Answer:
a.) 10Hz
b.) 0.1 s
c.) 187.4 m/s
d.) -412.6 m/s^2
Explanation:
Given that an object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in cm and t in seconds. Give decimal answers below.
(a) How many complete back-and-forth motions (from the origin to the right, back to the origin, to the left and finally back to the origin) does the object make in one second?
from the equation given, the angular speed w = 20π
but w = 2πf
where f = frequency.
substitute w for 20π
20π = 2πf
f = 20π/2π
f = 10 Hz
(b) What is t the first time that the object is at its farthest right?
since F = 1/T
T = 1 / f
T = 1/10
T = 0.1 s
Therefore, the t of first time that the object is at its farthest right is 0.1 s
(c) At the time found in part (b), what is the object's velocity?
The velocity can be found by differentiating the equation;
x(t) = 3sin(20πt)
dx/dt = 60πcos(20πt)
where dx/dt = velocity V
V = 60πcos(20π * 0.1)
V = 187.4 m/s
(d) At the time found in part (b), what is the object's acceleration?
to get the acceleration, differentiate equation V = 60πcos(20πt)
dv/dt = -1200πSin(20πt)
dv/dt = acceleration a
a = -1200πSin(20πt)
substitute t into the equation
a = -1200πSin(20π * 0.1)
a = - 412.6 m/s^2
Answer:
The answer is B) 3 seconds
Explanation:
I just took the test on 2020 edge and got it right
Answer:
Explanation:
The attached image shows the system expressed in the question.
We can define an expression for the system.
The equivalent equation for the system would be
so, the input signal could be expressed in dB terms
![P_{in} [dB] = 10 log_{10}(P_{in}) \\P_{in} [dB] = 10 log_{10}(0.4)\\P_{in} [dB] = -3.97 dB](https://tex.z-dn.net/?f=P_%7Bin%7D%20%5BdB%5D%20%3D%2010%20log_%7B10%7D%28P_%7Bin%7D%29%20%5C%5CP_%7Bin%7D%20%5BdB%5D%20%3D%2010%20log_%7B10%7D%280.4%29%5C%5CP_%7Bin%7D%20%5BdB%5D%20%3D%20-3.97%20dB)
(1)
so the output signal could be expressed as.
The gain should be expressed in dB terms and power in dBm terms so
using the (1) equation to find it in terms of Watts
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
Answer:
Explanation:
Heat capacity A = 3 x heat capacity of B
initial temperature of A = 2 x initial temperature of B
TA = 2 TB
Let T be the final temperature of the system
Heat lost by A is equal to the heat gained by B
mass of A x specific heat of A x (TA - T) = mass of B x specific heat of B x ( T - TB)
heat capacity of A x ( TA - T) = heat capacity of B x ( T - TB)
3 x heat capacity of B x ( TA - T) = heat capacity of B x ( T - TB)
3 TA - 3 T = T - TB
6 TB + TB = 4 T
T = 1.75 TB
