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lisabon 2012 [21]

2 years ago

9

You are hiking in Yellowstone National Park. A park ranger is speaking to students visiting a hot spring. Temperatures there can

reach as high as 456° Fahrenheit. She said some microscopic organisms can live in the hot springs. In what domain do these organisms most likely belong?
A.
Eubacteria

B.
Animal

C.
Archaebacteria

D.
Protist

1 answer:

Sholpan [36]2 years ago

3 0

I’m pretty sure it’s C Archaebacteria

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Which mineral is commonly pinkish and has light-colored, wavy lines? A) quartz B) potassium feldspar C) pyroxene D) amphibole D)

Nina [5.8K]

Answer:

option B

Explanation:

The correct answer is option B

the mineral which is pinkish and light in color with wavy lines is potassium feldspar .

pyroxene is a mineral is typically dark in color generally it's color is dark green.

amphibole is also dark in color an is in green color.

olivine is a mineral which is in olive green color.

Quartz is a mineral which is light colored and it is also pink in color but it does not have wavy line.

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2 years ago

A 755 N diver drops from a board 10.0 m above the water's surface. Find

MrRa [10]

Answer:

F = 755 N and h = 10.0 m and a = 9.8 m/s 2 and m=? and v = ?

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2 years ago

According to Care-stream's operating manual, what is the acceptable range of exposure indicators?

Aleks [24]

Answer:

correct option is C. 1700 - 2300

Explanation:

solution

according to Care stream, exposure indicators provide the technical adequancy incident radiations on the x ray

and we muse care stream formula that is

Exposure Index EI = 1000 × $log_{10} (mR)$ + 2000

here 1 mR = 2000 EI

and 2 mR = 2300 EI

and 3 mR = 1700 EI

so here acceptable range of exposure indicators is 1700 - 2300

correct option is C. 1700 - 2300

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2 years ago

A ball rolling down a hill was displaced 19.6 m while uniformly accelerating from rest. If the final velocity was 5.00m/s what w

poizon [28]

The rate of acceleration is $0.64 m/s^2$

Explanation:

Since the ball is moving at constant acceleration, we can solve the problem by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the ball in this problem,

u = 0

v = 5.00 m/s

s = 19.6 m

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{(5.00)^2-0}{2(19.6)}=0.64 m/s^2$

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7 0

2 years ago

A 2.0 kg block is pushed 1.0 m at a constant

svet-max [94.6K]

1) Work done by the force: 8.3 J

2) Work done by gravity: -19.6 J

3) Normal force: 16.3 N

Explanation:

1)

The work done by a force pushing an object is given by

$W=Fd cos \alpha$

where

F is the magnitude of the force

d is the displacement of the object

$\alpha$ is the angle between the direction of the force and of the displacement

First of all, we need to find the magnitude of the force F. Since the block is moving vertically at constant velocity (= zero acceleration), the equation of motion of the block along the vertical direction is:

$F sin \theta - \mu N - mg = 0$ (1)

Where

$\theta=27.0^{\circ}$

$\mu N$ is the force of friction, where

$\mu=0.40$ is the coefficient of friction

N is the normal reaction of the wall

(mg) is the weight of the block, where

m =2.0 kg is the mass of the block

$g=9.81 m/s^2$ is the acceleration of gravity

Along the horizontal direction, the equation of motion is:

$F cos \theta = N$ (2)

Substituting (2) into (1),

$F sin \theta - \mu F cos \theta - mg =0$

And solving for F,

$F(sin \theta - \mu cos \theta) = mg\\F=\frac{mg}{sin \theta - \mu cos \theta}=\frac{(2.0)(9.81)}{sin 27^{\circ} - 0.40 cos 27^{\circ}}=18.3 N$

Now we can calculate the work done by the force. Here we have

d = 1.0 m is the displacement

$\alpha = 90^{\circ} - 27^{\circ} = 63^{\circ}$ is the angle between the direction of the force and the displacement

Substituting,

$W=(18.3)(1.0)(cos 63^{\circ})=8.3 J$

2)

The work done by gravity is equal to:

$W_g = mg d cos \alpha$

where

m = 2.0 kg is the mass of the block

$g=9.81 m/s^2$

d = 1.0 m is the displacement

$\alpha=180^{\circ}$ , since the direction of gravity is downward while the displacement of the block is upward

Substituting,

$W=(2.0)(9.81)(1.0)(cos 180^{\circ})=-19.6 J$

3)

Equation (2) found in part 1) tells that

$N=F cos \theta$

which means that the normal force between the wall and the block is equal to the horizontal component of the pushing force on the block.

Here we have:

F = 18.3 N is the magnitude of the pushing force

$\theta=27.0^{\circ}$ is the angle of the force with the horizontal

Substituting, we find the normal force:

$N=(18.3)(cos 27^{\circ})=16.3 N$

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8 0

2 years ago