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AlladinOne [14]
2 years ago
6

Vector A⃗ has a magnitude of 3.00 and is directed parallel to the negative y-axis and vector B⃗ has a magnitude of 3.00 and is d

irected parallel to the positive y-axis. Determine the magnitude and direction angle (as measured counterclockwise from the positive x-axis) of vector C⃗ , if C⃗ =A⃗ −B⃗ .
C = 6.00; θ = 270˚
C = 3.00; θ = 270˚
C = 6.00; θ = 90˚
C = 3.00; θ = 90˚

Physics
2 answers:
Leona [35]2 years ago
5 0

The correct answer between all the choices given is the first choice, which is "<span>C = 6.00; θ = 270˚"</span><span>. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.</span>

tangare [24]2 years ago
3 0

<u>Answer</u>

C = 6.00; θ = 270˚

<u>Explanation</u>

A vector is described by giving both its magnitude and direction.

A = 3.00; 270°

B = -3.00; 90°

C = A - B. From this expression, it can be seen that we have already reversed the direction of B. So vector C is in the direction of A.

C = A - B

C = 3.00 - (-3.00)

= 6.00


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Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
Bricks and insulation are used to construct the walls of a house. The
ira [324]

Answer:

\dot Q=350.438\ W

Explanation:

Given:

<u>the thermal resistance in the form of </u>

R_1=\frac{x_1}{k_1} =0.095\ m^2.^{\circ}C.W^{-1}

R_2=\frac{x_2}{k_2} =0.704\ m^2.^{\circ}C.W^{-1}

where:

x_1\  \&\ x_2 are the thickness of the respective bricks

k_1\ \&\ k_2 are the respective coefficient of conductivity

temperature inside the house, T_h=24\ ^{\circ}C

temperature outside the house, \ T_c=10^{\circ}C

area of the wall, A=20\ m^2

Since the bricks and insulation are used to construct a wall then they must be used in series for better shielding.

<u>Using Fourier's law:</u>

\dot Q=k.A.\frac{dT}{x}

\dot Q={dT}\div {\frac{x}{k.A} }

in series the resistances get add up

\dot Q=dT\div (\frac{x_1}{k_1.A}+\frac{x_2}{k_2.A} )

\dot Q=(24-10)\div (\frac{0.095 }{20}+ \frac{0.704 }{20} )

\dot Q=350.438\ W

7 0
2 years ago
Dawn and Aram have stretched a slinky between them and begin experimenting with waves. As the frequency of the waves is doubled
m_a_m_a [10]

Answer:

halved

Explanation:

The velocity of the a wave is obtained by multiplying the frequency and wavelength.

v=f\lambda\\\Rightarrow f=\frac{v}{\lambda}\\\Rightarrow \lambda=\frac{v}{f}

Where

v = Velocity

f = Frequency

\lambda = Wavelength

The velocity here is constant. So, if the frequency is doubled the wavelength is halved.

6 0
2 years ago
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w
OLga [1]

Answer:

a). va=17.23 \frac{m}{s} or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65

The motion is an inelastic collision so

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

The force of the motion is contrarest by the force of friction so

F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s

So the velocity final can be find using this time

v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}

a).

Replacing in the first equation the final velocity can find the initial velocity

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

v_{b}=0

v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}

b).

35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

Ek=\frac{1}{2}*m*(v)^{2}\\  Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ

8 0
2 years ago
A moving roller coaster speeds up with constant acceleration for 2.3\,\text{s}2.3s2, point, 3, start text, s, end text until it
Ad libitum [116K]

Answer:

Δx=(v+v0/2)t

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In this problem, the target unknown is the initial velocity v_0v  

0

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v, start subscript, 0, end subscript of the roller coaster.

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