Answer:
Data
Explanation:
A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the driver notices a tree limb that has fallen on the road and brakes hard for t2 = 5 s with a constant acceleration of a2 = -5.93 m/s2.Write an expression for the car's speed just before the driver begins braking, v1.If the limb is on the road at a distance of 550 meters from where the car began its initial acceleration, will the car hit the limb?
Answer:
1) an expression for the car's speed is given as
v = u + at
where
v is the car's speed
u is the initial speed of the car
a is the car's acceleration
t is the time spent accelerating
2) The car does not hit the tree limb
Explanation:
The initial velocity of the car = 0 m/s (since it accelerates from rest)
acceleration of the car = 1.76 m/s
time spent accelerating = 20 s
For the car's speed just before the driver begins braking, we use the expression
v = u + at
where v is the final speed of the car just before the driver begins braking
u is the initial velocity with which the car starts moving
a is the acceleration of the car
t is the time spent accelerating from u to v
substituting values, we have
v = 0 + 1.76(20)
v = 0 + 35.2
the car's speed v = 35.2 m/s
In this time the car accelerates, the car moves a distance given by
s = ut + [tex]\frac{1}{2}[/tex]a[tex]t^2[/tex]
where s is the distance covered in this time
u is the initial speed of the vehicle
a is the acceleration
t is the time taken
substituting, we have
s = 0(20) + [tex]\frac{1}{2}[/tex](1.76)[tex]20^{2}[/tex]
s = 0 + 352
distance s = 352 m
When the driver brakes, we have
time spent braking = 5 s
acceleration = -5.93 m/s
and the distance to the limb = 550 m from where the car begun
to get the distance covered in this period, we use the expression
s = ut + [tex]\frac{1}{2}[/tex]a[tex]t^2[/tex]
where s is the distance traveled at this time
u is the speed of the car before it starts braking = 35.2 m/s
a is the acceleration at this point
t is the time taken to decelerate to a stop
substituting values, we have
s = 35.2(5) + [tex]\frac{1}{2}[/tex](-5.93 x [tex]5^2[/tex])
s = 176 - 74.125
s = 101.88 m
Total distance moved by the car = 352 m + 101.88 m = 453.88 m
Since the total distance traveled by the car is less than the distance from the starting point to the place where the tree limb is, the car does not hit the tree limb.
A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.500 m2 . At the window, the electric field of the wave has an rms value 0.0600 V/m .
How much energy does this wave carry through the window during a 30.0-s commercial? Express your answer with the appropriate units.
Answer:
The energy of the wave is 1.435 x 10⁻⁴ J
Explanation:
Given;
area of the window, A = 0.5 m²
the rms value of the field, E = 0.06 V/m
The peak value of electric field is given by;
[tex]E_o = \sqrt{2} *E_{rms}\\\\E_o = \sqrt{2}*0.06\\\\E_o = 0.0849 \ V/m[/tex]
The average intensity of the wave is given by;
[tex]I_{avg} = \frac{c \epsilon_o E_o^2 }{2}\\\\I_{avg} = \frac{(3*10^8)( 8.85*10^{-12}) (0.0849)^2 }{2}\\\\I_{avg} = 9.569*10^{-6} \ W/m^2[/tex]
The average power of the wave is given by;
P = I x A
P = (9.569 x 10⁻⁶ W/m²) (0.5 m²)
P = 4.784 x 10⁻⁶ W
The energy of the wave is given by;
E = P x t
E = (4.784 x 10⁻⁶ W)(30 s)
E = 1.435 x 10⁻⁴ J
Therefore, the energy of the wave is 1.435 x 10⁻⁴ J
Place gamma rays, infrared, microwaves, radio waves, ultraviolet, visible light, and x-rays in order from largest wavelength to smallest wavelength.
Answer:
Going by EM SPECTRUM WE HAVE
radio waves, microwaves, infrared, VISIBLE LIGHT, ultraviolet, X-rays, GAMMA RAYS
Explanation:
BECAUSE
V= WAVELENGTH/ FREQUENCY
AS FREQUENCY INCREASES WAVELENGTH DECREASE AN VICE VERSA
Given that the lines are parallel, what would be the value of angle "a" in the diagram found below?
Explanation:
Hey there!!
a = 40° { corresponding angles on a parallel lines are equal}.
As A and B are parallel, a and 40° are corresponding angles.
Hope it helps...
The value of angle "a" will be 40°
What is corresponding angle ?The angles which are on the same side of one of two lines cut by a transversal and are on the same side of the transversal is called corresponding angles.
since , angle a and angle B = 40° (given) are on the same relative position at the intersection where a straight line crosses two others and since both the lines are parallel to each other hence ,both the angles are said to be corresponding angles .This implies both the angles must be equal .
angle a = 40°
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what is the difference between each distance traveled and displacement travled
Displacement is a vector magnitude that depends on the position of the body which is individualistic for the trajectory.
While, Distance is a scalar magnitude that measures over the trajectory.
a car moves 8 m in 4 s at a comstant velocity. what is the cars acceleration
Answer:
Zero.
Explanation:
Acceleration is defined as change in velocity per unit time. For constant, the acceleration is zero.
[tex]\hat{a} =\frac{ \delta V}{t}[/tex]as [tex]\delta[/tex][tex]v\\[/tex] is zero.
The position of a particle is given by the function x = \left(2t^3 - 6t^2 + 12\right) m, where t is in s. Question:At what time is the acceleration zero? (with working out)
Answer:
t = 1secExplanation:
Given the position of a particle expressed by the equation x = (2t^3 - 6t^2 + 12)m, where t is in seconds, the acceleration function can be gotten by taking the second derivative of the function with respect to t as shown;
a = d/dt(dx/dt)
First let us get dx/dt
dx/dt = 3(2)t³⁻¹-2(6)t²⁻¹+0
dx/dt = 6t²-12t
a = d/dt(dx/dt)
a = d/dx(6t²-12t)
a = 2(6)t²⁻¹-12t¹⁻¹
a = 12t - 12t⁰
a = 12t-12
If the acceleration is zero, then;
12t-12 = 0
add 12 to both sides
12t-12+12 = 0+12
12t = 12
t = 12/12
t = 1sec
Hence the time when acceleration is zero is 1sec
Convert the following to SI units:_______. a. 9.12μs9.12μs b. 3.42 km c. 44 cm/ms d. 80 km/h
Answer:
Explanation:
A. We know that 1microsecond= 10^-6s
So = 9.12*10^-6=9.12*10^-6s
B 1km= 1000m
So 3.42km= 3240m
C.1ms= 10^-3s
And 1cm= 10^-2m
So 44*10^-2m/10^-3s=440m/s
D. 80km/hr= 80*1000/3600= 22.2m/s
Because 1km= 1000m
1hr= 3600s
How do you play Simon says?
Can air make shadows
Answer:no
Explanation:the way to see air is by steaming something sometimes you might not see the air but you can see the shadow
Give three examples of unbalanced forces in your everyday life. HELP FAST PLZ
Answer:
1. Kicking a soccer ball
2. Playing tug of war
3. Bouncing a Ball
Explanation:
The acceleration of a piston is given by : rw^2 cos(wt-pi/4) find amplitude.
Given :
The acceleration of a piston is given by :
[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]
To Find :
The amplitude of the acceleration of piston .
Solution :
Now , acceleration is :
[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]
Now , amplitude is maximum of any function .
So , maximum value of [tex]cos\ \theta[/tex] is equal to 1 .
Therefore , amplitude of the acceleration of piston is :
[tex]A=r\omega^2[/tex]
Hence , this is the required solution .
Force is the amount _____ or _____ on an object
Motion is the action of _____ from one place to another place.
Answer:
force is the amount of work or pressure given to an object
motion is the action of moving one place to another place
Part A. 13 in Express your answer to two significant figures and include the appropriate units. SubmitPrev Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Part B. 77 ft/s Express your answer to two significant figures and include the appropriate units ,, ? Value m/s
Answer:
Part A
[tex]x = 0.33 \ m[/tex]
Part B
[tex]y = 23.49 \ m/s[/tex]
Explanation:
From the question we are told to convert
13 inches to meters
Now
1 in = 0.0254 \ m
13 \ in = x m
=> [tex]x = \frac{13 * 0.0254 }{1}[/tex]
=> [tex]x = 0.33 \ m[/tex]
For part B we are told to covert 77ft/s to meters /srconds
So
1 ft/s = 0.305 \ m/s
77 ft/s = y
=> [tex]y = \frac{77 * 0.305 }{1}[/tex]
=> [tex]y = 23.49 \ m/s[/tex]
A ball easily moves but not a bus when we push them. why?
Answer: The amount of matter is higher in a bus than a ball.
Explanation:
μN/(kg⋅ns) in the correct SI of:________
Answer:
μN/ (kg ns) = 10³ N / (kg s)
Explanation:
In this exercise they ask us if the notation is correct. Let's write the different terms in the SI systems
force is N
time is in seconds
the unit given for the force is 1 N = 10⁶ μN
the unit of time is 1 s = 10⁹ ns
the correct way to give the answer should be: N / (kg s)
so the notation should be changed
μN /kg ns = μN / (kg ns) (1N / 10⁶ μN) (10⁹ ns / 1 s) =
μN/ (kg ns) = 10³ N / (kg s)
in an equation f = l^2-d^2/4l the intercept is
Answer:
the intercept is the orgin (0,0)
an object is accelerating if it is moving____. circle all that apply.
Answer:
With changing speed and/or in a circle
Circle all true statements:
a) Water at 90°C is warmer than water at 202°F
b) 40K corresponds to -40°C
c) Temperature differing by 25 on the Fahrenheit scale must differ by 45 on the Celsius scale.
d) Temperatures which differ by 10 on the Celsius scale must differ by 18 on the Fahrenheit scale.
e) 0°F corresponds to -32°C.
Answer:
The true statements are:
c) Temperature differing by 25 on the Fahrenheit scale must differ by 45 on the Celsius scale.
d) Temperatures which differ by 10 on the Celsius scale must differ by 18 on the Fahrenheit scale.
Explanation:
option a is false, 90°C is equal to 194°F which is not warmer than 202°F
option b is false, 40K is equal to -233°C
option e is false 0°F corresponds to -17.8°C.
Why si unit develop all over the world?
Answer:
SI unit is used in most places around the world, so our use of it allows scientists from disparate regions to use a single standard in communicating scientific data without vocabulary confusion.
Explanation:
hope it helps
Si unit is used almost all around the world, so our use of it allows
Scientist from disparate regions to use a single standered in communicating scientific data without vocabulary confusion
Aa commercial advertising a diet pill says it is scientifically proven to help you lose weight. It is recommended by a doctor who observed that some of his patients lost weight after taking the pill for a week. Why is this claim considered pseudoscience
Answer:
no more than 10 pills a day depending on what type it is
Explanation:
An object is moving in a negative direction with a negative acceleration, for example an object dropped out a window. What does it mean that both velocity and acceleration are negative?
Answer:
yes bc they are falling :]
Explanation:
Is the sinusoidal pattern on the string longer or shorter with a greater propagation velocity?
Answer:
Increase in velocity propagation would lead to a longer sinusoidal pattern.
Explanation:
The velocity of propagation has a relationship with the tension in a standing wave and is given by;
v = √(T/μ)
where:
μ is mass per unit length.
T is tension in string
For the sinusoidal pattern on the string to be longer or shorter, it means that the wavelength will be longer or shorter.
Now, relationship between wavelength and velocity and tension is;
v = λ/T
Where V is velocity of propagation, λ is wavelength and T is Tension.
So, λ = vT
From the earlier standing wave equation, we will see that if we increase the velocity, it means the Tension of the spring will increase as well.
Thus, from this wavelength equation, an increase in velocity means an increase in tension and which will in turn mean an increase in wavelength.
Therefore, an increase in velocity propagation means a longer sinusoidal pattern.
How do scientists use the evidence they gather?
Answer:
When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.
[tex]\huge{\gray{\sf Question:} }[/tex]
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm, (b) 3 cm and (c) 0 cm.
Answer:
ANSWER
A = 5 cm = 0.05 m
T = 0.2 s
ω=2π/T=2π/0.2=10πrad/s
plz give brainlist
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Answer:
[tex]here \: amplitude = 5cm (convert \: to \: si \: unit) = .05m \\ t = .2sec \\ omega = \frac{2\pi}{t} \\ = \frac{2\pi}{.2 } = 10\pi \frac{rad}{s} \\ we \: want \: find \: a \: and \: v \\ we \: know \: that \: a = - {omega}^{2} x \\ v = omega \sqrt{ {r}^{2} - {x}^{2} } \\(1)x = .05m \\ a = - {10\pi}^{2} \times .05 = - 5 {\pi}^{2} \frac{m}{ {s}^{2} } \\ v = 10\pi \sqrt{ {.05}^{2} - {.05}^{2} } = 0 \\ 2)x = 3cm = .03m \\ a = {(10\pi)}^{2} \times .03 = - 3 {\pi}^{2} \frac{m}{ {s}^{2} } \\ v = 10\pi \sqrt{ {.05}^{2} - {.03}^{2} } = 10\pi \times .04 = .4\pi \frac{m}{s} \\ 3)x = 0 \\ a = - {(10\pi)}^{2} \times 0 = 0 \\ v = 10\pi \times \sqrt{ {.05}^{2} - {0}^{2} } = - 10\pi \times .05 = .5\pi \frac{m}{s} \\ thank \: you[/tex]
Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge Q = 4.0 μC is located at x = 0.40 m, y = 0.What is the net force ((a)magnitude and (b)direction) on charge q1 exerted by the other two charges?
Answer:
F = 0.111015 N
Explanation:
For this exercise the force is given by Coulomb's law
F = k q₁q₂ / r₂₁²
we calculate the electric force of the other two particles on the charge q1
Charges q₁ and q₂
the distance between them is
r₁₂ = y₁ -y₂
r₁₂ = 0.30 + 0.30
r₁₂ = 0.60 m
let's calculate
F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2
F₁₂ = 1 10⁻¹ N
directed towards the positive side of the y-axis
Charges 1 and 3
Let's find the distance using the Pythagorean Theorem
r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]
r₁₃ = 0.50 m
F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²
F₁₃ = 1.697 10⁻² N
The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ 0.3 / 0.4
tea = 36.87º
The angle from the positive side of the x-axis is
θ ’= 180 - θ
θ ’= 180 - 36.87
θ ’= 143.13º
sin143.13 = F_13y / F₁₃
F_13y = F₁₃ sin 143.13
F{13y} = 1.697 10⁻² sin 143.13
F_13y = 1.0183 10⁻² N
cos 143.13 = F_13x / F₁₃
F₁₃ₓ = F₁₃ cos 143.13
F₁₃ₓ = 1.697 10⁻² cos 143.13
F₁₃ₓ = -1.357 10-2 N
Now we can find the components of the resultant force
Fx = F13x + F12x
Fx = -1,357 10-2 +0
Fx = -1.357 10-2 N
Fy = F13y + F12y
Fy = 1.0183 10-2 + 1 10-1
Fy = 0.110183 N
We use the Pythagorean theorem to find the modulus
F = Ra (Fx2 + Fy2)
F = RA [(1.357 10-2) 2 + 0.110183 2]
F = 0.111015 N
Let's use trigonometry for the angles
tan tea = Fy / Fx
tea = tan-1 (0.110183 / -0.01357)
tea = 1,448 rad
to find the angle about the positive side of the + x axis
tea '= pi - 1,448
Tea = 1.6936 rad
The magnitude of the net force on q1 exerted by the other two charges is 0.357 N.
The direction of the net force on q1 is 50⁰.
The given parameters;
q1 = 2.0 μC located at (0, 0.3) mq2 = 2.0 μC located at (0, -0.3) mq3 = 4.0 μC located at (0.4, 0) mThe force on q1 due to q2 occurs only in y-direction and can be calculated using Coulomb's law as shown below;
[tex]F_1_2 = \frac{kq^2}{r^2}j = \frac{(9\times 10^9) \times (2\times 10^{-6})^2 }{(0.3 +0.3)^2} j \\\\\F_{12} = (0.1)j[/tex]
The force on q1 due to q3 occurs both in x-direction and y-direction, and it is calculated as follows;
[tex]distance \ between \ q1 \ and \ q3\ , r_{13} = \sqrt{0.3^2 \ + \ 0.4^2} = 0.5 \ m\\\\F_{13} = \frac{kq^2}{r_{13}^2} (\frac{0.4i}{0.5} \ + \ \frac{0.3j}{0.5} )\\\\F_{13} = \frac{kq^2}{r_{13}^2} (0.8i + 0.6j)\\\\F_{13} = \frac{9\times 10^9 \times (2\times 10^{-6}) \times (4\times 10^{-6})}{0.5^2} (0.8i + 0.6j)\\\\F_{13} = 0.288(0.8i + 0.6j)\\\\F_{13} = 0.23i + 0.173j[/tex]
The net force is calculated as follows;
[tex]F_{net} = F_{12} \ + \ F_{13}\\\\F_{net} = (0.1j) \ + \ (0.23i + 0.173j)\\\\F_{net} = (0.23i + 0.273j)[/tex]
The magnitude of the net force on q1 is calculated as follows;
[tex]|F| = \sqrt{(0.23^2) \ + \ (0.273^2)} \\\\|F| = 0.357 \ N[/tex]
The direction of the net force on q1 is calculated as follows;
[tex]tan(\theta )= \frac{F_y}{F_x} \\\\\theta = tan^{-1} (\frac{0.273}{0.23} )\\\\\theta = 50^0[/tex]
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imagine imagining an imagination.
Answer:
We’re imagining imagining imagining an imagination...
What is the latitude of the vertical (direct) rays of the sun?
Answer:
23.5 degree North
Explanation:
The latitude of the vertical (direct) rays of the sun is 23.5 degree North. This is however due to the sun's vertical rays being located directly above 23.5 degree South which is the position of the Tropic of Capricorn.
The sun ray is highest at this pap tion and it occurs during Summer solstice which happens between June 21 / 22 of every year.
Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of magnitude of the distance from the sun to Earth is g
Answer:
The order of magnitude of the distance from the sun to Earth is 10⁸ km.
Explanation:
The order of magnitude of the distance from the sun to Earth can be calculated as follows:
[tex] c = \frac{x}{t} [/tex]
Where:
c: is the speed of light = 3x10⁸ m/s
t: is the time = 8 min
Hence, the distance is:
[tex] x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km [/tex]
Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.
I hope it helps you!
An isolated system contains two objects with charges q1q1 and q2q2. If the charge on object 1 is doubled, what is the charge on object 2? g
Answer:
The new charge on the second object is q₂ - q₁
Explanation:
Given;
charge on the first object, q₁
charge on the second object, q₂
Since there is no external force on isolated system, the total charge in the system is given by;
Q = q₁ + q₂
q₂ = Q - q₁
When the charge on object 1 is doubled, q₁' = 2q₁
let q₂' be the new charge on object 2
q₂' = Q - q₁'
q₂' = (q₁ + q₂) - 2q₁
q₂' = q₁ + q₂ - 2q₁
q₂' = q₂ - q₁
Therefore, the new charge on the second object is q₂ - q₁
The charge on object 2 will be "q₂ - q₁".
Charge on object:The charge's quantity on such an item represents the degree of imbalance between its electrons as well as protons.
Just to calculate the overall charge of a positively (+) charged item, divide the total no. of valence electrons by the overall number of protons.
According to the question,
Object 1's charge = q₁
Object 2's charge = q₂
Let,
New charge of object 2 be "q₂'".
The total charge in the system will be:
Q = q₁ + q₂
or,
q₂ = Q - q₁
hence,
The charge on object 2 be:
→ q₂' = Q - q₁'
By substituting the values,
= (q₁ + q₂) - 2q₁
= q₁ + q₂ - 2q₁
= q₂ - q₁
Thus the answer above is appropriate.
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