If someone drove for 20 minutes going 30 m/s, how far did they drive?

The focal length ____________ when the refractive index of the lens is increased and __________ when the curvature radius of the

Ivenika [448]

The focal length would likely decrease as the refractive index and the increase when the curvature radius of the lens increases. The decrease in focal length happens since a higher index of refraction would signify that the rays of the sun striking to an object would tend to bend more.

8 0

3 years ago

Sound travels through air at 343 m/s at 20 °C. A bat emits an ultrasonic squeak and hears the echo 0.05 second later . How far a

Rashid [163]

Answer:

Distance of the object is 8.6 m

Explanation:

As we know that the speed of sound at t degree C is given as

$v = 332 + 0.6 t$

here we know that the temperature is

t = 20 degree C

so we have

$v = 332 + 0.6(20)$

$v = 344 m/s$

now we know that bat heard the echo of sound in 0.05 s

so the to and fro distance of the object is d + d = 2d

so we have

$2 d = v\times t$

$2d = 344(0.05)$

$d = 8.6 m$

7 0

3 years ago

One-fourth of adults in the United States are not active at all.<br><br> true<br><br> false

Vera_Pavlovna [14]

Answer:

true 28%

Explanation:

;"/:;,*"# /hcrvuvf

3 0

3 years ago

Read 2 more answers

How long will it take a car to go from a complete stop to 44 m/s if they are

mafiozo [28]

Answer:

$\boxed{\sf Time \ taken \ (t) = 8.8 \ s}$

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 44 m/s

Acceleration (a) = 5 m/s²

To Find:

Time taken (t)

Explanation:

From equation of motion we have:

$\boxed{ \bold{v = u + at}}$

By substituting value of u, v & a in the equation we get:

$\sf \implies 44 = 0 + 5t \\ \\ \sf \implies 44 = 5t \\ \\ \sf \implies 5t = 44 \\ \\ \sf \implies t = \frac{44}{5} \\ \\ \sf \implies t = 8.8 \: s$

7 0

3 years ago

During an ultrasound, sound waves are sent by a transducer through muscle tissue at a speed of 1300 m/s. Some of the sound waves

Dominik [7]

Answer:

Distance of metal fragment is 0.0975 m

Explanation:

As we know that the ultrasound will move towards the metal fragment and reflected back to the transducer

So the time interval in which sound will go to the metal fragment and come back is given as

$t = 1.5 \times 10^{-4} s$

now we know that the speed of the wave in the metal is

v = 1300 m/s

now we know that total distance traveled by the ultrasonic wave is given as

$d = v t$

$d = 1300 (1.5 \times 10^[-4})$

$d = 0.195 m$

so the depth of metal fragment is half of total distance because in the ultraviolet sound move to and from between transducer and metal fragment so its distance is given as

$x = \frac{d}{2}$

$x = 0.0975 m$

8 0

3 years ago