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WARRIOR [948]
3 years ago
9

How long does it take an object to accelerate from 12 m/s to the right to 32 m/s to

Physics
1 answer:
Valentin [98]3 years ago
5 0

Answer:

4 seconds

Explanation:

Given:

v₀ = 12 m/s

v = 32 m/s

a = 5 m/s²

Find: t

v = at + v₀

32 m/s = (5 m/s²) t + 12 m/s

t = 4 s

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150 cm would be 1.5 m
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How to determine the amount of resistance when two or more resistors are in parallel.
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Answer: 1/R = 1/R1 + 1/R2+ ...+ 1/Rn

R is resistance of system in which there are resistors R1, R2 , ... Rn parallel.

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Ships A and B leave port together. For the next two hours, ship A travels at 40.0 mph in a direction 35.0° west of north while t
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Answer:

Explanation:

Given

Ship A velocity is 40 mph and is traveling 35 west of north

Therefore in 2 hours it will travel 40\times 2=80 miles

thus its position vector after two hours is

r_A=-80sin35\hat{i}+80cos35\hat{j}

similarly B travels with 20 mph and in 2 hours

=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}

Thus distance between A and B  is

r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}

|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}

|r_{AB}|=103.45 miles

Velocity of A

v_A=-40sin35\hat{i}+40cos35\hat{j}

Velocity of B

v_B=20sin80\hat{i}+20cos80\hat{j}

Velocity of A w.r.t B

v_{AB}=v_A-v_B

v_{AB}=\left ( -20sin80-40sin35\right )\hat{i}+\left ( 40cos35-20cos80\right )\hat{j}

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3 years ago
During a drive by golfer Tiger Woods, the angular velocity of his club is zero at the top of his backswing and 30 rad/s at the b
ddd [48]

Explanation:

Below is an attachment containing the solution

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4 years ago
A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un
Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within  1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

7 0
3 years ago
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