Will ice melt faster in salt, sugar or water?

20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?

Tresset [83]

Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance
so Work done = 0.085 x 20g x 28
= 466.956 J

7 0

3 years ago

Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone

anygoal [31]

Answer:

The fundamental frequency at which the sound of speakers at the microphone produce constructive interference is 801.076458 Hz

Explanation:

For a given arrangement having constructive interference, we have;

R₁ - R₂ = 2·x = 0 + n·λ

The distance from one speaker to the microphone. R₁ = 4.50 m

The distance between the two speakers = 2.00 m

The angle formed between the direction from the microphone to the speaker closest and the directional path between the speakers = 90°

Therefore, by Pythagoras's theorem, the distance from the speaker furthest from the microphone, to the microphone, 'R₂' is given as follows;

R₂ = √(4.50² + 2.00²) = √(24.25) = 4.9244289009 ≈ 4.924

∴ R₂ ≈ 4.9244289 m

R₂ - R₁ = 4.9244289 m - 4.5 m = 0.4244289 m

For constructive interference, R₂ - R₁ =0.4244289 m = n·λ

For n = 1, we have;

R₂ - R₁ =0.4244289 m = n·λ = 1 × λ = λ

λ = 0.4244589 m

f = v/λ = 340 m/sec/(0.4244289 m) ≈ 801.076458 Hz

Therefore the lowest possible fundamental frequency at which the speakers produce constructive interference, f = 801.076458 Hz

7 0

3 years ago

Assuming that each nucleus is roughlyspherical and that its mass is roughly equal to A (in atomic mass units {\rm u}), what is t

lara [203]

Answer:

ρ/ρ2 = 3 / R₀ the two densities are different

Explanation:

Density is defined as

ρ = M / V

As the nucleus is spherical

V = 4/3 π r³

Let's replace

ρ = A / (4/3 π R₀³)

ρ = ¾ A / π R₀³

b)

ρ2 = F / area

The area of a sphere is

A = 4π R₀²

ρ2 = F / 4π R₀²

Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.

Let's look for the relationship of the two densities

ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)

ρ /ρ2 = 3 (A / F) (1 / R₀)

In this case it does not say that the nucleon number is A (F = A), the relationship is

ρ/ρ2 = 3 / R₀

I see that the two densities are different

3 0

3 years ago

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

3 years ago

If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi

netineya [11]

Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

$F_s = \mu_s*N$

Where $F_s$ is the static friction force, $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, $N = 110\text{ N}$ , to make the puppy moving we need to use a force of 80 N, therefore, $F_s = 80 \text{ N}$ , so we can solve for the coefficient as shown below:

$80 = \mu_s*110\\\mu_s = \frac{80}{110} = 0.7273\\$

The coefficient of static friction between the puppy and the floor is 0.7273.

5 0

3 years ago