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Anni [7]
3 years ago
15

PHYSICS

Physics
1 answer:
quester [9]3 years ago
7 0

Answer:

c) 8.3 x 10^24 molecules

Explanation:

First of all, we convert the volume of the glass of water from mL to cm^3:

V=250 mL = 250 cm^3

The relationship between mass and volume is:

d=\frac{M}{V}

where

d=1.00 g/cm^3 is the density of water

M is the total mass of the glass of water

V=250 cm^3 is the volume of the water in the glass

Solving for M,

M=dV=(1.00)(250)=250 g = 0.250 kg

Now we know that the mass of a single molecule of water is

m=3.0\cdot 10^{-26}kg

The total mass of the water in the glass can be written as

M=Nm

where

N is the number of molecules in the glass of water

Solving for N, we find:

N=\frac{M}{m}=\frac{0.250}{3.0\cdot 10^{-26}}=8.3\cdot 10^{24}

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The watch hand covers an angular displacement of 2π radians in 60 seconds.

ω = 2π/60
ω = 0.1 rad/s

v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s
4 0
3 years ago
8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer
Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \  by \ the \ effort}{disatnce \ moved \  by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m

Therefore, the distance that the object is raised above its initial position is 5.625 m.​

3 0
3 years ago
A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.
alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

F_g = F_n + F_a

given that

F_g = mg

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

F_n = 150 N

now from above equation

30*10 = 150 + F_a

F_a = 300 - 150 = 150 N

So force applied by the person must be 150 N

7 0
3 years ago
Lucia raced her car on a raceway
denpristay [2]

Answer:

Good question to ask in physics, sir maam

8 0
3 years ago
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What is the equation for frequency, wavelength, and speed of a wave?
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Answer:

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