All categories

3 years ago

The speed of a wave is 95 m/sec. If the wavelength of the wave is 0.3 meters, what is the frequency of the wave?

Physics

1 answer:

Dimas [21]3 years ago

8 0

<u>Answer</u>:

316.67 Hz

<u>Explanation</u>:

Given:

Wave velocity ( v ) = 95 m / sec

wavelength ( λ ) = 0.3 m

We have to calculate Frequency ( f ) :

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > f = v / λ

Putting values here we get:

= > f = 95 / 0.3 Hz

= > f = 950 / 3 Hz

= > f = 316.67 Hz

Hence, frequency of sound is 316.67 Hz.

You might be interested in

a body that has a mass of 0.04 k g is thrown up with a speed of 60m/s. Kinetic energy? a) in the begining,b)after 6 seconds

Katena32 [7]

Answer:

rhtnyhjukjhngccdvfnrtj tv drbrbfntjjz

8 0

3 years ago

Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus

tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0

3 years ago

if the angular momentum of a rigid body is changing, does that mean that there must be a net torque acting on the body?

masha68 [24]

The attribute of any rotating object determined by the product of the moment of inertia and the angular velocity is known as angular momentum.

<h3>What is Angular Momentum?</h3>

Without a kickstand, attempting to balance while getting on a bicycle will definitely result in you falling off. However, these wheels gain angular momentum once you begin pedaling. They're going to be resistant to change, which will make balance simpler.
The definition of angular momentum is: any rotating object's characteristic determined by moment of inertia times angular velocity.
It is a characteristic of rotating bodies determined by the sum of their moment of inertia and angular velocity. Since it is a vector quantity, the direction must also be taken into account in addition to the magnitude.

Angular Momentum Examples : We encounter this property frequently, whether knowingly or unknowingly.
The following provides some examples : Ice-skater
In order to begin a spin, an ice skater starts with her hands and legs spread widely from the center of her body. She moves her hands and leg closer to her body when she needs to spin with more angular velocity, though.
As a result, she conserves angular momentum and spins faster.

To Learn more About angular momentum refer to :

brainly.com/question/26889176

#SPJ4

6 0

1 year ago

Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration

eimsori [14]

The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

<em />

<u>Solution:</u>

1) Acceleration: $-2336 m/s^2$ on the hardwood floor, $-382 m/s^2$ on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

$v^2 - u^2 = 2ad$

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

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

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

$v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s$

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

$d = 1.8 mm = 0.0018 m$

So we can find the acceleration by using again the equation

$v'^2 - v^2 = 2ad$

Solving for a,

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2$

For the carpeted floor instead,

$d=1.1 cm = 0.011 m$

therefore the acceleration is

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2$

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

$a=\frac{v'-v}{t}$

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

$a=-2336 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms$

For the carpeted floor,

$a=-382 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms$

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

8 0

3 years ago

Radio telescopes are said to "listen" to space because...

almond37 [142]

Answer:

2 they detect radio waves

3 0

3 years ago