All categories

2 years ago

Please answer of this question

Physics

1 answer:

Bogdan [553]2 years ago

8 0

Answer:

$\frac{\pi }{15}$ m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

Angle between the two hands = 2/12 * 360

= 60°

Arc Length = $2\pi r(\frac{60}{360} )$

= $2\pi (0.2)(\frac{1}{6} )\\= \frac{\pi }{15} m$

You might be interested in

Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction betw

jarptica [38.1K]

Answer:

a) W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d) the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

θ = 90 cos 90 = 0

W=0

In conclusion the work is zero

b) The friction force opposes the displacement whereby the angle is

θ = 180 cos 190 = -1

W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

m g sin θ L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀= + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

ΔU = -U₀

the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

W_Total = W_weight - W_roce

as the body moves down

W_Total> 0

Therefore the total work is positive

3 0

2 years ago

Calculation using Ohm's Law: If a circuit has a voltage of 500 V and a resistance of 250 ohms, what is the current?

konstantin123 [22]

I = V / R

Current = (voltage) / (resistance)

Current = (500 V) / (250 ohms)

Current = (500/250) Amperes

<em>Current = 2 Amperes</em>

6 0

2 years ago

Read 2 more answers

Consider the following arrangement with a frictionless/massless pulley. Determine the force F required to move block A if the co

Delicious77 [7]

Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.

Explanation: Hope this helped :)

6 0

2 years ago

A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

centripetal force acting on satellite = $\dfrac{mv^2}{r}$

gravitational force = $\dfrac{GMm}{r^2}$

equating both the above equation

$\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$

$v = \sqrt{\dfrac{GM}{r}}$

$v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}$

v = 5578.5 m/s

b) $T= \dfrac{2\pi\ r}{v}$

$T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}$

T = 14416.92 s

$T = \dfrac{14416.92}{3600}\ hr$

T = 4 hr

c) gravitational force acting

$F = \dfrac{GMm}{r^2}$

$F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}$

F = 5202 N

4 0

3 years ago

S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum

erastovalidia [21]

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be $\frac{4.83598}{c^{\frac{1}{3} } }$ .