How do particles move when a surface wave passes through a medium?

A perfectly still 0.01 kg marble is hit with a force of 50 N by a pencil. According to Newton's third law, which describes the f

WARRIOR [948]

During the time that the pencil is exerting 50N of force on the marble, the marble is exerting 50N of force in exactly the opposite direction on the pencil. <em>(D) </em>The mass of the marble doesn't matter, and it doesn't even matter whether the marble is moving or perfectly still.

As soon as the pencil stops exerting any force on the marble, the marble immediately stops exerting any force on the pencil.

5 0

3 years ago

An object is propelled straight up from ground level with an initial velocity of 48 feet per second. Its height at time t is mod

Ket [755]

Answer:

Explanation:

For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:

v₀ = 48 ft/s

a = -32 ft/s/s

v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:

v² = v₀² + 2aΔx and filling in:

$0=48^2+2(-32)$ Δx and

0 = 2300 - 64Δx and

-2300 = -64Δs so

Δx = 36 feet.

Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:

v = v₀ + at and filling in:

0 = 48 - 32t and

-48 = -32t so

t = 1.5 sec. That's part a. Onto part b:

The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in:

$0=48t+\frac{1}{2}(-32)t^2$ and

$0=48t-16t^2$ and

0 = 16t(3 - t) so

t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:

We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:

$48t+\frac{1}{2}(-32)t^2 >32$ and get everything on one side and factor it again:

$-16t^2+48t-32>0$ and we find that

1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.

8 0

3 years ago

g An object with mass m=2 kg is completely submerged, and tethered, to the bottom of a large body of water. If the density of th

Anit [1.1K]

Answer:

The tension in the rope is 20 N

Solution:

As per the question:

Mass of the object, M = 2 kg

Density of water, $\rho_{w} = 1000\ kg/m^{3}$

Density of the object, $\rho_{ob} = 500\kg/m^{3}$

Acceleration due to gravity, g = $10\ m/s^{2}$

Now,

From the fig.1:

'N' represents the Bouyant force and T represents tension in the rope.

Suppose, the volume of the block be V:

V = $\frac{M}{\rho_{ob}}$ (1)

Also, we know that Bouyant force is given by:

$N = \rho_{w}Vg$

Using eqn (1):

$N = \rho_{w}\frac{M}{\rho_{ob}}g$

$N = 1000\frac{2}{500}\times 10 = 40\ N$

From the fig.1:

N = Mg + T

40 = 2(10) + T

T = 40 - 20 = 20 N

$N = \rho_{w}Vg$

3 0

3 years ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a

ValentinkaMS [17]

Answer:

a. 11.5kv/m

b.102nC/m^2

c.3.363pF

d. 77.3pC

Explanation:

Data given

$area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v$

to calculate the electric field, we use the equation below

V=Ed

where v=voltage, d= distance and E=electric field.

Hence we have

$E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m$

b.the expression for the charge density is expressed as

σ=ξE

where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2

If we insert the values we have

$8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}$

c.

from the expression for the capacitance

$C=eA/d$

if we substitute values we arrive at

$C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF$

d. To calculate the charge on each plate, we use the formula below

$Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC$

8 0

3 years ago

When a satellite is a distance R from the center of the Earth, the force of gravity on the satellite is F. What is the gravitati

Lubov Fominskaja [6]

Gravitational force on a satellite is given by the formula

$F = \frac{GMm}{r^2}$

now here we know that force on the satellite is F when its distance from center of Earth is R

Now the distance from the center of earth will be 3R so the force is given as

$F' = \frac{GMm}{(3R)^2}$

$F' = \frac{GMm}{9R^2}$

so if we compare it with initial value of force then it is

$F' = \frac{F}{9}$

so correct answer is

$(1) F/9$

7 0

3 years ago

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