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3 years ago

5 cm ______ 500 mm.

Physics

2 answers:

cestrela7 [59]3 years ago

5 0

Answer:

$c)Is \: equal \: to$

Yuri [45]3 years ago

4 0

Answer:b

Explanation: each cm is 10mm so 5cm is 50mm that is less than 500mm

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The Sun appears to move relative to the stars. For example, right now the Sun is in front of the constellation Leo, but in a mon

Crank

Answer:

0.98°

Explanation:

The Sun appears to move across various constellation because of the Earth's revolution around it. This apparent motion is essentially same as the actual motion of the Earth. The orbit of Earth around the Sun is almost circular.

Time taken to complete one revolution = 365 days

Degrees traveled in 365 Days = 360°

The motion per day is

$=\frac{360}{365}$

= 0.98° per day

7 0

3 years ago

wo parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude

melisa1 [442]

Answer:

-18896.49 V/m

Explanation:

Distance between the two plates = 10 cm = 10 x $10^{-2}$ m = 0.1 m

Also, one of the plates is taken as zero volt.

a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V

b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V

Potential field strength = -dV/dx

where dV is voltage difference between these points,

dx is the difference in distance between these points

For the first case above,

potential field strength = -393/0.0705 = -5574.46 V/m

For the second case ,

potential field strength = -393/0.0295 = -13322.03 V/m

Magnitude of the field strength across the plates will be

-5574.46 + (-13322.03) = -5574.46 + 13322.03 = -18896.49 V/m

6 0

4 years ago

A U-shaped valley is most often formed by

Elza [17]

formed by the process of glaciation.

5 0

3 years ago

Read 2 more answers

A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.325-kg particle slides around the inside edge o

aleksandrvk [35]

Answer:

a) W = - 6.825 J, b) θ = 1.72 revolution

Explanation:

a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle

W = ΔK

W = K_f - K₀

W = ½ m v_f² - ½ m v₀²

W = ½ 0.325 (5.5² - 8.5²)

W = - 6.825 J

b) find us the coefficient of friction

Let's use Newton's second law

fr = μ N

y-axis (vertical) N-W = 0

fr = μ W

work is defined by

W = F d

the distance traveled in a revolution is

d₀ = 2π r

W = μ mg d₀ = -6.825

μ = $\frac{ -6.825}{d_o \ mg}$

The total work as the object stops the final velocity is zero v_f = 0

W = 0 - ½ m v₀²

W = - ½ 0.325 8.5²

W = - 11.74 J

μ mg d = -11.74

we subtitle the friction coefficient value

( $\frac{-6.8525 }{d_o mg}$ ) m g d = -11.74

6.825 $\frac{d}{d_o}$ = 11.74

d = 11.74/6.825 d₀

d = 1.7201 2π 0.400

d = 4.32 m

this is the total distance traveled, the distance and the angle are related

θ = d / r

θ = 4.32 / 0.40

θ = 10.808 rad

we reduce to revolutions

θ = 10.808 rad (1rev / 2π rad)

θ = 1.72 revolution

3 0

3 years ago

A small pebble and one large boulder start at the same height and begin rolling down the side of a mountain. Which object would

vitfil [10]

Answer:

The small pebble

Explanation:

Since the potential energy, P.E lost equals kinetic energy, K.E gained,

P.E = K.E

P.E = mgh = K.E

So, K.E = mgh where g = acceleration due to gravity and h = height of drop

Since h and g are constant

K.E ∝ m

So, the kinetic energy of the object is directly proportional to its mass. Thus, the object with the smaller mass has the lesser kinetic energy.

Since the object with the smaller mass is the small pebble, so the small pebble would have less kinetic energy as it crashes on the road at the bottom of the mountain.

8 0

3 years ago