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Ierofanga [76]
2 years ago
11

A student must design an experiment to determine the gravitational mass of an object. Which of the following experiments could t

he student use? Select two answers.
Physics
1 answer:
Elena L [17]2 years ago
5 0

Answer: a. Place the object on one side of a lever at a known distance away from the fulcrum. Place known masses on the other side of the fulcrum so that they are also paced on the lever at known distance from the fulcrum. Move the known masses to a known distance such that the lever is in static equilibrium.

d. Place the object on the end of a vertically hanging spring with a known spring constant. Allow the spring to stretch to a new equilibrium position and measure the distance the spring is stretched from its original equilibrium position.

Explanation:

The options are:

a. Place the object on one side of a lever at a known distance away from the fulcrum. Place known masses on the other side of the fulcrum so that they are also paced on the lever at known distance from the fulcrum. Move the known masses to a known distance such that the lever is in static equilibrium.

b. Place the object on a surface of negligible friction and pull the object horizontally across the surface with a spring scale at a non constant speed such that a motion detector can measure how the objects speed as a function of time changes.

c. Place the object on a surface that provides friction between the object and the surface. Use a surface such that the coefficient of friction between the object and the surface is known. Pull the object horizontally across the surface with a spring scale at a nonconstant speed such that a motion detector can measure how the objects speed as a function of time changes.

d. Place the object on the end of a vertically hanging spring with a known spring constant. Allow the spring to stretch to a new equilibrium position and measure the distance the spring is stretched from its original equilibrium position.

Gravitational mass simply has to do with how the body responds to the force of gravity. From the options given, the correct options are A and D.

For option A, by balancing the torque, the mass can be calculated. Since the known mass and the distance has been given here, the unknown mass can be calculated.

For option D, here the gravitational force can be balanced by the spring force and hence the mass can be calculated.

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A ball bearing of radius of 1.5 mm made of iron of density
Serjik [45]

Answer:

\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine (\sf \eta)

Explanation:

\boxed{ \bold{v =  \frac{2}{9}  \frac{( {r}^{2} ( \rho -  \sigma)g)}{ \eta} }}

\sf \implies \eta =  \frac{2}{9}  \frac{( {r}^{2}( \rho -  \sigma)g )}{v}

Substituting values of r, ρ, σ, v & g in the equation:

\sf \implies \eta =  \frac{2}{9}  \frac{( {(0.15)}^{2}  \times  (7.85 - 1.25) \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times  \frac{145.6191}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times 64.7196

\sf \implies \eta =  2 \times 7.191

\sf \implies \eta =  14.382 \: poise

6 0
2 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Juliette [100K]

Answer:

24.348mm

Explanation:

NB: I'll be attaching pictures so as to depict missing mathematical expressions or special characters which are not easily found on keyboards

K = d / €^n

Note : d represents the greek alphabet epsilion.

K = 345 / 0.02⁰.²² = 816mPa

The true strain based upon the stress of 414mPa =

€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576

However the true relationship between true strain and length is given by

€ = ln(Li/Lo)

Making Li the subject of formula by rearranging,

Li = Lo.e^€

Li = 520e⁰.⁰⁴⁵⁷⁶

Li = 544.348mm

The amount of elongation can be calculated from

Change in L = Li - Lo = 544.348 - 520 change in L = 24.348mm.

8 0
2 years ago
Two objects, X and Y, move toward one another and eventually collide. Object X has a mass of 2M and is moving at a speed of 2v0
Sergeu [11.5K]

Answer:

c. The force exerted by X on Y is F to the right, and the force exerted by Yon X is F to the left.

Explanation:

If we take both objects as one single system, during the collision, assuming no external forces acting, the only forces present are the one that object X exerts on object Y, and the force that object Y does on object X.

These two forces, according to Newton's 3rd Law, form an action-reaction pair, and consequently, are equal in magnitude, acting in opposite directions.

As the object X is moving to the right, the force that produces, (F), is in the same direction (on Y), while for object Y, moving to the left, the force that produces (F also in magnitude) is in the same direction (on X), so the right answer is c.

The effect of the forces is different, due to masses are different, according Newton's 2nd Law.

6 0
2 years ago
Tex, an 85.0 kilogram rodeo bull rider is thrown from the bull after a short ride. The 520. kilogram bull chases after Tex at 13
Julli [10]
The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
                 m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
                    (85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s
3 0
2 years ago
Technician a says that using a pressure transducer and lab scope is a similar process to using a vacuum gauge. technician b says
Phantasy [73]

Answer: Both Technician A and B

Explanation:

There is a similar process in using a pressure transducer and lab scope to using a vacuum gauge.

And also, the pressure transducer can be used to tie any issues to individual cylinders if paired with a second trace consisting of the ignition pattern. Therefore, both Technician A and B are correct.

7 0
2 years ago
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