The period of a wave is 10 seconds and it’s wavelength is 2 meters. What is the wave’s speed?

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vert

WARRIOR [948]

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

$\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h$

$\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h$

$h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}$ (1)

Where:

$h$ - Maximum height of the wood block, in meters.

$v$ - Initial speed of the block, in meters per second.

$\mu_{k}$ - Kinetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$m$ - Mass, in kilograms.

$s$ - Distance travelled by the wood block along the wooden ramp, in meters.

$\theta$ - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that $v = 10\,\frac{m}{s}$ , $\mu_{k} = 0.20$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the height reached by the block above its starting point is:

$h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h = 4.249\,m$

The wood block reaches a height of 4.249 meters above its starting point.

5 0

2 years ago

Two manned satellites approach one another at a relative velocity of v = 0.150 m/s, intending to dock. The first has a mass of m

Aleksandr [31]

Answer:

= - 0.41m/s

Explanation:

Velocity of first satellite

$V_1 = \frac{m_1 -m_2}{m_1 + m_2} u$

$V_1 = \frac{(4 \times 10^3) - (7.5 \times 10^3)}{(4 \times 10^3) + (7.5 \times 10^3)} \times 0.15\\\\= -0.30435$

Velocity of the second satellite

$V_2 = \frac{2m_1}{m_1 + m_2} u$

$V_2 = \frac{2 \times 4 \times 10^3}{4 \times 10^3 + 7.5 \times 10^3} \\\\= 0.10435$

Final velocity = V(1) - V(2)

$V = -0.30435 - 0.10435\\\\= -0.4087$

≅ -0.41m/s

8 0

3 years ago

Read 2 more answers

Two objects are thrown vertically upward, first one, and then, a bit later, the other. Is it (a) possible or (b) impossible that

Studentka2010 [4]

Answer:

No, it is impossible

Explanation:

Kinematics equation:

$Vf^{2} =Vo^{2} -2gy$

if height is maximum:

y=H and Vf=0

so:

Analysis: From the last equation we see that the maximum height depends ONLY on the initial speed. This means that if both objects reach the same maximum height, then they necessarily need to have the SAME initial velocity. If they have the same initial velocity and in order to reach the maximum height at the SAME time the only way is that they are released at the SAME TIME.

5 0

2 years ago

What is the length x of the side of the triangle below? (Hint: use the cosine function.)

Firlakuza [10]

A because that is the answer

7 0

2 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 122m in 27s?

Stolb23 [73]

Speed = distance/time
speed= 122÷27=4.52m/s (3sf)

7 0

2 years ago