Can somebody help me with this quiz please

What is the frequency of a wave that has a wavelength of 0.39 m and a speed

gogolik [260]

Answer:

<h3>The answer is option B</h3>

Explanation:

The frequency of a wave can be found by using the formula

$f = \frac{c}{ \lambda} \\$

where

c is the velocity

From the question

wavelength = 0.39 m

c = 86 m/s

We have

$f = \frac{86}{0.39} \\ = 220.512820...$

We have the final answer as

Hope this helps you

7 0

4 years ago

A certain sprinter has a top speed of 10.5 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able

Svet_ta [14]

Answer:

Total time = 10.667 seconds

Explanation:

For the first 12m

V² = u² + 2as

10.5² = 0² + 2a(12)

110.25 = 24a

a = 4.59375 m/s²

V = u + at

10.5 = 0 + 4.59375t

t = 10.5/4.59375

t = 2.286 seconds

For the remaining race

100-12 = 88m

He travels at a constant speed for 88m

S = ut + 1/2(at²)

But a = 0

S = ut

88= 10.5t

t = 88/10.5

t = 8.38.seconds

Total time = 2.286 + 8.381

Total time = 10.667 seconds

4 0

4 years ago

A diver, of mass 40kg, climbs up to a diving platform 1.25m high. What’s her KE

Leona [35]

Explanation:

The formula for KE ( Kinetic Energy ) is : 1/2 mv²

Solution:

First we solve for v ( velocity ) using the formula 2gh, where g is gravity [ 9.8 m/s² ] and h is height [ 1.25 m ]

= 2 ( 9.8 m/s² × 1.25 m ) = 24.5 m/s

The velocity is 24.5 m/s.

Now solve for KE using the formula 1/2 mv²

= 1/2 ( 40 kg × 24.5m/s ² )

= 1/2 ( 40 kg × 600.25 )

= 1/2 ( 24,010 )

= 12,005 Joules

The diver's kinetic energy is 12,005 J.

6 0

2 years ago

Why are some forces considered to be noncontact forces? A. Objects must be far apart in order to exert a force. B. Objects do no

kompoz [17]

Answer:

B. Objects do not have to touch each other to experience a force.

Explanation:

For example ..One of the noncontact forces is magnetic force whereby a magnetic object will be attracted to another magnetic object of oppsite charged particles, through waves called electromagnetic waves. On the other hand, the two magnetic objects of similar charged particles can repel through electromagnetic waves..

8 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$