Hi there!
a.
We know that:
[tex]\Sigma F_y = 0 \\\\\Sigma F_x = 0[/tex]
Begin by determining the forces in the vertical direction:
W = weight of traffic light
T₁sinθ = vertical component of T₁
T₂sinθ = vertical component of T₂
b.
The ropes provide a horizontal force:
T₁cosθ = Horizontal component of T1
T₂cosθ = Horizontal component of T2
Thus:
0 = T₁cosθ - T₂cosθ
T₁cosθ = T₂cosθ
T₁ = T₂
c.
Since the angles for both ropes are the same, we can say that:
T₁ = T₂
Sum the forces:
ΣFy = T₁sinθ + T₁sinθ - W = 0
2T₁sinθ = W
d.
Now, we can begin by solving for the tensions:
2T₁sinθ = W
[tex]T_1 = T_2 = \frac{W}{2sin\theta} = \frac{100}{2sin(37)} = \boxed{83.08 N}[/tex]
Describe a vibration that is not periodic. NO LINKS PLEASE
Answer:
1)The position change of almost any manually operated room light switch.
2) Sunlight striking a point on the ground on a partly cloudy and windy day
Explanation:
Dagmar says that diffusion happens really quickly. Is he right or wrong? Explain.
Answer:
Diffusion in gases is quick because the particles in a gas move quickly. It happens even faster in hot gases because the particles of gas move faster.
define parking orbit?
Answer:
An orbit of a spacecraft from which the spacecraft or another vehicle may be launched on a new trajectory.
In a police ballistics test, 2.00-g bullet traveling at 700 m/s suddenly hits and becomes embedded in a stationary 5.00-kg wood block. What is the speed of the block immediately after the bullet has stopped moving relative to the block
Answer:
Here we use the conservation of momentum theorem.m stands for mass, and v stands for velocity. The numbers refer to the respective objects.
m1v1 + m2v2 = m1vf1 + m2vf2
Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.
m1v1 + m2v2 = vf(m1 + m2)
Let’s substitute in our givens.
(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)
I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.
Note that I have considered the bullet’s velocity to be in the positive direction,
The answer is vf = 0.280 m/s
Can someone help label these?
Can you solve this question?
Hi there!
In this instance, the object's centripetal force is provided by the horizontal component of the tension, so:
Tsinθ = mv²/r
**We use sine because in this situation, the angle is with the vertical**
We can plug in the known values for tension and theta:
60sin(60) = mv²/r
51.96 = mv²/r
The radius is equivalent to the sine of the string in respect to theta:
sin(60) = O/H = r/L
2sin(60) = 1.732 m
Now, solve for the velocity:
51.96 = mv²/r
51.96r / m = v²
51.96(1.732)/.400 = v²
v² = 225
v = 15 m/s
If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
[tex]\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time, t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0 +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}[/tex]
What is the net force here?
11 N left
6 N right
1 N right
4 N right
answer = 6n to the right
Explanation:
2n plus 4n equals 6n
since 6n is more than 5n it goes 6n to the right
A disgruntled physics student, frustrated with
finals, releases his tensions by bombarding the
adjacent building, 13.5 m away, with water
balloons. He fires one at 38◦
from the horizontal with an initial speed of 23.6 m/s.
The acceleration of gravity is 9.8 m/s
2
.
For how long is the balloon in the air?
Answer:
Explanation:
The balloon would require a time of
t = d/v = 13.5/ (23.6cos38) = 0.7259...s
to travel the horizontal distance.
the vertical position relative to the throw point at that time is
h = 0 + (23.6sin38)(0.7259) + ½(-9.8)(0.7259²)
h = 7.9652...
so as long as the adjacent building is at least 8.0 m higher than the student position, the balloon is in the air for 0.726 s.
If the building is shorter than 8.0 m above the student, the balloon will land on the building roof and will be in the air for a longer period of time
A spring in a dart gun is compresscht a distance of 0.05 m. The spring has a spring constant
of 1,115 N/m. If the dart has a mass of 0.025 kg, determine the velocity of the dart as it
leaves the dart gun.
Answer:
Explanation:
ASSUMING that the dart is fired horizontally so that gravity potential energy considerations are not needed. Also ignoring friction work.
The spring potential will convert to kinetic.
KE = PS
½mv² = ½kx²
v = [tex]\sqrt{kx^2/m}[/tex]
v = [tex]\sqrt{1115(0.05^2)/0.025}[/tex]
v = 10.55935...
v = 11 m/s
A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the ground). Neglect drag and the initial height of the football.
How long does the football need to rise?
What height will the football reach?
With what speed does the punter need to kick the football?
At what angle (θ), with the horizontal, does the punter need to kick the football?
Answer:
Explanation:
How long does the football need to rise?
4.70/3 = 2.35 s
What height will the football reach?
h = ½(9.81)2.35² = 27.1 m
With what speed does the punter need to kick the football?
vy = g•t = 9.81(2.35) = 23.1 m/s
vx = d/t = 56.0/4.70 = 11.9 m/s
v = √(vx²+vy²) = 26.0 m/s
At what angle (θ), with the horizontal, does the punter need to kick the football?
θ = arctan(vy/vx) = 62.7°
A block of mass m = 3.0 kg is pushed a distance d = 2.0 m along a frictionless horizontal table by
a constant applied force of magnitude F= 20.0 N directed at an angle 0= 30.0° below the horizontal
as shown in Figure. Determine the work done by (a) the applied force, (b) the normal force exerted
by the table, and (d) the net force on the block.
Explanation:
We apply the definition of work by a constant force in the first three parts, but then in the fourth part we add up the answers. The total (net) work is the sum of the amounts of work done by the individual forces, and is the work done by the total (net) force. This identification is not represented by an equation in the chapter text, but is something you know by thinking about it, without relying on an equation in a list.
The definition of work by a constant force is W=FΔrcosθ.
(a) The applied force does work given by
W=FΔrcosθ=(16.0N)(2.20m)cos25.00=31.9J
(b), (c) The normal force and the weight are both at 900 to the displacement in any time interval. Both do 0 work.
(d) ∑W=31.9J+0+0=31.9J
Understanding what motivates anyone is not easy because each individual has different
Find the net torque .
Answer:
Explanation:
I will ASSUME this means torque about the dot.
3) 20(3) + 10(6) - 30(4) = 0 N•m
4) 10(0.5) - 6sin45(1) = -0.7573593... or about 0.76 N•m CCW
5) 25(3) - 40sin30(4) = -5 N•m or 5 N•m CCW
6) 15(3) - 12(2) - 10sin45(6) = -21.4264068... or about 21 N•m CCW
In a Little League baseball game, the 145 g ball enters the strike zone with a speed of 17.0 m/s . The batter hits the ball, and it leaves his bat with a speed of 20.0 m/s in exactly the opposite direction. Part A What is the magnitude of the impulse delivered by the bat to the ball
Hi there!
Impulse = Change in momentum
I = Δp = mΔv = m(vf - vi)
Where:
m = mass of object (kg)
vf = final velocity (m/s)
vi = initial velocity (m/s)
Begin by converting grams to kilograms:
1 kg = 1000g ⇒ 145g = .145kg
Now, plug in the given values. Remember to assign directions since velocity is a vector. Let the initial direction be positive and the opposite be negative.
I = (.145)(-20 - 17) = -5.365 Ns
The magnitude is the absolute value, so:
|-5.365| = 5.365 Ns
The mass of fifteen washers is _____ kg, which exerts a force of _____ N
Answer:
It could be related with the lesson from which this question belongs as far we did not read the lesson
Sorry
what is the force that every mass experts on every other mass called?
Answer: The forces of gravity
Explanation: The consequence of this phenomenon is that every mass exerts a so-called "force of mutual attraction" on every other mass. The attractive force that the celestial bodies exert on other masses by virtue of their total mass is called the force of gravity.
Hope this helps
59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determine the acceleration of the crate as it slides down the plane. (b) If the crate starts from rest 8.15 m up along the plane from its base, what will be the crate's speed when it reaches the bottom of the incline?
Explanation:
a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:
[tex]x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)[/tex]
[tex]y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)[/tex]
From Eqn(2), we see that
[tex]N = mg\cos 25°\;\;\;\;\;\;\;(3)[/tex]
so using Eqn(3) on Eqn(1), we get
[tex]mg\sin 25° - \mu_kmg\cos 25° = ma[/tex]
Solving for the acceleration, we see that
[tex]a = g(\sin 25° - \mu_k\cos 25°)[/tex]
[tex]\;\;\;\;= 2.45\:\text{m/s}^2[/tex]
b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation
[tex]v^2 = v_0^2 + 2ax[/tex]
Since the crate started from rest, [tex]v_0 = 0.[/tex] Thus our equation reduces to
[tex]v^2 = 2ax \Rightarrow v = \sqrt{2ax}[/tex]
[tex]v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}[/tex]
[tex]\;\;\;\;= 6.32\:\text{m/s}[/tex]
True or False If the mass of the object increases, then the potential energy of the object decreases.
what torque required stopping awheel of moment of inertia 6 × 10^-3kgm2 from speed of 40rad/s in 20 sec.
solution:
the formula is T = F * r * sin(theta) so just input the numbers and solve it.
Explanation:
Torque is the twisting force that tends to cause rotation. The point where the object rotates is known as the axis of rotation. Mathematically, torque can be written as T = F * r * sin(theta), and it has units of Newton-meters
An object of mass 6.36 kg is released from rest and drops 2.05 m to the floor. The collision is completely inelastic. How much kinetic energy is lost during the collision
Answer:
Essentially all of it
Explanation:
The potential energy was
PE = mgh = 6.36(9.81)(2.05) = 127.90278 = 128 J
ignoring air resistance, this PE converts to KE
With no rebound final velocity is zero, so Kinetic energy lost = 128 J
what is the pressure exerted by a force of 25 N on an area of 5m square
Answer:
pressure = force / area
then pressure = 25 / 5 = "5" N/m^2
I need ideas of what kind of simple motor i can build and how i can build it. The simple motor MUST spin without using your own force. What materials would i use and how would i create it. what would i create
Answer:
i don't know but my father i think he can't answer this
An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time
Answer:
[tex]\huge\boxed{\sf a = 1200\ m/s\²}[/tex]
Explanation:
Given Data:
Initial Velocity = Vi = 40 m/s
Final Velocity = Vf = 80 m/s
Distance = S = 200 m
Required:
Acceleration = a = ?
Formula:
2aS = Vf² - Vi² (THIRD EQUATION OF MOTION)
Solution:
2a (200) = (80)² - (40)²
400a = 6400 - 1600
400a = 4800
Divide 400 to both sides
a = 4800 / 400
a = 1200 m/s²
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807AnswAnswer This!!!!!!
I'll give brainliest to whoever gets it right.
Name the energy possessed by hot air
Answer:
geothermal energy
Explanation:
the energy is obtained from the heat within the surface of earth
Answer:
heat energy
Explanation:
CAN SOMEONE PLZ HELP
Answer:
magnetic force.
Explanation:bc it makes sense, and can i please get brainliest answer i never asked. its ok if you say no. have a great day <3.
what type of data do you need to collect in a ADI
if the momentum of a 1,400 kg car is the same as the truck in question 17, what is the velocity of the car?
Answer:
Explanation:
momentum is mass times velocity
p = mv
so take the momentum of the truck in question 17 and divide by the mass of this car
v = p/m = p / 1400
A 5.0 m length of rope, with a mass of 0.52 kg, is pulled taut with a tension of 46 N. Find the speed of waves on the rope
Answer:
Speed of waves on the rope is 21 m/s
Explanation:
Length of the rope (l) = 5.0 m
Mass of the rope (m) = 0.52 kg
Tension in the rope (T) = 46 N
Formula of speed of waves on the rope:
[tex] \bold{v = \sqrt{\dfrac{T}{\mu}}} [/tex]
[tex] \mu [/tex] = Mass per unit length of the rope (m/l)
By substituting the values in the formula we get:
[tex] \implies \rm v = \sqrt{\dfrac{T}{ \dfrac{m}{l} }} \\ \\ \implies \rm v = \sqrt{\dfrac{Tl}{m}} \\ \\ \implies \rm v = \sqrt{ \dfrac{46 \times 5}{0.52} } \\ \\ \implies \rm v = \sqrt{ \dfrac{230}{0.52} } \\ \\ \implies \rm v = \sqrt{442.3} \\ \\ \implies \rm v = 21 \: m {s}^{ - 1} [/tex]
Speed of waves on the rope (v) = 21 m/s