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3 years ago

Asap PLEASE HELP!!!! Match these vocabulary words with their definitions.

Physics

1 answer:

lubasha [3.4K]3 years ago

3 0

1. Isotherm
2. Labrador current
3. Longitude
4. Gulf stream
5. Horse latitudes
6. Maritime
7. Middle latitude climates
8. North Atlantic drift

hope this helps

mark me brainliest please!

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Quanto vale a resultante de duas forças de mesmo módulo mesma direção e sentidos opostos atuando sobre um corpo de massa igual a

Crank

A resultante das duas forças será zero, já q os sentidos são opostos e sãos iguais em módulo, elas se anulam. Logo, se a força resultante é zero, e F=ma, aceleração também será igual a zero.

6 0

3 years ago

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The tape in a videotape cassette has a total

Softa [21]

Answer:

1.37 rad/s

Explanation:

Given:

Total length of the tape is, $d= 297$ m

Total time of run is, $t = 2.1$ hours

We know, 1 hour = 3600 s

So, 2.1 hours = 2.1 × 3600 = 7560 s

So, total time of run is, $t= 7560$ s

Inner radius is, $r = 10\ mm = 0.01\ m
$

Outer radius is, $R = 47\ mm = 0.047\ m
$

Now, linear speed of the tape is, $v =\frac{d}{t}=\frac{297}{7560}=0.039\ m/s
$

Let the same angular speed be $\omega$ .

Now, average radius of the reel is given as the sum of the two radii divided by 2.

So, average radius is, $R_{avg}=\frac{R+r}{2}=\frac{0.047+0.01}{2}=\frac{0.057}{2}=0.0285\ m
$

Now, common angular speed is given as the ratio of linear speed and average radius of the tape. So,

$\omega=\dfrac{v}{R_{avg}}\\\\\\\omega=\dfrac{0.039}{0.0285}\\\\\\\omega=1.37\ rad/s
$

Therefore, the common angular speed of the reels is 1.37 rad/s.

5 0

3 years ago

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity o

MariettaO [177]

Complete question:

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?

Answer:

The frequency reflected by the stationary wall to the bat is 41 kHz

Explanation:

Given;

frequency emitted by the bat, = 39 kHz

velocity of the bat, $v_b$ = 8.32 m/s

speed of sound in air, v = 340 m/s

The apparent frequency of sound striking the wall is calculated as;

$f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz$

The frequency reflected by the stationary wall to the bat is calculated as;

$f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz$

$f_s\approx 41 \ kHz$

3 0

3 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$

If we manipulate the equation we have that:

$k=\frac{2mgh}{d^{2} }$

$k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}$

$k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}$

$k=243\ \frac{N}{m}$

5 0

3 years ago

Help!!

scZoUnD [109]

If Fg=mg=ma and, Fg(planetX)=1/5Fg(earth)

then the time would be 5x of the time as gravity is acceleration. So 3.9s*5=19.5s

As the force of gravity is less, then the acceleration of masses is also less, therefore it will take more time for the object to fall by the factor of the force of gravity difference

7 0

3 years ago

Read 2 more answers