PHYSICS COLLEGE

In a region where there is a uniform electric field that is upward and has magnitude 3.80x104 N/C a small object is projected upward with an initial speed of 2.32 m/s The object travels upward a distance of 5.98 cm in 0 200 s. What is the object's charge-to-mass ratio q/m (magnitude and sign)?Assume g 9.80 m/s and ignore air resistance E3? C/kg q/m

Answers

Answer 1

Answer:

Answer:

6.03 x 10^-3 C/Kg

Explanation:

E = 3.8 x 10^4 N/C, u = 2.32 m/s, s = 5.98 cm = 0.0598 m, t = 0.2 s, g = 9.8 m/s^2

Acceleration on object is a .

Use second equation of motion.

S = u t + 1/2 a t^2

0.0598 = 2.32 x 0.2 + 0.5 x a x 0.2 x 0.2

0.0598 = 4.64 + 0.02 x a

a = - 229 m/s^2

Now, F = ma = qE

q / m = a / E = 229 / (3.8 x 10000)

q / m = 6.03 x 10^-3 C/Kg

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"At time t = 0 a 2330-kg rocket in outer space fires an engine that exerts" an increasing force on it in the +x-direction. This force obeys the equation Fx=At2, where t is time, and has a magnitude of 781.25 N when t = 1.27 s .What impulse does the engine exert on the rocket during the 1.50- s interval starting 2.00 s after the engine is fired?

Answers

Answer:

$Imp = 5626.488\,(kg\cdot m)/(s)$

Explanation:

First, it is required to model the function that models the increasing force in the +x direction:

$a =(781..25\,N)/((1.27\,s)^(2))$

$a = 484 (N)/(s^(2))$

The equation is:

$F_(x) = 484\,(N)/(s^(2))\cdot t^(2)$

The impulse done by the engine is given by the following integral:

$Imp=484\,(N)/(s^(2)) \int\limits^(3.50\,s)_(2\,s) {t^(2)} \, dt$

$Imp = 161.333\,(N)/(s^(2))\cdot [(3.50\,s)^(3)-(2\,s)^(3)]$

$Imp = 5626.488\,(kg\cdot m)/(s)$

A galilean telescope adjusted for a relaxed eye is 36cm long. If the objective lens has a focal length of 40cm, what is the magnification?

Answers

For this problem, we use the mirror equation which is expressed as:

1/di + 1/f = 1/d0

Magnification is expressed as the ratio of di and d0.

Manipulating the equation, we will have:

M = di/f +1
M = 36/40 + 1
M = 1.9

Hope this answers the equation.

A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turned on, the unit reaches its rated speed in 5 s, and when the power is turned off, the unit coasts to rest in 70 s. Assuming uniformly accelerated motion, determine the number of revolutions that the motor executes (a) in reaching its rated speed, (b) in coasting to rest.

Answers

Answer:

(a) θ1 = 942.5rad, (b) θ2 = 13195 rad

Explanation:

(a) Given

ωo = 0 rad/s

ω = 3600rev/min = 3600×2(pi)/60 rad/s

ω = 377rad/s

t1 = 5s

θ1 = (ω + ωo)t/2

θ1 = (377 +0)×5/2

θ1 = 942.5 rads

(b) ωo = 377rad/s

ω = 0 rad/s

t2 = 70s

θ2 = (ω + ωo)t/2

θ2 = (377 +0)×70/2

θ2 = 13195 rad

A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants appropriate units. Determine the magnetic field vector inside this wire.

Answers

Answer:

$B = \mu_0((1)/(3) (J_0)/(r_0) r^2)$

Explanation:

As the current density is given as

$J = (J_0)/(r_0)r$

now we have current inside wire given as

$i = \int J(2\pi r)dr$

$i = \int (J_0)/(r_0) r(2\pi r)dr$

$i = 2\pi (J_0)/(r_0) \int r^2 dr$

$i = (2)/(3) \pi (J_0)/(r_0) r^3$

Now by Ampere's law we will have

$\int B. dl = \mu_0 i$

$B. (2\pi r) = \mu_0((2)/(3) \pi (J_0)/(r_0) r^3)$

$B = \mu_0((1)/(3) (J_0)/(r_0) r^2)$

A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse through a high-powered telescope he owns. You are curious what the eclipse might look like from different perspectives in space. If the moon has a diameter of 2,159.14 miles, what is the maximum distance that it could be observed by the naked eye with enough detail that you could distinguish it from other celestial bodies (assuming that you have 20/20 vision)

Answers

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

θ = 1.22 550 10⁻⁹ / 0.002

θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

tan θ = y / L

y = L tan θ

y = 2,389 10⁵ tan 3,355 10⁻⁴

y = 8.02 10¹ mi

y = 80.2 mille

This is the smallest size of an object seen directly by the eye

Final answer:

An individual with 20/20 vision can observe the moon from a maximum distance of around 6200 km or 3850 miles. Beyond this distance, it might be difficult to distinguish the moon from other celestial objects without using a telescope. The use of a telescope can expand this range significantly.

Explanation:

The detailed observation of a lunar eclipsed, when viewed without any form of optical aid like a telescope, is contingent on many factors, one of which is the human eye's angular resolution—the eye's ability to differentiate between two separate points of light. For an average human eye with 20/20 vision, the angular resolution is approximately 0.02 degrees.

To calculate the maximum distance at which the moon could be observed clearly with the eye, the formula for small angle approximation can be used, which in this context is: Distance = Size / Angle = (2159.14 miles) / (0.02 degrees in radians). This calculates to a distance of approximately 6200 km or 3850 miles.

Beyond this distance, distinguishing the moon from other celestial bodies might be challenging using just the eye. Utilizing a high-powered telescope would significantly extend this range by magnifying the image, allowing clearer detail over much greater distances.

Learn more about Observing Moon here:

brainly.com/question/34031036

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1. If a net force of 412 N is required to accelerate an object at 5.82 m/s2, what must theobject's mass be?

Answers

Answer:

The mass of the object is approximately 70.79 kilograms

Explanation:

We use Newton's second law to solve this problem. This law states that the net force on an object equals the product of its mass times the acceleration:

$F_(net)=m\,a$

Therefore, for this case, since the net force on the object and its acceleration are given, we can use the equation above to solve for the unknown mass:

$F_(net)=m\,a\n412\,N=m\,(5.82\,(m)/(s^2) )\nm=(412\,N)/(5.82\,(m)/(s^2) ) \nm=70.79\.kg$

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