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Fittoniya [83]
3 years ago
15

"Today you will use this chart to help you identify some common minerals," instructed Mr. Grant, Jimmy's teacher. Each group of

students was given three mineral samples to identify. Jimmy and his partners were puzzled by one sample. It scratched glass, so it was pretty hard. One student thought it was feldspar; another thought it was quartz. What other properties could the group use to identify the mineral?
Physics
2 answers:
Zigmanuir [339]3 years ago
7 0

The correct answer to the question is luster and color

Brainliest plz!


scoray [572]3 years ago
4 0

The properties of minerals that might help Jimmy and his partners to identify the sample other than Hardness (scratch test) are:

1. Color

2. Crystalline Structure

3. Cleavage or Fracture

4. Transparency

5. Tenacity





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Two ice skaters, paula and ricardo, push off from each other. ricardo weighs more than paula. part a which skater, if either, ha
ahrayia [7]

Apply the law of conservation of momentum for this situation. The law states that the momentum of a system is constant (in absence of external forces acting on it).

The 'system' in this case are the two skaters. There is no external force on the skaters. Suppose the skaters are initially standing still. The momentum in the system is 0. This value will need to remain constant, even after the mutual push (which is a set of forces from <em>inside</em> the system). So we know that

(total momentum before) = (total momentum after)

Indexing the masses and velocities by the first letter of the skaters' names:

0 = m_P\vec v_P+m_R\vec v_R\\m_P\vec v_P = m_R(-\vec v_R)

From the last row, you can see that the skaters will have momentum of same magnitude but opposite direction, after the push off. That answers the first question: neither will have a greater momentum (both will have one of same magnitude).

Since Ricardo is heavier, from the above equality it follows that

m_R>m_P\implies|\vec v_R|

In words, Paula has the greater speed, after the push-off.

7 0
3 years ago
The wavelength of a wave is the horizontal distance from a ________ to an adjacent ________.
goldenfox [79]
<span>A. crest, crest

hope im right!(: </span>
5 0
3 years ago
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An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.
CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the heigth of the image

h_o is the height of the object

q=36.0 cm

p=16.0 cm

Solving the formula for h_i, we find

h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of h_i is positive, the image is erect

- When the sign of h_i is negative, the image is inverted

In this case, h_i is negative, so the image is inverted.

4 0
3 years ago
MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle
labwork [276]

Answer:

(a)  y_{max}=0.423m

(b)  \alpha =64.3^{o}

Explanation:

Given data

v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle

Solution

For Part (a)

As the velocity component in direction of y is given by:

v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s

The maximum displacement is given by:

v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m

For Part (b)

To reach y=46cm =0.46m apply:

0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha  =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}

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3 years ago
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Which example is not a part of the electromagnetic spectrum
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There no examples given.
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