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4 years ago

5

Where is a water or a mechanical wave in which the particles in the medium move in a direction perpendicular to the direction of

energy transport identified another name for this type of wave motion

2 answers:

horrorfan [7]4 years ago

6 0

transverse waves. ripples on water surface

astra-53 [7]4 years ago

4 0

Answer:Transverse waves

Explanation: Its on USA TEST PREP

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The thermal energy of what system does change

Inessa05 [86]

Answer:

there is the increase the temperature of cold body and decrease the temperature of hot body

4 0

2 years ago

An automobile tire having a temperature of 6.7 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 25 lb/in2 . What

OlgaM077 [116]

Answer: The gauge pressure in the tire when its temperature rises to 33◦C will be 28.7 lb/in2

Explanation: Please see the attachments below

3 0

3 years ago

A 94 g particle undergoes SHM with an amplitude of 8.3 mm, a maximum acceleration of magnitude 7.8 x 103 m/s2, and an unknown ph

Lelechka [254]

Answer:

a) T = 6.49*10^-3 s

b) v = 8 m/s

c) E = 3 J

d) F = 733 N

e) F = 366.5 J

Explanation:

Given

Mass of particle, m = 94 g = 0.094 kg

Amplitude of the particle, A = 8.3 mm = 8.3*10^-3 m

Maximum acceleration of particle, a = 7.8*10^3 m/s²

the equation describing Simple Harmonic Motion is given as

x = A cos (wt +φ)

To fond the acceleration of this relationship, we would have to integrate. Twice, the first would be a Velocity, and the second acceleration that we need.

Velocity = dx/dt = -Aw sin(wt + φ)

Acceleration = d²x/dt = -Aw² cos(wt + φ)

From the question, we were given, magnitude of acceleration to be 7.8*10^3 m/s²

Aw² = 7.8*10^3

w² = 7.8*10^3 / A

w² = 7.8*10^3 / 8.3*10^-3

w² = 939759

w = √939759

w = 969

Recall, T = 2π/w, so that

T = (2 * 3.142) / 969

T = 6.49*10^-3 s

Maximum speed = Aw

Maximum speed = 8.3*10^-3 * 969

Maximum speed = 8.0 m/s

Total mechanical energy oscillator =

mgx + 1/2mx² =

1/2mv(max)² =

1/2 * 0.094 * 8² =

3 J

Maximum displacement

x = A cos(wt + φ)

For x to be maximum here, then cos(wt + φ) Must be equal to 1

Acceleration = d²x/dt² = -Aw²

And force = mass * acceleration

Force = 0.094 * 7.8*10^3

Force = 733 N

x = A cos(wt + φ), where cos(wt + φ) = 1/2

d²x/dt² = -Aw² * 1/2

d²x/dt² = 733 * 0.5

= 366.5 N

7 0

3 years ago

A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic fr

Ksivusya [100]

a) The initial speed of the bullet is 330.5 m/s

b) The collision is inelastic

c) The impulse is -2.97 kg m/s

Explanation:

a)

The energy lost by the block while sliding (which is equal to the work done by friction) is equal to the kinetic energy of the block after the bullet has been embedded into it, therefore we can write

$KE=W$

$\frac{1}{2}(M+m)v^2=(\mu (M+m)g) d$

where

M = 1.20 kg is the mass of the block

m = 9.00 g = 0.009 kg is the mass of the bullet

v is the combined speed of bullet+block after the collision

$\mu = 0.20$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration of gravity

d = 0.310 m is the distance through which the block slides

Solving for v,

$v=\sqrt{2gd}=\sqrt{2(9.8)(0.310)}=2.46 m/s$

This is the final velocity of the block+bullet after the collision.

Now we can apply the law of conservation of momentum: in fact, the total momentum of the system before the collision must be equal to the total momentum after the collision, so we get

$mu+MU = (m+M)v$

where

u is the initial velocity of the bullet

U = 0 is the initial velocity of the block (initially at rest)

v = 2.46 m/s

Solving for u,

$u=\frac{(m+M)v}{m}=\frac{(0.009+1.20)(2.46)}{0.009}=330.5 m/s$

b)

To check whether the collision is elastic or inelastic, we just need to compare the total kinetic energy before and after the collision.

Before the collision, we have:

$K_i = \frac{1}{2}mu^2 = \frac{1}{2}(0.009)(330.5)^2=491.5 J$

While after the collision

$K_f = \frac{1}{2}(m+M)v^2 = \frac{1}{2}(0.009+1.20)(2.46)^2=3.7 J$

We see that the final kinetic energy is less than the initial kinetic energy: therefore, the collision is inelastic, since part of the energy has been converted into other forms of energy (e.g. thermal energy).

c)

The impulse of the block is equal to its change in momentum, so:

$I=\Delta p =p_f - p_i = (m+M)v'-(m+M)v$

where

v' = 0, since the block comes to a stop

v = 2.46 m/s is the velocity of the block just after the collision

Substituting,

$I=0-(0.009+1.20)(2.46)=-2.97 kg m/s$

And the impulse is negative, because its direction is opposite to the direction of motion of the block (this means that the force exerted on the block, which is the force of friction, acts in the direction opposite to the motion of the block).

Learn more about kinetic energy and momentum:

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3 0

4 years ago

Two 2.5 kg bowling balls are 0.50 m apart. What is the force of gravity of the first bowling ball due to the

Makovka662 [10]

Answer:The answer is A

Explanation:

To determine the force, we use the equation

Fg= G m1 m2/r2

, where Fg is the force of gravity, G is the gravitational constant,

m1

is the mass of the first body,

m2

is the mass of the second body, and r is the radius. Hence, we have

Fg=6.67×10−11 N×m2/kg2×2.5 kg×2.5 kg/(0.50 m)2=1.7×10−9N

. Since this is a gravitational problem, the force on one body has the same force as the other body, but the directions are opposite.

5 0

3 years ago