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2 years ago

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A student sorted mineral samples into two groups: dull and shiny. Which of the following properties did the student use to sort

the mineral samples into groups? a
streak
b
luster
c
cleavage
d
color free brainliest

2 answers:

Katyanochek1 [597]2 years ago

7 0

Answer:

luster

Explanation:

xenn [34]2 years ago

4 0

The answer would be luster

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HELP ME 20 POINTS!!!Assume that the average volume of an adult human body is one-tenth

Agata [3.3K]

Answer:

,Assume that the average volume of an adult human body is one-tenth

cubic meter (0.10 m) and that there are two billion (2.0 x 109)

adults in the world.

a. What would be the total volume of all the adults in the world?

b. Compute the length of one edge of a cubic container that has a

volume equal to the volume of all the adults in the world.

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2 years ago

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago

A ball thrown in to the air starts out with an initial upward velocity but eventually slows and stops before falling back to the

Mashutka [201]

Answer:

b) The downward force of gravity

Explanation:

The gravity force has the biggest influence on the deceleration of the ball because no matter how much force you applied on the ball, it will eventually go down again, according to newton's second law:

$\sum F=m.a$

right after you throw the ball, only the force exerted by the gravity will affect the ball (neglecting air resistance):

$m.g=m.a\\a=g$

so the object will eventually be going down again.

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2 years ago

A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find

Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

m = mass of the ladder, 7.98 kg

L = length of the ladder, 4.15 m

On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

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3 years ago

The fluid friction that opposes the motion of objects through air is known as what?

PSYCHO15rus [73]

The answer is Air Resistance

6 0

3 years ago