What is the wave speed if a wave has a frequency of 12 hz and a wavelength of 3.0 m?

A solid disk of radius 5.50 cm and mass 1.25 kg , which is rolling at a speed of 1.50 m/s , begins rolling without slipping up a

shtirl [24]

Answer:

Explanation:

Deceleration of solid disk = g sin10/1 + k²/r² = g sin 10 / 1 + 1/2 = g sin 10 x 2/3

[ k is radius of gyration of disk which is equal to( 1/√2)x r ]

deceleration a = -1.1345 m/s²

v = u - at , t = u / a = 1.5 / 1.1345 = 1.322 s.

5 0

3 years ago

A dog running to the right at 4 m/s sees a ball and accelerates steadily to catch it. The dog accelerates to the right at a rate

antoniya [11.8K]

Answer:

D.-4.798m/s

Explanation:

Greetings !

Given values

$u= 4ms \\ a = 0.21ms {}^{2} \\ t = 3.8sec$

Solve for V of the given expression

Firstly, recall the velocity-time equation

$v = u + at$

plug in known values to the equation

$v = (4) + (0.21)(3.8)$

solve for final velocity

$v = 4.792ms$

Hope it helps!

6 0

1 year ago

Which is a sub-atomic particle?

Natalka [10]

A particle that is smaller than an atom or a cluster of particles.

4 0

3 years ago

Could a nucleus that has one proton but no neutrons exist?

diamong [38]

Yes, but there is only 1 atom like that and is is hydrogen. Hydrogen is the only element that could have a nucleus with one proton and no neutrons exist.

8 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

2 years ago