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ella [17]
3 years ago
5

a train leaves a station heading east at 50 mph from that same station a car drives north at a speed of 30mph after 3 hours how

far apart are the two vehicles?
Physics
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

d = 175 miles

Explanation:

Train is moving towards East with constant speed of 50 mph

While car is moving at speed of 30 mph

so after t = 3 hours

the distance moved by the train is given as

d_1 = vt

d_1 = (50)(3) = 150 miles

at the same time the distance moved by the car is given as

d_2 = 30(3) = 90 miles

now we know that both car and train is moving perpendicular to each other

So the distance between them after t = 3 hours is given as

d= \sqrt{d_1^2 + d_2^2}

d = \sqrt{150^2 + 90^2}

d = 175 miles

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A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the
Scorpion4ik [409]

Answer:

(a) Current is 2831.93 A

(b) 8.40A/m^2

(c) \rho =15.52\times 10^{-9}ohm-m

Explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2

Resistance R=11.9mohm=11.9\times 10^{-3}ohm

Potential difference V = 33.7 volt

(A) current is equal to

i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A

(B) Current density is equal to

J=\frac{i}{A}

J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2

(c) Resistance is equal to

R=\frac{\rho l}{A}

11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}

\rho =15.52\times 10^{-9}ohm-m

4 0
3 years ago
A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp
amm1812

Answer:

the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = -\beta\frac{dx}{dt} = \frac{-10dx}{dt}

\beta = 10

By Newtons second law ,

The diffrential equation of motion with damping is given by

m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}

substitute the value of m =1kg, k = 21N/M, and \beta = 10

1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}

\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0

suppose the equation of the form x =e^m^t,

and the auxilliary equation is given by

m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3

The general solution for the above differential equation is

x(t) =C_1e^{-3t}+C_2e^{-7t}

Derivate with respect to t

x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

x'(0) = 0

substitute the initial condition

C_1 +C_2 = 1

-3C_1-7C_2=0

Solve the above equation to get C₁ and C₂

C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}

substitute for C₁ and C₂ in general solution

x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

Thus the required value is x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}

3 0
3 years ago
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lara31 [8.8K]

Answer:

Check the explanation

Explanation:

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Kindly check the attached image below to get the step by step explanation to the question above.

8 0
3 years ago
Please please help me out I’m lost
disa [49]

Answer:

Option B

The correct answer is 2

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Nonrenewable resources are natural resources that are limited in supply and cannot be replaced as quickly as they are used up.
zysi [14]

Answer:

True

Explanation:

Nonrenewable resources ARE limited in supply. They don't get replaced at the speed they get made. For example: we pump crude oil from the ground at a rate that makes it impossible for crude oil to be replaced. Crude oil takes millions of years to produce

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