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3 years ago

2. A mass-spring system oscillates with a frequency of 20 Hz. What is the period?

Physics

1 answer:

Alex_Xolod [135]3 years ago

5 0

Answer:

0.05s

Explanation:

f=20Hz

we know,

time period=t=1/f

t=1/20

=0.05s

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A lorry of mass 12,000kg travelling at a velocity of 2 m/s collides with a stationary car of mass 1500kg. The vehicles move toge

Lubov Fominskaja [6]

Answer:

Approximately $1.8\; \rm m \cdot s^{-1}$ . Assumption: the friction between the two vehicles and the ground is negligible.

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$ :

$p = m \cdot v$ .

Momentum should be conserved in this collision if there's no external force on these two vehicles. In other words, the sum of the momentum of the lorry and the car should be the same before and after the collision.

Mass of the lorry: $m_1 = 12000\; \rm kg$ .

Velocity of the lorry right before the collision: $2\; \rm m \cdot s^{-1}$ .

Therefore, the momentum of the lorry right before the collision would be $12000\; \rm kg \times 2\; \rm m \cdot s^{-1} = 24000\; \rm kg \cdot m\cdot s^{-1}$ .

Mass of the car: $m_2 = 1500\; \rm kg$ .

Velocity of the car before the collision: $0\; \rm m \cdot s^{-1}$ .

Therefore, the velocity of the car before the collision would be $1500\; \rm kg \times 0\; \rm m \cdot s^{-1} = 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of the lorry and the car right before the collision:

$24000\; \rm kg \cdot m \cdot s^{-1} + 0\; \rm kg \cdot m \cdot s^{-1} = 24000\; \rm kg \cdot m \cdot s^{-1}$ .

Under the assumption that there is no external forces on the lorry and the car, the sum of the momentum of the lorry and the car right after the collision should also be $24000\; \rm kg \cdot m \cdot s^{-1}$ .

Let $v\; \rm m \cdot s^{-1}$ be the velocity of the lorry and the car after the collision. (The velocity of the two vehicles should be the same because they were moving together.)

Right after the collision, the momentum of the lorry (with a mass of $12000\; \rm kg$ ) would be ${12000\; {\rm kg} \times v\; \rm m \cdot s^{-1}} = 12000\, v \; \rm kg \cdot m \cdot s^{-1}$ .

Similarly, right after the collision, the momentum of the car (with a mass of $1500\; \rm kg$ ) would be ${1500\; \rm kg} \times v\; {\rm m \cdot s^{-1}} = 1500\, v\; \rm m \cdot s^{-1}$ .

The sum of the velocities of the two vehicles right after the collision would be:

$\left(12000\, v \; {\rm kg \cdot m \cdot s^{-1}} + 1500\, v \; {\rm kg \cdot m \cdot s^{-1}}\right)$ .

That quantity should match the momentum right before the collision. In other words:

${12000\; \rm kg \cdot m \cdot s^{-1}} = \left(12000\, v \; {\rm kg \cdot m \cdot s^{-1}} + 1500\, v \; {\rm kg \cdot m \cdot s^{-1}}\right)$ .

Solve this equation for the velocity of the two vehicles after the collision:

$\displaystyle v = \frac{12000}{12000 + 1500}\approx 1.8$ .

Therefore, the velocity of the two vehicles right after the collision would be approximately $1.8\; \rm m \cdot s^{-1}$ .

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2 years ago

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3 years ago

Read 2 more answers

A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius o

Setler79 [48]

Answer:

21.8 m/s

Explanation:

At the top of the hill (crest), there are two forces acting on the motorcycle:

- The reaction force of the road, N (upward)

- The force of gravity, mg (downward)

Since the motorcycle is moving by circular motion, the resultant of these forces will give the centripetal force, so:

$mg-N = m\frac{v^2}{r}$

where the direction of the weight (mg) is equal to that of the centripetal force, and where

m is the mass of the cycle

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

v is the speed

r = 48.6 is the radius of the hill

The cycle loses contact with the road when the reaction force becomes zero:

N = 0

Substituting into the equation, we therefore find the maximum speed that is allowed for the cycle before losing constact:

$mg = m\frac{v^2}{r}\\v=\sqrt{gr}=\sqrt{(9.8)(48.6)}=21.8 m/s$

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2 years ago

A cube of water 10 cm on a side is placed in a microwave beam having Ea = 11 kV/m‘ The microwaves illuminate one faceofthe cube,

Shtirlitz [24]

Answer:

The time is 133.5 sec.

Explanation:

Given that,

One side of cube = 10 cm

Intensity of electric field = 11 kV/m

Suppose How long will it take to raise the water temperature by 41°C Assume that the water has no heat loss during this time.

We need to calculate the rate of energy transfer from the beam to the cube

Using formula of rate of energy

$P=(0.80)IA$

$P=0.80\times\dfrac{c\mu_{0}E^2}{2}\times A$

Put the value into the formula

$P=0.80\times\dfrac{3\times10^{8}\times8.85\times10^{-12}\times(1.1\times10^{4})^2}{2}\times(10\times10^{-2})^2$

$P=1285.02\ W$

We need to calculate the amount of heat

Using formula of heat

$E =mc\Delta T$

$E =\rho Vc\Delta T$

Put the value into the formula

$E=1000\times(0.10)^3\times4186\times41$

$E=171626\ J$

We need to calculate the time

Using formula of time

$t=\dfrac{E}{P}$

Put the value into the formula

$t=\dfrac{171626}{1285.02}$

$t=133.5\ sec$

Hence, The time is 133.5 sec.

3 0

3 years ago

A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of

Dennis_Churaev [7]

Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = $85\frac{1000}{3600}=23.61\ m/s$

Initial velocity = u = 0

Equation of motion

$v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s$

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s$

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2$

Emergency acceleration is 2.99 m/s² (magnitude)

6 0

2 years ago