Answer:
Option - CHope it helps youWhich statement is supported by this scenario?
Yanni is sitting in the stands watching a football game.
The quarterback runs forward toward the end zone
and throws a pass straight downfield to a player for a
touchdown. The quarterback ran at a speed of 8 m/s
and threw the ball with a speed of 15 m/s.
For Yanni, the speed of the ball is 23 m/s, and for
the quarterback, the speed of the ball is 15 m/s.
For Yanni, the speed of the ball is 15 m/s, and for
the quarterback, the speed of the ball is 8 m/s.
For both Yanni and the quarterback, the speed of
the ball is 23 m/s.
For both Yanni and the quarterback, the speed of
the ball is 15 m/s.
Answer:
A
For Yanni, the speed of the ball is 23 m/s, and for the quarterback, the speed of the ball is 15 m/s.
Explanation:
Right on edge 2021
The statement supported by this scenario is, for Yanni, the speed of the ball is 23 m/s, and for the quarterback, the speed of the ball is 15 m/s.
Relative speed of the ballThe relative speed of the ball is determined with respect to frame of reference.
For Yanni at rest ( zero speed);Vr/b = va + vb
Vr/b = 8 + 15 = 23 m/s
For quarterback in motionVr/b = va + vb
Vr/b = 0 + 15 = 15 m/s
Thus, the statement supported by this scenario is, for Yanni, the speed of the ball is 23 m/s, and for the quarterback, the speed of the ball is 15 m/s.
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You are an electrician working on a house. What type of circuit should you use for the house so that the owners don’t call to complain about their wiring? Why use this circuit?
Answer: Parallel
Explanation: Parallel because you don’t want all the lights to go out because of one light.
The electrical wiring and safety devices applied have to be perfect for non complainable circuit system in houses. One thing have to ensure that all the electrical components needs to follow the requirements set by national electrical code.
What are circuits?Circuits are way for the passage of electrical current which consists of electrical wires and devices such as bulbs, fan etc. The circuit system have to be properly designed so that no overload or short circuit can be occured.
Proper thermal insulations for cables and wires have to be chosen. Moreover, all the electrical components should follow the standard requirements set by local bodies or national electrical code.
Safety devices such as circuit breaker or switches like fuse of standard quality must be applied to protect the devices. Circuit breakers are more better than fuses because they don't need replacements.
Similarly, the selection of wires of cables have to careful and need to check the quality of plastic sheaths on them.
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If an 800.0 Kg sports car slows to 8.0 m/s in the forward direction to check out an accident scene and a 1200.0 kg pick-up truck behind them continues traveling at 15.0 m/s forward, with what velocity will the two move if they lock bumpers after a rear-end collision
Answer:
12.2m/sExplanation:
Step one:
given data
mass m1=800kg
velocity v1=8m/s
mass m2= 1200kg
velocity v2=15m/s
Note: the two bodies move with the same velocity after collision, hence the common velocity is v=?
Step two:
the expression for the total momentum is
m1v1+m2v2=v(m1+m2)
substituting we have
800*8+1200*15=v(800+1200)
6400+18000=v(2000)
24400=2000v
divide both sides by 2000 to find v
v=24400/2000
v=12.2m/s
Their common velocity is 12.2m/s
What is the horizontal component of a ball thrown at a 27 degree angle at 16 m/s?
Answer:
14.25 m/s
Explanation:
In this problem, we need to find the horizontal component of a ball thrown at a 27 degree angle at 16 m/s.
It can be given by :
[tex]v_x=v\cos\theta\\\\=16\times \cos(27)\\\\=14.25\ m/s[/tex]
So, the horizontal component of the ball is 14.25 m/s.
A bike traveling initially at a speed of 32 m/s accelerates
uniformly at the rate of 3 m/s2 for a distance of 40 meters. The
bike's velocity after covering this distance is _m/s.
We are given:
Initial velocity (u) = 32 m/s
Acceleration (a) = 3 m/s²
Displacement (s) = 40 m
Final Velocity (v) = v m/s
Solving for the Final Velocity:
from the third equation of motion:
v² - u² = 2as
replacing the variables
v² - (32)² = 2(3)(40)
v² = 240 + 1024
v² = 1264
v = √1264
v = 35.5 m/s
Therefore, the velocity of the bike after travelling 40 m is 35.5 m/s
The bike's velocity after covering 40 m distance is 35.55 m/s.
Explanation:
Given:
The initial velocity of the bike = 32 m/s
The rate of acceleration of the bike = 3 ms^2
Distance covered by the bike = 40 m
To find:
The final velocity of the bike.
Solution:
The initial velocity of the bike = u = 32 m/s
The rate of acceleration of the bike = a = 3 ms^2
Distance covered by the bike = s = 40 m
The final velocity of the bike = v
Using the third equation of motion, which is written as :
[tex]v^2-u^2=2as\\v^2-(32m/s)^2=2\times 3m/s^2\times 40m\\v^2-1024 m^2/s^2=240 m^2/s^2\\v^2=240 m^2/s^2+1024 m^2/s^2\\v^2=1264 m^2/s^2\\v=35.55 m/s[/tex]
The bike's velocity after covering 40 m distance is 35.55 m/s.
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Create a hypothesis for this testable question: How does price affect the amount of chocolate people buy?
Answer:
IF the price of chocolate increases, THEN the amount of chocolate people buy decreases
Explanation:
In a scientific investigation, an observation is first made. Based on this observation, a scientific question is then asked. However, a HYPOTHESIS is given next to explain the question asked. A hyothesis is a testable explanation given to solve an observed problem or provide a possible answer to a question.
The hypothesis must be subject to testing via EXPERIMENTATION. A hypothesis usually goes in an IF, THEN format. In this case with the testable question: How does price affect the amount of chocolate people buy?
A hypothesis that can explain this question is: IF the price of chocolate increases, THEN the amount of chocolate people buy decreases.
PLEASE HELP.
How does the motion of our solar system compare to the motion of the Milky Way?
Which chemical equation is balanced?
O Na + O2--> Na20
O 2Na + 202 --> 2Na20
O 2Na2 + 02 --> 2Na20
4Na + 02 --> 2Na20
G
The gravitational force between two asteroids is 2.59 × 10 (exponent)-6 N. The centers of mass are 2000 meters away and their masses are equal. What is the mass of each asteroid?
Answer:
[tex]2.79 \times 10^5 \ \text{kg}[/tex]
Explanation:
Newton's Law of Universal Gravitation:
[tex]$F= G\frac{m_1 m_2}{r^2}[/tex] F = force of gravity (N)G = gravitational constant [tex](6.67 \times 10^-^1^1 \ N\frac{m^2}{kg^2})[/tex][tex]m_1[/tex] = mass of Object 1 (kg)[tex]m_2[/tex] = mass of Object 2 (kg)r = distance between the center of mass (m)Let's convert our given information to scientific notation:
[tex]2000 \ m \rightarrow 2.0 \times 10^3 \ m[/tex]Now using the gravitational force and the distance between centers of mass that are given, we can plug these into Newton's law:
[tex]2.59 \times 10^-^6 $\ N = 6.67 \times 10^-^1^1 \ N \frac{m^2}{kg^2} \times \frac{m_1 m_2}{(2.0 \times 10^3 \ m)^2}[/tex]Remove the units for better readability.
[tex]2.59 \times 10^-^6=6.67 \times 10^-^1^1 \frac{m_1m_2}{(2.0 \times 10^3)^2}[/tex]Divide both sides of the equation by the gravitational constant G.
[tex]\frac{2.59 \times 10^-^6}{6.67 \times 10^-^1^1} =\frac{m_1m_2}{(2.0 \times 10^3)^2}[/tex]Distribute the power of 2 inside the parentheses.
[tex]\frac{2.59 \times 10^-^6}{6.67 \times 10^-^1^1} =\frac{m_1m_2}{2.0 \times 10^6}[/tex]If we evaluate the left side of the equation, we get:
[tex]3.88305847 \times 10^4 = \frac{m_1m_2}{2.0 \times 10^6}[/tex]Multiply both sides of the equation by r.
[tex]7.76611694 \times 10^1^0= m_1m_2[/tex]In order to find the mass of one asteroid, we can use the fact that both asteroids have the same mass, therefore, we can rewrite [tex]m_1m_2[/tex] as [tex]m^2[/tex].
[tex]7.76611694 \times 10^1^0= m^2[/tex]Square root both sides of the equation.
[tex]m=\sqrt{7.76611694 \times 10^1^0}[/tex][tex]m=2.78677536 \times 10^5[/tex] [tex]m=2.79 \times 10^5[/tex]Since m is in units of kg, we can state that the mass of each asteroid is 2.79 * 10⁵ kg.
A hunter is practicing hitting a target that is down range. If the arrow leaves the bow at a velocity of 30ms at an angle of 40° above the horizontal.
a.
Find out how far down range the arrow goes.
b. What is the maximum altitude the arrow reaches?
Answer:
Explanation:
range of projectile = u² sin2θ / g
u = 30 m /s
θ = 40°
range = 30² x sin 80 / 9.8
= 90.44 m
b )
maximum altitude H = u² sin²θ / 2 x g
= 30² sin²40 / 9.8
= 37.94 m .
A 180-kg load is hung on a wire of length of 4.70 m, cross-sectional area 2.000 10-5 m2, and Young's modulus 8.00 1010 N/m2. What is its increase in length?
i saw it online once but i cant find it so i would look this up on maybe quizlet i use it a lot and it helps
A box is pressed against a vertical wall by a force F which is directed horizontally. If the magnitude of F is the minimum required to hold the box at rest, what is its value
Answer:
F = mg /μ
Explanation:
For this exercise we define a coordinate system with the horizontal x axis and the vertical y axis, let's write the equilibrium equations for each axis
X axis
F -N = 0
F = N
Y axisy
fr - W = 0
fr = W (2)
the force and touch have the expression
fr = μ N
we substitute
fr = μ F
we substitute in 2
μ F = m g
F = mg /μ
A major artery with a 1.7 cm2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area of 0.4 cm2. By what factor is the average velocity of the blood reduced when it passes into these branches
Answer:
0.14
Explanation:
Flow rate is the volume flowing through a point at a particular time, in calcuing flow rate we have
Q= v*t
it in terms of Area, we have Q= A*v
Where A= area
v= velocity.
Solving the question , flow rate is constant then
A*v= constant
A(i) v(i)= A(f) v(f)
Where A(i)= initial area= 1.00cm^2
A(f)= final area= 0.400cm^2
V(i) and V(f) are the initial and final velocity respectively and the ratio of the two will gives us the factor
Substitute the values into the equation we have
1 V(i)= 4 V(f)
But we were told that the cross sectional area of 1.00cm^2 branches into 18 smaller arteries.
Then
1 V(i)=0.4 V(f)*(18)
1 V(i)=7.2V(f)
Then if we find the ratio of the velocity, we will get the factor.
V(f)/V(i)= 1/7.2
V(f)/V(i)=0.14
Hence, the factor of the average velocity of the blood reduced when it passes into these branches is 0.14
A train of mass 3.3 × 10^6 kg is moving at a constant speed up a slope inclined at an angle of 0.64°
to the horizontal. The engine of the train is producing a useful output power of 14 MW.
Assume that there are no frictional forces opposing the motion of the train.
What is the speed of the train?
A 0.43 m s–1 B 4.2 m s–1 C 39 m s–1 D 380 m s–1.
Ans is C how?
ref: Q14 - 9702_s18_qp_11
Answer:
C. 39 m/s
Explanation:
First we need to calculate the total force required to move the train along the inclined plane. So, it is clear that the work done will be equal to the component of the weight that is parallel to the inclined plane, because there is no frictional force present:
Force = F = mg Sin θ
where,
m = mass of train = 3.3 x 10⁶ kg
g = 9.8 m/s²
θ = Angle of Inclination = 0.64°
Therefore,
F = (3.3 x 10⁶ kg)(9.8 m/s²)Sin 0.64°
F = 3.612 x 10⁵ N
Now, the formula for power is:
P = FV
V = P/F
where,
V = Velocity of Train = ?
P = Power of Engine = 14 MW = 1.4 x 10⁷ W
Therefore,
V = 1.4 x 10⁷ W/3.612 x 10⁵ N
V = 38.75 m/s
which is approximately equal to:
C. 39 m/s
The speed of train is 39 m/s. Hence, option (C) is correct.
Given data:
The mass of train is, [tex]m = 3.3 \times 10^{6} \;\rm kg[/tex].
The angle of inclination is, [tex]\theta = 0.64^ {\circ}[/tex].
The useful power output value is, [tex]P= 14 \;\rm MW = 14 \times 10^{6} \;\rm W[/tex].
The work done will be equal to the component of the weight that is parallel to the inclined plane, because there is no frictional force present. So, the total force required to move the train along the inclined plane is given as,
[tex]F = mg sin \theta\\\\F = 3.3 \times 10^{6} \times 9.8 \times sin0.64\\\\F = 361233.75 \;\rm N[/tex]
Now, use the formula of the power to obtain the value of speed as,
[tex]P = F \times v\\\\14 \times 10^{6} =361233.75 \times v\\\\v \approx 38.75 \;\rm m/s[/tex]
Thus, the speed of train is 39 m/s. Hence, option (C) is correct.
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QUESTION If the angular acceleration were doubled for the same duration, by what factor would the angular displacement change
Answer:
By a factor of 2
Explanation:
The angular displacement would change by a factor of 2.
The angular acceleration is represented by the formula
α = Δw / Δt, where
α = angular acceleration
Δw = change in velocity.
Δt = change in time taken
The angular displacement is given by the relation
s = rθ
s = arc length
r = distance
θ = angular displacement
We all know that velocity is the ratio of displacement and time, thus,
v = s/t
Angular acceleration on the other hand says that
α = w/t, substitute w for v, we have
α = s/t ÷ t
α = s/t * 1/t
α = s/t²
We see here that multiplying the acceleration by 2 will only be balanced by increasing the displacement by the same number, 2 in this case
The volume of a cube is given by the formula V = s3, where s is the length of one side. What is the volume of the cube including the planet? First, justify your answer using the Product of Powers Property. Then justify your answer using the Power of a Product Property.
Answer:
1000 cm³
Explanation:
The product of powers property is a rule that helps in simplifying the difficulties that comes with multiplying powers of numbers. It states that when multiplying two powers having the same base, we take one of the base, and then just add the exponents.
This is illustrated in the example.
S¹ * S². The product of powers asks us to pick one of the bases, S, and then add up the powers, thus
S¹ * S² = S^(¹+²)
S¹ * S² = S³
Now, using this in our question, volume of the cube is S³. Since we aren't given a length, I'm going to assume a length.
L = 10 cm. We all know that all sides are equal in a cube,
So, S = 10 cm.
S³ = volume of the cube
V = 10³
V = 1000 cm³
Now, to test the property of products, we say
S = 10, even though it doesn't show any visible exponent, we know that it's raised to the power of 1, and thus
S = 10¹
V = S * S * S
V = 10¹ * 10¹ * 10¹
V = 10^(¹+¹+¹)
V = 10³
V = 1000 cm³
Justified.......
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HELPPP MEEE PLEADEEEE How could a cow have more momentum than a tiger ? Explain
Answer:
If a cow has a greater mass than the tiger, it has greater momentum, but only if they are moving at the same speed, or the tiger is going slower (i think)
Explanation:
the quantity of motion of a moving body, measured as a product of its mass and velocity.
A penny and a quarter are embedded in the concrete bottom of a swimming pool filled with water. Which of these coins experiences the greater downward force due to water pressure acting on it? Explain.
Answer: The quarter, because it has a larger surface area.
Explanation:
remember the equation:
Pressure = Force/Area
Now, if we isolate Force, we get:
Force = Pressure*Area.
The pressure will be the same for both coins, but the area is not.
Then the one with a larger surface area will suffer a greater downward force. (We can consider the area as the area of the top or bottom of the coin)
The area of a circle of diameter d, is:
A = pi*(d/2)^2
where pi = 3.14
The penny has a diameter of 19.05 mm, then its area is:
A = 3.14*(19.05mm/2)^2 = 284.87mm^2
The quarter has a diameter of 24.26mm, then the area is:
A = 3.14*(24.26mm/2)^2 = 462 mm^2
Then the quarter has a larger area, this means that the quarter will experience the greater downward force due to water pressure.
A hydrogen atom in an excited state absorbs a photon of wavelength 411 nm. What were the initial and final states of the hydrogen atom?
Answer:
The initial and final states of the hydrogen atom were n=2 and n=6 respectively.
Explanation:
We must first obtain the energy of the photon;
E= hc/λ
where;
h= Plank's constant = 6.6 * 10^-34 JS
c= speed of light = 3* 10^8 m/s
λ = wavelength of light= 411 nm = 411* 10^-9 m
Substituting values;
E = 6.6 * 10^-34 * 3* 10^8 / 411* 10^-9
E = 4.8 * 10^-19 J or 3.0 eV
But ;
En = 13.6/n^2
So E = En final - En initial
3.0 = -13.6(1/n^2final - 1/n^2initial)
If we substitute n^2final = 6 and n^2 initial = 2 then the RHS becomes approximately equal to the LHS
Therefore the initial and final states of the hydrogen atom were n=2 and n=6 respectively.
How fast does water flow from a hole at the bottom of a very wide, 7.1 mm deep storage tank filled with water
Answer:
The water will move fasterExplanation:
This happens as part of the consequences of the continuity of an incompressible fluid, of which water is an example.
so as the water emerges from the wider section the flow velocity increases and vice versa.
Hence the flow velocity in indirectly proportional to the area
The continuity equation is given below for our analysis
[tex]A1V1=A2V2\\[/tex]-------------------1
where
A1= the area of the wider section of the tank
V1= the flow velocity from the wider section
A2= the area of the hole at the bottom
V2= velocity of flow at the bottom.
The shortest wavelength of visible light is approximately 400 nm express the wavelength in centimeters
Answer:
4x [tex]10^{-5}[/tex] cm
Explanation:
400nm = 400 x [tex]10^{-9}[/tex] m
1m = 100 cm
400 x [tex]10^{-9}[/tex] m = 400 x [tex]10^{-9}[/tex] x 100 cm
ans : wavelength of light = 4x [tex]10^{-5}[/tex] cm
Answer:
[tex]4 * 10^{5}[/tex]
Explanation:
[tex]1c m = 1*10^-{7}nm[/tex]
An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a less dense liquid. What would you observe?
Answer:
The fraction of its volume inside liquid is increased .
Explanation:
According to principle pf floatation , an object floats on the surface of water
when the weight of liquid displaced by it becomes equal to weight of the object . weight of the liquid depends upon the density of the liquid .
In the second case , when the body is dipped into liquid of lesser density , in order to balance the weight of body , more volume of liquid will be displaced so that weight of displaced liquid becomes equal to object's weight . So the body floats with greater depth inside liquid . The fraction of its volume inside liquid is increased .
What is the momentum of a 533 kg blimp moving east at +75 m/s
Answer:
39975kgm/s due east
Explanation:
Given parameters:
Mass of the blimp = 533kg
Velocity = +75m/s due east
Unknown:
Momentum of the body = ?
Solution:
The momentum of a body is the amount of motion it posses.
Momentum is the product of mass and velocity;
Momentum = mass x velocity
Insert the parameters and solve;
Momentum = 533 x 75 = 39975kgm/s
The momentum of the body is 39975kgm/s due east
The momentum of a 533 kg blimp moving east at +75 m/s is 39,975kgm/s.
MOMENTUM:Momentum of a body can be calculated by multiplying the mass of the substance by its velocity. That is;
Momentum (p) = mass (m) × velocity (v)
According to this question, a 533 kg blimp is said to be moving east at +75 m/s. The momentum is calculated thus:
Momentum = 533 × 75
Momentum = 39,975kgm/s.
Therefore, momentum of a 533 kg blimp moving east at +75 m/s is 39,975kgm/s
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Aman tosses a dart upward with a velocity of 14.1 m/s at 60° angle.
How much time is it aloft?
What is its max height and range?
Angle of projection of the dart = 60°
initial velocity of the dart = 14.1 m/s
Horizontal and vertical component of the Dart:Using basic trigonometry, we can see that:
Vertical component of the Dart = 14.1*Sin 60° = 12.2 m/s
Horizontal component of the Dart = 14.1*Cos 60° = 7.05 m/s
Time the Dart is Aloft:
Explaining the concept:
Once the dart is in mid-air, there will be no force that will reduce its horizontal velocity
So, the dart will be moving at a constant horizontal velocity of 7.05 m/s
but the force of gravity will be applied on the dart while mid-air and will pull it downwards with an acceleration of 9.8 m/s²
Hence, the time the dart will be aloft is the same as the time taken by the dart to reach the ground if we threw it vertically upwards at a velocity of 12.2 m/s
Solving for the time taken for the dart to reach the ground:
initial velocity = 12.2 m/s
velocity at max height = 0 m/s
acceleration = -9.8 m/s
time taken to reach max height:
v = u + at using the first equation of motion
replacing the variables
0 = 12.2 + (-9.8)t
t = -12.2 / -9.8
t = 1.24 seconds
Time taken to reach the ground:
time taken to reach the ground = 2 * time taken to reach max height
(Since the dart will take the same time to come back down)
Time taken to reach the ground = 2 * 1.24
Time taken to reach the ground = 2.48 seconds
Therefore, the dart will hit the ground and stop 2.48 seconds after throwing
Max height of the Dart:
Explaining the concept:
From newton's second equation of motion
s = ut + 1/2 at²
Since we need the max vertical height, we need the vertical component of the values,
so this equation can be rewritten for vertical height as :
s(vertical) = u(vertical)* t + 1/2 * a(vertical) * t²
from the last section, we know that the dart reaches its maximum height at t = 1.24 seconds
replacing the values in this equation:
s(vertical) = (12.2 * 1.24) + 1/2 * (-9.8) * (1.24)²
s(vertical) = 15.128 + (-7.5)
s(vertical) = 7.63 m
Therefore, the maximum height of the dart is 7.63 m
The range of the Dart:
Explaining the concept:
The range of the dart is the horizontal distance covered by the dart
from the second section, we know that the dart travels horizontally with a constant velocity of 7.05 m/s
from the third section, we know that the total time taken by the dart to hit the ground is 2.48 seconds
Since the horizontal velocity of the dart is constant, we can say that it moved horizontally for 2.48 seconds at a constant velocity of 7.05 m/s
Solving for the range:
s = ut + 1/2 at² From the second equation of motion
this equation can be rewritten for horizontal distance as:
s(horizontal) = u(horizontal)* t + 1/2 (a)(t)²
The dart is moving at a constant velocity, that means that its acceleration is 0
s(horizontal) = u(horizontal)* t + 1/2 (0)(t)²
s(horizontal) = u(horizontal)* t
replacing the variables
s(horizontal) = 7.05 * 2.48 [here, 2.48 is the time taken by the dart to reach the ground and stop]
s(horizontal) = 17.484 m
The horizontal distance covered by the dart is 17.484 m
Therefore, the range of the dart is 17.484 m
Answer:
time = 2.49 s
max height = 7.61 m
range = 17.57 m
See below for exact values.
Explanation:
Vertical/Horizontal Components:Since we are given the angle at which the object is tossed, this object is in projectile motion.
We are given the initial velocity of the dart, and we can use the angle to solve for its horizontal and vertical components.
Horizontal component:
[tex](v_i)_x =v_i \cdot cos(\theta)[/tex] [tex](v_i)_x=14.1\cdot cos(60)[/tex]Vertical component:
[tex](v_i)_y =v_i \cdot sin(\theta)[/tex] [tex](v_i)_y=14.1 \cdot sin(60)[/tex] Time In The Air:In order to find the time that the dart stays in the air, we can use the constant acceleration equation that does not include displacement. This equation is:
[tex]v_f=v_i+at[/tex]Since we solve for time using the vertical motion of the projectile, we will use this equation in terms of the y-direction.
[tex](v_f)_y = (v_i)_y+a_yt[/tex]We don't know what the final velocity of the dart when it reaches the ground is, but we do know that its final velocity when it reaches its maximum height is 0 m/s.
Therefore, we can solve for half of the full time that it takes the object to reach the ground, and double this value at the end.
Let's set the downwards direction to be negative and the upwards direction to be positive.
Using our knowledge and previous calculations, we know that:
[tex](v_f)_y = 0[/tex] [tex](v_i)_y=14.1\cdot sin(60)[/tex] [tex]a_y=-9.8[/tex]The acceleration of gravity is in the y-direction and facing downwards, so that's why it is -9.8 m/s².
Substitute these values into the constant acceleration equation.
[tex]0=14.1\cdot sin(60) + (-9.8)t[/tex]Subtract 14.1 * sin(60) from both sides of the equation.
[tex]-14.1\cdot sin(60)=-9.8t[/tex]Divide both sides of the equation by -9.8 to solve for t.
[tex]\displaystyle{\frac{-14.1\cdot sin(60)}{-9.8} =t}[/tex] [tex]t=1.246016142[/tex]Now, this is only the time for half of the object's trajectory. Double it to find the total time the dart is aloft:
[tex]2t = 2.492032284[/tex]The dart is in the air for a total of 2.49 seconds.
Maximum Height:We can find the maximum height of the dart by using another constant acceleration equation.
Since we don't have the final velocity of the object, we can use this equation:
[tex]x_f=x_i+v_it+\frac{1}{2}at^2[/tex]Subtract [tex]x_i[/tex] from both sides to get the change in position, or delta x.
[tex]\triangle x=v_it+\frac{1}{2}at^2[/tex]In order to find the maximum height, we need to use this equation in terms of the y-direction.
[tex]\triangle x_y=(v_i)_yt+\frac{1}{2}a_yt^2[/tex]Remember that time is the same regardless of the x- or y- direction.
Now, we can solve for the displacement in the y-direction by plugging in the values that we know.
[tex](v_i)_y=14.1\cdot sin(60)[/tex] [tex]t=1.246016142[/tex] [tex]a_y=-9.8[/tex]Note that we are using half of the time t, since this is where the maximum height occurs.
Plug these known values into the constant acceleration equation:
[tex]\triangle x_y=[14.1 \cdot sin(60)] (1.246016142) + \frac{1}{2}(-9.8)(1.246016142)^2[/tex] [tex]\triangle x_y=(15.21505102) + (-4.9)(1.552556226)[/tex] [tex]\triangle x_y=15.21505102-7.607525508[/tex] [tex]\triangle x_y=7.60752551[/tex]The maximum height of the dart is 7.61 meters.
Range:Finding the range of the object in projectile motion involves the same constant acceleration equation, but this time we are solving for the displacement in the horizontal direction (x-direction). Therefore:
[tex]\triangle x_x=(v_i)_xt+\frac{1}{2}a_xt^2[/tex]We know the horizontal component of the initial velocity vector, which is what we will be using.
The acceleration, of an object in projectile motion, in the x-direction is always 0 m/s².
We will use the full time of the object since we want to find the entire horizontal distance that the object covers. We have:
[tex](v_i)_x=14.1\cdot cos(60)[/tex] [tex]t=2.492032284[/tex] [tex]a_x=0[/tex]Plug these values into the constant acceleration equation.
[tex]\triangle x=[14.1\cdot cos(60)](2.492032284)+\frac{1}{2} (0)(2.492032284)^2[/tex] [tex]\triangle x=[14.1\cdot cos(60)](2.492032284)[/tex] [tex]\triangle x=17.5688276[/tex]The range of the dart is 17.57 meters.
A ball is projected horizontally from the top of a 92.0-meter high cliff with an initial speed of 19.8m/s. Determine the horizontal displacement.
Answer:
85.14 m
__________________________________________________________
(y) denotes "in the vertical direction"
(x) denotes "in the horizontal direction"
We are given:
Initial Horizontal velocity of the Ball (u(x)) = 19.8 m/s
Initial height of the ball (s(y)) = 92 m
Initial Vertical velocity of the Ball (u(y)) = 0 m/s
Time taken to reach the ground:
taking downwards direction as positive
Since the horizontal velocity is not opposed by any force, it will be the same until the ball reaches the ground
The vertical velocity will be increasing at a rate of (10 m/s)/s until the ball hits the ground
ay = 10 m/s²
So, while calculating the time. we can just ignore the horizontal velocity
Solving for the time taken:
s(y) = u(y)t + 1/2a(y)t² [second equation of motion]
92 = (0)(t) + 1/2(10)(t)² [replacing the variables]
92 = 5t²
t² = 92/5 [dividing both sides by 5]
t = √18.4 [taking the square root of both sides]
t = 4.3 seconds
So, it took the ball 4.3 seconds to reach the ground
Horizontal Distance travelled by the ball:
We know that the ball will reach the ground in 4.3 seconds
Since the horizontal velocity will not change, the ball will move with a constant velocity of 19.8 m/s in the horizontal direction
Horizontal distance travelled:
s(x) = u(x)t + 1/2a(x)t² [second equation of motion]
s(x) = (19.8)(4.3) + 1/2(0)(t)² [replacing the variables]
s(x) = 85.14 m
Hence, the ball travels 85.14 m horizontally
Can anyone do this for me thanks
Answer:
Question 3: Shortest Wavelenght
Question 4: Longes Wavelenghth
Explanation:
Use the table below to answer the questions concerning absolute and comparative advantages.
Use the table below to answer the questions concerning absolute and comparative advantages.
Lumber
Automobiles
United States
8
8
Canada
4
2
Which country has an absolute advantage in producing automobiles?
Lumber
Automobiles
United States
8
8
Canada
4
2
2. Which country has an absolute advantage in producing automobiles?
Answer:
United States
Explanation:
A bug crawled 35 meters in 20 seconds. What was the bugs velocity ?
(EXPLAIN)
The United States and South Korean soccer teams are playing in the first round of the World Cup. An American kicks the ball, giving it an initial velocity of 4.3 m/s. The ball rolls a distance of 5.0 m and is then intercepted by a South Korean player. If the ball accelerates at −0.50 m/s2 while rolling along the grass, find its velocity at the time of interception.
Answer:
Vf = 3.67 [m/s]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f} ^{2} =v_{i} ^{2} -(2*a*x)[/tex]
where:
Vf = final velocity [m/s]
Vi = initial velocity = 4.3 [m/s]
a = acceleration or desacceleration = 0.5 [m/s²]
x = distance = 5 [m]
Note: The negative sign in the above equation means that the velocity of the ball is decreasing (desacceleration).
Now replacing:
Vf² = (4.3)² - (2*0.5*5)
Vf² = 18.49 - 5
Vf² = 13.49
using the square root, we have.
Vf = 3.67 [m/s]
In the early 1900s many scientists thought that an atom consisted of a positive substance with negative charges scattered throughout the substance. Then Ernest Rutherford completed an experiment that changed the concept of an atom. His discovery led to the understanding that an atom consists mostly of empty space with —
Answer:
Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus. Based on these results, Rutherford proposed the nuclear model of the atom.
In the early 1900s, scientists thought that an atom consisted of a positive substance with negative charges scattered throughout the substance, but Rutherford explains that the positive charge is present at the center while the negative charges are present around it in the form of clouds.
What is the Rutherford gold foil experiment?There are various models that tried to explain the atomic model, and one of that Rutherford model is one in which he explains the model using the gold foil, which he then bombarded with the alpha particles. some of the particles that came back, while some passed through it by making some angles, and from this he concluded that the positive charge remains in the center while the negative is displaced.
Hence, in the early 1900s, scientists thought that an atom consisted of a positive substance with negative charges scattered throughout the substance, but Rutherford explains that the positive charge is present at the center while the negative charges are present around it in the form of clouds.
Learn more about the Rutherford gold foil experiment here.
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