel flight.
rad/s2

(c) Find the linear acceleration of the airplane tangent to itsflight path.
m/s2