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egoroff_w [7]
3 years ago
10

a car traveling north at 17.7 m/s. after 12 seconds its velocity is 14.1 m/s in the same direction. find the magnitude and direc

tion of the car's average acceleration
Physics
1 answer:
elixir [45]3 years ago
7 0
Acceleration is equal to the change in velocity per unit time. In this case, velocity has changed from 17.7 m/s to 14.1 m/s, which is a decrease of 3.6 m/s. This takes place in 12 seconds, so the average acceleration is 3.6 m/s / 12s = 0.3 m/s^2. The direction of this acceleration is south, since the car's northbound speed is decreasing.
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To answer the question, the lower mass limit for a main sequence star is about 0.08. If the mass of a main sequence star is lower than the above-mentioned value, there would be a deficit or insufficiency of gravitational force to generate a standard temperature for hydrogen core fusion to take place and the underdeveloped star would form into a brown dwarf instead.
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2 years ago
An ideal gas is enclosed in a cylinder that has a movable piston on top. The piston has a mass m and an area A and is free to sl
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given mass of piston \left ( m\right )

no. of moles =n

Given Pressure remains same

Temperature changes from T_1 to T_2

Work done\left ( W\right ) is given by=\int_{V_1}^{V_2}PdV

W=P\left ( V_2-V_1\right )

also PV_1=nRT_1

PV_2=nRT_2

W=nR\left ( T_2-T_1\right )

4 0
3 years ago
How would you obtain a mean value for the specific heat<br> capacity of a material?
lidiya [134]
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8 0
2 years ago
Explain why a 10 kg stone and 1 kg stone dropped from the same height
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GIVE ME POINTS
3 0
3 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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