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Vinvika [58]
3 years ago
14

How much work is needed to stop a 1,110 kg car that is moving straight down the

Physics
1 answer:
schepotkina [342]3 years ago
3 0

Answer:

382.74 kJ.

Explanation:

The work that must be done to stop an 1100 kg car travelling at 59  km/h is  - 382.74 kJ.

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Solve each of these problems, remembering to include formula, calculations, and units!
KATRIN_1 [288]

Answer:

I nEeD heLp

Explanation:

HelP

5 0
3 years ago
A measure of randomness or disorder
vodka [1.7K]
It is simply called Entropy.

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3 years ago
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Calculate the percentage of an iceberg submerged beneath the surface of the ocean given that the density of ice is 916.3kg/m3 an
pogonyaev

Answer:

The percentage of an iceberg submerged beneath the surface of the ocean = 89.67%

Explanation:

Let V be the total volume of the iceberg

Let x be the volume of iceberg submerged

According to Archimedes principle,

weight of the iceberg = weight of the water displaced (that is, weight of x volume of water)

Weight of the iceberg = mg= ρ(iceberg) × V × g

Weight of water displaced = ρ(fluid) × x × g

We then have

ρ(iceberg) × V × g = ρ(fluid) × x × g

(x/V) = ρ(iceberg) ÷ ρ(fluid) = 916.3 ÷ 1021.9 = 0.8967 = 89.67%

Hope this Helps!!!!

6 0
3 years ago
Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32 .1 kj/mol. de
atroni [7]
Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c
5 0
4 years ago
A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,
gavmur [86]

Answer:

a. ρ_\beta=1.996J/m^3

b. U_E=9.445x10^{-15} J/m^3

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

i=14A, d=2.5mm, R=3.3Ω, l=1 km, E_o=8.85x10^{-12}F/m, u_o=4*x10^{-7}H/m

ρ_\beta=\frac{\beta^2}{2*u_o}

ρ_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2

ρ_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}

ρ_\beta=1.996J/m^3

b. The electric field can be find using the equation:

U_E=\frac{1}{2}*E_o*E^2

E=(\frac{i*R}{l})^2

U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2

U_E=9.445x10^{-15} J/m^3

4 0
3 years ago
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