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3 years ago

Which would not provide a useful measurement of temperature?

Physics

1 answer:

STatiana [176]3 years ago

6 0

It could maybe be D which is the expansion of a liquid inside a closed tube

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2. We know that the Earth is moving around the sun. When we look at the sky we see

marin [14]

Answer:

Answer: The Sun, the Moon, the planets, and the stars all rise in the east and set in the west. And that's because Earth spins -- toward the east. ... Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.

6 0

3 years ago

Betty (mass 40 kg), standing on slippery ice, catches her leaping dog (mass 15 kg) moving horizontally at 3.0 m/s. Show that the

MA_775_DIABLO [31]

Answer:

v = 2.18m/s

Explanation:

In order to calculate the speed of Betty and her dog you take into account the law of momentum conservation. The total momentum before Betty catches her dog must be equal to the total momentum after.

Then you have:

$Mv_{1o}+mv_{2o}=(M+m)v$ (1)

M: mass Betty = 40kg

m: mass of the dog = 15kg

v1o: initial speed of Betty = 3.0m/s

v2o: initial speed of the dog = 0 m/s

v: speed of both Betty and her dog = ?

You solve the equation (1) for v:

$v=\frac{Mv_{1o}+mv_{2o}}{M+m}=\frac{(40kg)(3.0m/s)+(15kg)(0m/s)}{40kg+15kg}\\\\v=2.18m/s$

The speed fo both Betty and her dog is 2.18m/s

7 0

2 years ago

A person wants to fire a water balloon cannon such that it hits a target 100m100m away. if the cannon can only be launched at 45

vladimir2022 [97]

<span>31.3 m/s Since the water balloon is being launched at a 45 degree angle, the horizontal and vertical speeds will be identical. Also the time the balloon takes to reach its peak altitude will match the time it takes to fall. So let's create a few expressions about what we know. Distance the water balloon travels at velocity v for time t d = vt Total time required for the entire trip is double since the balloon goes up, then goes down t = 2v/a Now let's plug in the numbers we have, assuming the acceleration due to gravity is 9.8 m/s^2 t = 2v/9.8 100 = vt Substitute 2v/9.8 for t in the 2nd formula 100 = v(2v/9.8) Solve for v. 100 = v(2v/9.8) 100 = 2v^2/9.8 980. = 2v^2 490 = v^2 22.13594 = v So we now know that both the horizontal velocity and vertical velocity needed is 22.13594 m/s. Let's verify that 2*22.13594 / 9.8 = 4.51754 So it will take 4.51754 second for the balloon to hit the ground after being launched. 4.51754 * 22.13594 = 100 And during that time it will travel 100 meters horizontally. But we need to know the total velocity. And the Pythagorean theorem comes to the rescue. Just square the 2 velocities, add them together, and take the square root. We already know the square is 490 from the work above, so sqrt(490+490) = sqrt(980) = 31.30495 m/s</span>

3 0

3 years ago

What is the speed of an object that travels 60 metres in 4 seconds

Zina [86]

Answer:

$s = \frac{d}{t} = \frac{60}{4} \\ \boxed{speed = 15m. {sec}^{ - 1} }$

4 0

3 years ago

Read 2 more answers

A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

bija089 [108]

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$