All categories

4 years ago

Which statement correctly compares the motion of atoms in liquid mercury and solid copper?

Physics

2 answers:

vesna_86 [32]4 years ago

7 0

Answer:Copper atoms vibrate around fixed positions, whereas mercury atoms move around while staying close together.

Explanation:

Copper atoms vibrate around fixed positions, whereas mercury atoms move around while staying close together.

vovikov84 [41]4 years ago

4 0

Answer:

the copper vibrates

You might be interested in

In an experiment, a shearwater (a seabird) was taken from its nest, flown a distance 5220 km away, and released. It found its wa

Korolek [52]

Answer:

4.6834625323 m/s

0 m/s

Explanation:

s = Displacement

t = Time

Velocity is given by

$v=\dfrac{s}{t}\\\Rightarrow v=\dfrac{5220000}{12.9\times 24\times 60\times 60}\\\Rightarrow v=4.6834625323\ m/s$

The bird's average velocity for the return flight is 4.6834625323 m/s

In the whole episode the bird went 5220 km away from its nest and came back. This means the displacement is zero.

Hence, the average velocity for the whole episode is 0 m/s

7 0

3 years ago

Read 2 more answers

Mathew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the bal

Cloud [144]

When It begins to drop because that when gravity will have its strongest pull on the object.

8 0

3 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

3 years ago

A skater makes one revolution per second, f = w/(2p) = 1, with her arms and leg stretched (left). By lowering her leg and bringi

Firlakuza [10]

Answer:

Answer is given in the attachment.

Explanation:

Download pdf

4 0

3 years ago

Read 2 more answers

A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from

Pepsi [2]

Answer: WAIT WHATTTT i have that same test due today and the answer is in explanation

Explanation:

Bike, truck train. we are in the same school i think. Its imma say the incisal JMES Im Lusi i used to help in the library

4 0

3 years ago