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tankabanditka [31]
3 years ago
10

Amit has found a rock in his backyard and would like to measure its mass, volume, and density. What tools can he use to perform

these measurements? meter stick, balance, and graduated cylinder graduated cylinder and balance meter stick and balance graduated cylinder and meter stick
Physics
1 answer:
pishuonlain [190]3 years ago
6 0

Answer: a graduated cylinder and balance

A rock has an irregular shape. so, it would be difficult to measure using a meter stick. most convenient method to find the volume would be using a graduated cylinder. we can take a fixed volume of water in the cylinder. then drop rock into it. note the new raised volume. Subtract the initial volume from final to find the volume of rock.

Using balance, mass of the rock can be measured.

we can calculate the density using the following formula:

Density=\frac{Mass}{Volume}

Therefore, we need only two tools, a graduated cylinder and a balance to measure mass, volume and density of a rock.

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One student did an experiment with two unknown minerals, Mineral 1 and Mineral 2. The hardness scale shown below was used for th
Umnica [9.8K]
Scale: (soft 1-->6 hardest)
<span>1=Talc 2=Gypsum 3=Calcite 4=Fluorite 5=Apatite 6=Orthoclase

</span>Mineral #1 can only scratch two other minerals therefore it must have a hardness level of 2+1=3 which is Calcite. (scratches talc, gypsum)

Mineral #2 can scratch four other minerals therefore it must have a hardness level of 4+1=5 which is Apatite. (scratches all but Apatite, Orthoclase)

Looking through the possible conclusions.. It looks like answer is D.


3 0
3 years ago
Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
loris [4]

Answer: 36.86\°

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: d=2.4 m

Ferdinand's initial velocity: V_{o}=5 m/s

Time it takes a jump: t=0.6 s

We need to find the angle \theta at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

V_{x}=\frac{d}{t} (1)

V_{x}=\frac{2.4 m}{0.6 s} (2)

V_{x}=4 m/s (3)

On the other hand, the x-component of the velocity is expressed as:

V_{x}=V_{o}cos\theta (4)

Substituting (3) in (4):

4 m/s=5 m/s cos\theta (5)

Clearing \theta:

\theta=cos^{-1} (\frac{4 m/s}{5 m/s})

\theta=36.86\° This is the angle at which Ferdinand the frog jumps between lily pads

4 0
3 years ago
A 580-turn solenoid is 18 cm long. The current in it is 36 A. A straight wire cuts through the center of the solenoid, along a 2
Karolina [17]

Answer:

F = 0.078N

Explanation:

In order to calculate the magnitude of the force on the wire you first calculate the magnitude of the magnetic field generated by the solenoid, by using the following formula:

B=\frac{\mu_oNi}{L}         (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

N: turns of the solenoid = 580

i: current in the solenoid = 36A

L: length of the solenoid = 18cm = 0.18m

You replace the values of all parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(580)(36A)}{0.18m}=0.145T

Next, you calculate the force exerted on the wire, by using the following formula:

F=iLBsin\theta         (2)

i: current in the wire = 27A

L: length of the wire that perceives the magnetic field (the same as the radius of the solenoid) = 2.0 cm = 0.02m

θ: angle between wire and the direction of B

B: magneitc field in the solenoid = 0.145T

The direction of the wire are perpendicular to the direction of the magnetic field, hence, the angle is 90°.

You replace the values of the parameters in the equation (2):

F=(27A)(0.02m)(0.145T)sin90\°=0.078N

The magnitude of the force on the wire is 0.078N

8 0
3 years ago
Read 2 more answers
Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

3 0
3 years ago
An object whose specific gravity is 0.850 is placed in water. What fraction of the object is below the surface of the water?
Fynjy0 [20]

Answer:

The fraction of the object that is below the surface of the water is ¹⁷/₂₀

Explanation:

Given;

specific gravity of the object, γ = 0.850

Specific gravity is given as;

specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3

Fraction of the object's weight below the surface of water is calculated as;

= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}

Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀

8 0
3 years ago
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