All categories

2 years ago

What is the period of a wave if the wavelength is 100 m and the speed is 200 m/s?

Physics

2 answers:

Stells [14]2 years ago

8 0

This is simple question and easy to understand.
so first you use a formula to calculate frequency of the wave.
v _ stands for velocity a.k.a speed
f_ stands for frequency which you have to find.
(pronounced lemda)_ stands for wavelength.
Then you gonna place all your values according to you have in the formula as shown then solve for f.
Finally, you use the time formula
where
T_ is period a.k.a time which equal to the reciprocal of frequency or equal to 1/f.
Simply input your values according to the formula and you your answer as 0.5s

dsp732 years ago

5 0

You might be interested in

A heat pump absorbs heat from the cold outdoors at 3Â°c and supplies heat to a house at 20Â°c at a rate of 30,000 kj/h. if the p

Mazyrski [523]

A heat pump absorbs heat from the cold outdoors at 3 C and supplies heat to a
house at 20 C at a rate of 30,000 kJ/h. If the power consumed by the heat pump
is 3 kW, find the coefficient of performance of the heat pump.

6 0

2 years ago

If a rock is thrown upward on the planet mars with a velocity of 14 m/s, its height (in meters) after t seconds is given by h =

crimeas [40]

Answer:

Velocity of rock after 2 seconds = 6.56 m/s

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Here height of rock in meters, h = $14t-1.86t^2$

Comparing both the equations

We will get initial velocity = 14 m/s(already given) and $\frac{1}{2} a = -1.86$

So, Acceleration, a = -3.72 $m/s^2$

Now we have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

When time is 2 seconds we need to find final velocity.

v = 14 - 3.72 * 2 = 6.56 m/s.

So, Velocity of rock after 2 seconds = 6.56 m/s

6 0

2 years ago

A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an

Whitepunk [10]

Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

F = Force exerted by empty rack = Weight of Empty Rack = 40 lb

A = Base Area = 452.4 in²

Therefore,

P₁ = 40 lb/452.4 in²

P₁ = 0.088 psi

Now, we calculate the pressure exerted by the rack along with the coat.

P₂ = F/A

where,

P₂ = Pressure exerted by rack filled with coats= ?

F = Force exerted by filled rack = Weight of Filled Rack = 65 lb

A = Base Area = 452.4 in²

Therefore,

P₂ = 65 lb/452.4 in²

P₂ = 0.144 psi

Now, the difference between both pressures is:

ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

ΔP = 0.056 psi

8 0

2 years ago

The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction betwe

Novosadov [1.4K]

Answer:

a) a= 8.33 m/s², T = 12.495 N , b) a = 2.45 m / s²

Explanation:

a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.

We apply Newton's second law to the lower charge

fr₁ + fr₂ = ma

The equation for the force of friction is

fr = μ N

Y Axis

N - W₁ –W₂ = 0

N = W₁ + W₂

N = (m₁ + m₂) g

Since the beams are the same, it has the same mass

N = 2 m g

We replace

μ₁ 2mg + μ₂ mg = m a

a = (2μ₁ + μ₂) g

a = (2 0.30 + 0.25) 9.8

a= 8.33 m/s²

Let's look for cable tension with beam 2

T = m₂ a

T = 1500 8.33

T = 12.495 N

b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal

fr = m₂ a₂

N- w₂ = 0

N = W₂ = mg

μ₂ mg = m a₂

a = μ₂ g

a = 0.25 9.8

a = 2.45 m / s²

4 0

2 years ago

A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

Answer:

Cannonball will be in flight before it hits the ground for 2.02 seconds

Explanation:

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So it will take 2.02 seconds to reach ground.

5 0

2 years ago