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olganol [36]
3 years ago
14

A 0.358 kg particle slides around a horizontal track. The track has a smooth, vertical outer wall forming a circle with a radius

of 1.66 m. The particle is given an initial speed of 8.19 m/s. After one revolution, its speed has dropped to 6.29 m/s because of friction with the rough floor of the track. Calculate the energy loss due to friction in one revolution.
Physics
1 answer:
oee [108]3 years ago
4 0

Answer:

The energy loss due to friction in one revolution is 4.92 J.

Explanation:

Given that,

Mass of particle = 0.358 kg

Radius = 1.66 m

Initial speed =8.19 m/s

Final speed = 6.29 m/s

We need to calculate the kinetic energy

Using formula of kinetic energy

\Delta K.E=\dfrac{1}{2}m(v_{f}^2-v_{i}^2)

Put the value into the formula

\Delta K.E=\dfrac{1}{2}\times0.358\times(8.19^2-6.29^2)

\Delta K.E=4.92\ J

Hence, The energy loss due to friction in one revolution is 4.92 J.

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A 46 g domino slides down a 30 degrees incline at a constant speed. What is the coefficient of friction?
blondinia [14]

Answer:

40

Explanation:

30

6 0
2 years ago
a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/
LiRa [457]

Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\

we isolate th (needed to reach the maximum height hmax)

th = \frac{v_{0}*sin(\alpha)}{g}

The formula describing vertical distance is:

y = Vy * t-g* t^{2} / 2

So, given y = hmax and t = th, we can join those two equations together:

hmax = Vy* th-g*th^{2}/2

hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)

if we launch a projectile from some initial height h all you need to do is add this initial elevation

hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)

hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft

6 0
3 years ago
Explain how grality and electric charge are different
stich3 [128]
A difference between the two forces is the fact that gravitation only attracts, while electrical forces attract when the electrical charges are opposite and repel if the charges are similar. Thus, gravitation is considered a monopole force, while electrostatics is a dipole force. Please mark me as a brainiest!!!
6 0
3 years ago
A steady beam of alpha particles (q = + 2e, mass m = 6.68 × 10-27 kg) traveling with constant kinetic energy 22 MeV carries a cu
cluponka [151]

Answer:

Explanation:

q = 2e = 3.2 x 10^-19 C

mass, m = 6.68 x 10^-27 kg

Kinetic energy, K = 22 MeV

Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A

(a) time, t = 2.8 s

Let N be the alpha particles strike the surface.

N x 2e = q

N x 3.2 x 10^-19 = i t

N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8

N = 2.36 x 10^12

(b) Length, L = 16 cm = 0.16 m

Let N be the alpha particles

K = 0.5 x mv²

22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²

v² = 1.054 x 10^15

v = 3.25 x 10^7 m/s

So, N x 2e = i x t

N x 2e = i x L / v

N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)

N = 4153.85

(c) Us ethe conservation of energy

Kinetic energy = Potential energy

K = q x V

22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V

V = 1.17 x 10^7 V

5 0
3 years ago
When the mass of the bottle is 0.125 kg, the KE is______ kg m2/s2.
DiKsa [7]
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2


Explanation:

(1) Given mass = 0.125 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2

(2) Given mass = 0.250 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2

(3) Given mass = 0.375 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2

(4) Given mass = 0.500 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
3 0
3 years ago
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