How can we get a clue as to the temperature of a star

Which of these best shows how waves can affect matter? A) the moon’s gravity pulling the ocean a few feet up and down each day B

qaws [65]

Answer:

D: a buoy bobbing up and down in place with ocean waves

Explanation:

7 0

4 years ago

As ocean waves approach shore, their velocity decreases. How does a decrease in velocity affect the frequency and wavelength of

timama [110]

Explanation:

As ocean waves approach shore, their velocity decreases. The relation between the velocity, wavelength and the frequency of a wave is given by :

$v=f\times \lambda$

f = frequency

$\lambda$ = wavelength

$f=\dfrac{v}{\lambda}$ ..............(1)

So, as the velocity of the wave increases, the wavelength and frequency also increases as per equation (1). When the velocity decreases, the frequency and the wavelength also decreases.

3 0

3 years ago

Read 2 more answers

On dry concrete, a car can decelerate at a rate of 7.00 m/s2 , whereas on wet concrete it can decelerate at only 5.00 m/s2 . Fin

MariettaO [177]

Answer:

Explanation:

a ) Let the distance required in former case be d₁ .

initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 7.00 m /s²

v² = u² - 2 a s

0 = 30² - 2 x 7 x d₁

d₁ = 64.28 m

b) initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 5.00 m /s²

v² = u² - 2 a s

0 = 30² - 2 x 5 x d₂

d₂ = 90 m

c)

t = .5 s

s₁ = ut - .5 at²

= 30 x .5 - .5 x 7 x .5²

= 15 - .875

= 14.125 m

t = .5 s

s₂ = ut - .5 at²

= 30 x .5 - .5 x 5 x .5²

= 15 - .625

= 14.375 m

3 0

4 years ago

A 30 kg box sits on the floor it requires 275N of force to get it moving once it is moving it only takes 225N of force.what are

gavmur [86]

1) The coefficient of static friction is 0.935

2) The coefficient of kinetic friction is 0.765

Explanation:

1)

When a force is applied on a box sitting on the floor, the force that must be applied in order to make the box moving is equal to the maximum force of static friction between the floor and the box, which is:

$F = \mu_s mg$

where

$\mu_s$ is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

Here we have:

m = 30 kg

$g=9.8 m/s^2$

F = 275 N

Therefore, the coefficient of static friction is

$\mu_s = \frac{F}{mg}=\frac{275}{(30)(9.8)}=0.935$

2)

Once the box is in motion, the force that must be applied in order to make the box moving at constant velocity is equal to the force of kinetic friction between the floor and the box, which is:

$F = \mu_k mg$