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Debora [2.8K]
3 years ago
8

Which of the following experiments could be used to determine the inertial mass of a block? A. Place the block on a rough horizo

ntal surface. Lift one end of the surface up and measure the angle the surface makes with the horizontal at the moment the block begins to slide. B. A Drop the block from different heights and measure the time of fall from each height. C. Place the block on a rough horizontal surface. Give the block an initial velocity and then let it come to rest. Measure the initial velocity and the distance the block moves in coming to rest. D. Use a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force.
Physics
1 answer:
valentinak56 [21]3 years ago
3 0

Answer:

D, using a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force

Explanation:

For this you would use the net force equation acceleration=net force * mass however you will want to isolate mass so it would be acceleration/ net force to get mass. Then process of elimination comes to play.

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The speed of a wave on a violin A string is 288 m/s and on the G string is 128 m/s. The force exerted on the ends of the string
Katyanochek1 [597]

Answer:

\dfrac{\mu_A}{\mu_G}=0.197

Explanation:

given,

Speed of a wave on violin A = 288 m/s

Speed on the G string = 128 m/s

Force at the end of string G  = 110 N

Force at the end of string A = 350 N

the ratio of mass per unit length of the strings (A/G). = ?

speed for string A

 v_A = \sqrt{\dfrac{F_A}{\mu_A}}.......(1)

speed for string G

 v_G = \sqrt{\dfrac{F_G}{\mu_G}}........(2)

Assuming force is same in both the string

now,

dividing equation (2)/(1)

\dfrac{v_G}{v_A}=\dfrac{\sqrt{\dfrac{F_G}{\mu_G}}}{\sqrt{\dfrac{F_A}{\mu_A}}}

\dfrac{v_G}{v_A}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}

\dfrac{128}{288}=\dfrac{\sqrt{\mu_A}}{\sqrt{\mu_G}}

\dfrac{\mu_A}{\mu_G}=0.197

5 0
3 years ago
you want to get more physically fit but often make excuses for why you don't have time to be active what strategy can help with
Anna35 [415]
<span>Find an activity you enjoy and make it a priority in your schedule.</span>
6 0
3 years ago
You heat up a 3.0 kg aluminum pot with 1.5 kg of water from 5 to 90o Celsius. The specific heat capacity of Aluminum is 900 J/kg
Evgen [1.6K]

Answer:

<em>765,000Joules or 765kJ</em>

Explanation:

The Quantity of heat required is expressed as;

Q = (mcΔt)al + (mcΔt)water

m is the mass

c is specific heat capacity

Δt is the change in temperature

Q = (3(900)(90-5)) + (1.5(4200)(90-5))

Q = 2700*85 + 6300*85

Q = (2700+6300)85

Q = 9000*85

<em>Q = 765,000</em>

<em>Hence the amount of energy needed is 765,000Joules or 765kJ</em>

8 0
2 years ago
Does the size of a paper airplane affect how far it flies
astra-53 [7]
No the only thing that affects it is how it is built
4 0
3 years ago
A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnet
Fiesta28 [93]

Answer:

the charge of the particle is 2.47 x 10⁻¹⁹ C

Explanation:

Given;

mass of the particle, m = 6.64 x 10⁻²⁷ kg

velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

where;

q is the charge of the particle

a is the acceleration of the charge in the circular path

a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C

Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C

6 0
3 years ago
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