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3 years ago

Which three quantities can be used to calculate acceleration?

1 answer:

4 0

D is the correct answer, assuming that this is the special case of classical kinematics at constant acceleration. You can use the equation V = Vo + at, where Vo is the initial velocity, V is the final velocity, and t is the time elapsed. In D, all three of these values are given, so you simply solve for a, the acceleration.
A and C are clearly incorrect, as mass and force (in terms of projectile motion) have no effect on an object's motion. B is incorrect because it is not useful to know the position or distance traveled, unless it will help you find displacement. Even then, you would not have enough information to use a kinematics equation to find a.

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Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87

marshall27 [118]

Answer:

a) $\lambda=0.935\ \textup{m}$

b) $f=36.19\approx 36\ \textup{Hz}$

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

$L= n\frac{\lambda}{2}$

where,

$\lambda$ = wavelength

on substituting the values, we get

$1.87= 4\times \frac{\lambda}{2}$

$\lambda=0.935\ \textup{m}$

b) Speed of the wave (v) in the string is given as:

$v =f\lambda$

also,

$v=\sqrt\frac{T}{(\frac{m}{L})}$

equating both the formula for 'v' we get,

$f\lambda=\sqrt\frac{T}{(\frac{m}{L})}$

on substituting the values, we get

$f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}$

$f=\frac{33.84}{0.935}$

$f=36.19\approx 36\ \textup{Hz}$

5 0

2 years ago

Light of wavelength 500nm is passed through a diffraction grating which has 400 lines per mm. What is the angular spectrum betwe

Citrus2011 [14]

Answer:

because of the gravity of the earth

8 0

1 year ago

A football player, with a mass of 69.0 kg, slides on the ground after being knocked down. At the start of the slide, the player

White raven [17]

Answer:

(a) -472.305 J

(b) 1 m

Explanation:

(a)

Change in mechanical energy equals change in kinetic energy

Kinetic energy is given by $0.5mv^{2}$

Initial kinetic energy is $0.5\times 69\times 3.7^{2}=472.305 J$

Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero

Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence

0-472.305 J=-472.305 J

(b)

From fundamental kinematic equation

$v^{2}=u^{2}+2as$

Where v and u are final and initial velocities respectively, a is acceleration, s is distance

Making s the subject we obtain

$s=\frac {v^{2}-u^{2}}{-2a}$ but a=\mu g hence

$s=\frac {v^{2}-u^{2}}{-2\mu g}=\frac {0^{2}-3.7^{2}}{-2*0.7*9.81}=0.996796272\approx 1 m$

7 0

2 years ago

A block with mass 0.5kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2m. whe

dangina [55]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
Read more on Brainly.com - brainly.com/question/1581851#readmore

4 0

3 years ago

Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of

Radda [10]

According to the information provided to define an average density, it is necessary to use the concepts related to mass calculation based on gravitational constants and radius, as well as the calculation of the volume of a sphere.

By definition we know that the mass of a body in this case of the earth is given as a function of

$M = \frac{gr^2}{G}$

Where,

g= gravitational acceleration

G = Universal gravitational constant

r = radius (earth at this case)

All of this values we have,

$g = 9.8m/s^2\\G = 6.67*10^{-11} m^3/kg*s^2\\r = 6378*10^3 m$

Replacing at this equation we have that

$M = \frac{gr^2}{G} \\M = \frac{(9.8)(6378*10^3)^2}{6.67*10^{-11}} \\M = 5.972*10^{24}kg$

The Volume of a Sphere is equal to

$V = \frac{4}{3}\pi r^3\\V = \frac{4}{3} \pi (6378*10^3)^3\\V = 1.08*10^{21}m^3$

Therefore using the relation between mass, volume and density we have that

$\rho = \frac{m}{V}\\\rho = \frac{5.972*10^{24}}{1.08*10^{21}}\\\rho = 5.52*10^3kg/m^3$

6 0

2 years ago