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stellarik [79]
4 years ago
13

(a) A small object with mass 3.75 kg moves counterclockwise with constant speed 1.55 rad/s in a circle of radius 2.55 m centered

at the origin. It starts at the point with position vector 2.55 i m. Then it undergoes an angular displacement of 8.95 rad.
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis?
(c) What is its velocity?
Physics
1 answer:
Eddi Din [679]4 years ago
8 0

Answer:3.95 m/s

Explanation:

Given

mass of object m=3.75 kg

\omega =1.55 rad/s

radius of circle =2.55 m

initial Position r=2.55 \hat{i}

angular displacement \theta _0=8.95 rad

8.95 radian can be written as

1.42 (2\pi )

i.e. Particle is at first quadrant with \theta =0.4242\pi \times \frac{180}{\pi }

\theta =76.36^{\circ}

(c)velocity is v=\omgea \times r

v=1.55\times 2.55=3.95 m/s

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Two small spheres spaced 20.0 cm apart have equal charge. How many excess electrons must be present on each sphere if the magnit
Snowcat [4.5K]

Answer:

894 electrons

Explanation:

The electrostatic force between the two charges is given by:

F=\frac{k q_1 q_2}{r^2}

where we have

F=4.57\cdot 10^{-21} N is the force

k is the Coulomb's constant

q1 = q2 =q is the magnitude of the charge on each sphere

r = 20.0 cm = 0.20 m is the distance between the two spheres

Substituting and solving for q, we find the charge on each sphere:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(4.57\cdot 10^{-21} N)(0.20 m)^2}{9\cdot 10^9 Nm^2C^{-2}}}=1.43\cdot 10^{-16} C

And since each electron has a charge of

e=1.6\cdot 10^{-19}C

the net charge on each sphere will be given by

q=Ne

where N is the number of excess electrons; solving for N,

N=\frac{q}{e}=\frac{1.43\cdot 10^{-16}C}{1.6\cdot 10^{-19}C}=894

7 0
4 years ago
Each roller under a conveyor belt has a radius of 0.5 meters. the rollers turn at a rate of 30 revolutions per minute. what is t
svp [43]
Distance in a minute=<span>0.5 times 30=15 meters
distance in a second</span><span>=15 divided by 60=0.25 meters per second
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3 0
3 years ago
Read 2 more answers
A roller coaster starts from rest at point A. What is its speed at point C if the track is frictionless.
OleMash [197]
At A, coaster is only associated with potential energy.
At B, coaster is associated with kinetic as well as potential energy.
Since the track is frictionless, no energy will be lost when coaster reaches from point A to point B. Therefore, according to conservation of energy, total energy at A should be equal to total energy at B.
Total energy at A = mgh = mg(12) 
Total energy at B = mgh+ mv²/2 = mg(2) + mv²/2
∴12mg = 2mg + mv²/2
∴(12g-2g)×2 = v²
∴v² = 20g
∴v = 14m/s.

Again conserving energy at points B and C.
Total energy at B = 2mg + m(14)²/2 
Total energy at C = 4mg + mv²/2
∴2mg + m(14²)/2 = 4mg + mv²/2
Solving this you get,
v = 12.52 m/s.
Therefore, speed of roller coaster at point C is 12.52 m/s.
8 0
4 years ago
Sabe-se que um alqueire paulista equivale a 24200 metros quadrados. Uma chácara retangular tem um alqueire e mede 100m de frente
sattari [20]

Answer:

b = 242 m

Explanation:

A = 24200 m²

a = 100 m

b = ?

A seguinte fórmula é aplicada

A = a*b

⇒  b = A / a

⇒  b = (24200 m²) / (100 m)

⇒  b = 242 m

5 0
3 years ago
A force of 20 N acts over an area of 2 m2. What is the pressure? ​
ycow [4]

\text{Given that,}\\\\\text{Force, F = 20 N.}  ~ \\\\ \text{Area , A = 2 m}^2 . \\\\ \text{Pressure, P =? }\\\\P = \dfrac FA = \dfrac{20}2 = 10~~ \text{Nm}^{-2}

5 0
3 years ago
Read 2 more answers
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