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4 years ago

Which statements accurately describe the Doppler effect?

Physics

2 answers:

Mnenie [13.5K]4 years ago

8 0

Answer:

A. It is used to learn about the behavior of the universe.

B. It was used by Hubble to measure velocities of galaxies.

C. It indicates galaxy motion through light wavelengths.

chubhunter [2.5K]4 years ago

5 0

The doppler effect is the increase or decrease in the frequency of sound, light, or other waves as the source and observer move toward or away from each other.

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A bullet with a mass mb=11.5 g is fired into a block of wood at velocity ????b=265 m/s. The block is attached to a spring that h

Alla [95]

Answer:

$m=0.496 kg$

Explanation:

Knowing the potential energy in the spring is equal to the initial kinetic energy so:

$F_s=E_K$

$\frac{1}{2}*K*x^2=\frac{1}{2}*m_b*v_f^2$

$\frac{1}{2}*205n/m*(0.35m)^2=\frac{1}{2}*m_b*v_f^2$

Now using the conservation of momentum

$P_i=P_f$

$m_b*v_i=(m_b+m_w)*v_f$

$11.5x10^3kg*265m/s=(11.5x10^3kg+m)*v_f$

$v_f=\frac{3.0475 kg*m/s}{(11.5x1063kg+m)}$

Replacing in initial equation so:

$12.5556J=\frac{1}{2}*11.5x10^3kg*(\frac{3.0475 kg*m/s}{11.5x10^3kg+m})^2$

Solve to m

$m=\frac{6.23}{12.556}$

$m=0.496 kg$

6 0

3 years ago

A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0Â° with the horizontal. The coeff

natulia [17]

Answer:

$H =212.6 \ J$

$v = 7.647 \ m/s$

Explanation:

From the question we are told that

The child's weight is $W_c = 287 \ N$

The length of the sliding surface of the playground is $L = 7.20 \ m$

The coefficient of friction is $\mu = 0.120$

The angle is $\theta = 31.0 ^o$

The initial speed is $u = 0.559 \ m/s$

Generally the normal force acting on the child is mathematically represented as

=> $N = mg * cos \theta$

Note $m * g = W_c$

Generally the frictional force between the slide and the child is

$F_f = \mu * mg * cos \theta$

Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as

$F =m* g sin(\theta) - F_f$

Here F is the resultant force and it is represented as $F = ma$

=> $ma = m* g sin(31.0) - \mu * mg * cos (31.0)$

=> $a = g sin(31.0)- \mu * g * cos (31.0)$

=> $a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)$

=> $a = 4.039 \ m/s^2$

$F_f = 0.120 * 287 * cos (31.0)$

=> $F_f = 29.52 \ N$

Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as

$H = F_f * L$

=> $H = 29.52 * 7.2$

=> $H =212.6 \ J$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

=> $v^2 = 0.559^2 + 2 * 4.039 * 7.2$

=> $v = \sqrt{0.559^2 + 2 * 4.039 * 7.2}$

=> $v = 7.647 \ m/s$

6 0

3 years ago

What causes the earths crust to move ?

tangare [24]

Upwelling currents in the molten material beneath the crust.

The rocks that make up the crust are light, compared with the metal-rich material beneath. The crust floats on top like an iceberg. Slow-moving currents underneath propel the continents around the surface

8 0

4 years ago

Read 2 more answers

a 1150 kg car is on a 8.70 hill. using x-y axis tilted down the plane, what is the x-component of the normal force(unit=N)

rodikova [14]

The x-component of the normal force is equal to <u>1706.45 N.</u>

Why?

To solve the problem, and since there is no additional information, we can safely assume that the x-axis is parallalel to the hill surface and the y-axis is perpendicular to the x-axis. Knowing that, we can calculate the components of the normal force (or weight for this case), using the following formulas:

$N_{x}=W*Sin(\alpha)=mg*Sin(\alpha)\\\\N_{y}=W*Cos(\alpha)=mg*Cos(\alpha)$

Now, using the given information, we have:

$mass=m=1150Kg\\\alpha=8.70\°\\g=9.81\frac{m}{s^{2}}$

Calculating, we have:

$N_{x}=mg*Sin(\alpha)$

$N_{x}=1150Kg*9.81\frac{m}{s^{2}}*Sin(8.70\°)\\\\N_{x}=11281.5\frac{Kg.m}{s^{2} }*Sin(8.70\°)=1706.45\frac{Kg.m}{s^{2} }=1706.45.23N$

Hence, we have that the x-component of the normal force is equal to <u>1706.45 N.</u>

Have a nice day!

3 0

4 years ago

A bucket of warm water can be considered a closed system for a very short period of time. Which best explains this phenomenon? A

zlopas [31]

The reason is that a closed system has nothing going in or out and water retains energy very well also water it polar so it is attracted to itself. It doesnt like contact with other things.
So it's likely to be C

3 0

3 years ago