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nekit [7.7K]
3 years ago
13

Can waves move or transport matter? Why or why not?

Physics
1 answer:
OLga [1]3 years ago
6 0
They can.
Surfers use waves to move themselves around.
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Mercury is in the 80th position in the periodic table. How many protons does it have?
aalyn [17]
Mercury has 80 protons.
4 0
3 years ago
Read 2 more answers
Which of the following material offers the lowest resistivity?
Yuki888 [10]
Lower resistivity means higher conductivity: among these materials, the best conductor is silver, therefore it must be the material with lowest resistivity among those. We can also have a look at the value of resistivity of the different materials listed in the problem, to check our answer. The resistivities of the materials are the following:
Wood: 1.00\cdot 10^{3} \Omega \cdot m
Salt water: 2.00\cdot 10^{-1} \Omega \cdot m
Silver: 1.59\cdot 10^{-8} \Omega \cdot m
Lead: 2.20\cdot 10^{-7} \Omega \cdot m
<span>We see that the material with lowest resistivity among those is silver, therefore the correct answer is silver.</span>
7 0
3 years ago
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A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
4 years ago
A 1.50 kg rock is thrown up into the air from ground level, reaches a maximum height of 7.00 m, then returns to the ground. Calc
Minchanka [31]

Answer:

17.565 kgm/s

Explanation:

Momentum = mass × velocity

I = mv..................... Equation 1

But we can calculate the value of v  using the equation of motion under gravity.

v² =  u²+2gs............. Equation 2

Where u = initial velocity, s = maximum heigth, g = acceleration due to gravity.

Given: u = 0 m/s (at the maximum heigth), s = 7.0 m.

Constant: g = 9.8 m/s²

Substitute these values into equation 2

v² = 0²+ 2×7×9.8

v² = 137.2

v = √137.2

v = 11.71 m/s.

Also given: m = 1.50 kg

substitute these values into equation 1

Therefore,

I = 1.5×11.71

I = 17.565 kgm/s

3 0
3 years ago
Which statement is the best interpretation of the ray diagram shown below?
kicyunya [14]
This correct answer would be C
5 0
3 years ago
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