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3 years ago

WILL GIVE BRAINLIEST !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

6 0

Gold is a soft metal with a number of properties..It is both malleable and ductille,Also it's a heavy metal ,It is soft,Yellow....etc. Hope that helps

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In some applications of ultrasound, such as its use on cranial tissues, large reflections from the surrounding bones can produce

worty [1.4K]

Answer:

b. 0.75 mm

Explanation:

The distance between antinodes d is half the wavelength $\lambda$ . We can obtain the wavelength with the formula $v=\lambda f$ , where f is the frequency given ( $f=1MHz=1\times10^6Hz$ ) and v is the speed of sound in body tissues (v=1540m/s), so putting all together we have:

$d=\frac{\lambda}{2}=\frac{v}{2f}=\frac{1540m/s}{2(1\times10^6Hz)}=0.00077m=0.77mm$

which is very close to the 0.75mm option.

4 0

4 years ago

In your own words explain the importance of the cycles to an ecosystem and how the cycles of matter differ from the flow of ener

Rina8888 [55]

The three main cycles of an ecosystem are the water cycle, the carbon cycle and the nitrogen cycle.

8 0

3 years ago

Until a train is a safe distance from the station it must travel at 5 m/s. Once the train is on open track it can speed up

kirza4 [7]

Answer:

I believe the answer is b

Explanation:

5 0

4 years ago

A rock weighs 110 N in air and has a volume of 0.00337 m3 . What is its apparent weight when submerged in water? The acceleratio

pickupchik [31]

Explanation:

It is given that,

Weight of the rock in air, W = 110 N

Since, W = mg

$m=\dfrac{W}{g}$

$m=\dfrac{110\ N}{9.8\ m/s^2}$

m = 11.22 kg

We need to find the apparent weight of the rock when it is submerged in water. Apparent weight is equal to the weight of liquid displaced i.e.

$M=d\times V$

d is the density of water, $d=1000\ kg/m^3$

V is the volume of rock, $V=0.00337\ m^3$

$M=1000\ kg/m^3\times 0.00337\ m^3$

M = 3.37 kg

The apparent weight in water, W = m - M

$W=7.85\ kg\times 9.8\ m/s^2$

W = 76.93 N

So, the apparent weight of the rock is 76.93 N. Hence, this is the required solution.

4 0

3 years ago

A baseball player slides into third base with an initial speed of 4.0 m/s . if the coefficient of kinetic friction between the p

satela [25.4K]

The work done by the friction force to stop the player is equal to his loss of kinetic energy:
$W= \Delta K$

The work done by the friction force is the magnitude of the force $\mu m g$ times the distance covered by the player, d:
$W= \mu mg d$

The loss in kinetic energy is simply equal to the initial kinetic energy of the player, since the final kinetic energy is zero (the player comes to rest):
$\Delta K=K_i = \frac{1}{2}mv^2$

Substituting into the first equation, we get:
$\mu m g d= \frac{1}{2}mv^2$
from which we find d, the distance covered by the player:
$d= \frac{v^2}{2 \mu g}= \frac{(4.0 m/s)^2}{2 \cdot 0.4 \cdot 9.81 m/s^2}=2.04 m$

4 0

3 years ago