Answer:
A. ) F₁ₓ = -45.0 N F₁y = 53.6 N
B.) F₂ₓ = 3.48 N F₂y = -19.7 N
C.) Fₓ = -41.5 N Fy = 33.9 N
D) F = 53.6 N
E) θ = -39. 2º (320.8º)
Explanation:
A)
Applying simple trig, like definitions of cos and sin of an angle, we can get the x- and y- components of F₁, as follows:[tex]F_{x1} = F_{1} * cos (130) = 70 N * cos (130) = -45 N (1)\\F_{y1} = F_{1} * sin (130) = 70 N * sin (130) = 53.6 N (2)[/tex]
B)
Repeating for F₂:[tex]F_{x2} = F_{2} * cos (280) = 20 N * cos (280) = 3.48 N (3)\\F_{y2} = F_{2} * sin (280) = 20 N * sin (280) = -19.7 N (4)[/tex]
C)
The x- and y- components of the resultant force, are just the algebraicsum of the x- and - y components of F₁ and F₂:
Fₓ = Fₓ₁ + Fₓ₂ = -45 N + 3.48 N = -41.5 N (5)By the same token, Fy can be written as follows:Fy = Fy₁ + Fy₂ = 53.6 N + (-19.7 N) = 33.9 N (6)D)
The magnitude of the resultant force can be obtained applying the Pythagorean Theorem to Fx and Fy, as follows:[tex]F_{t} =\sqrt{F_{x} ^{2} + F_{y} ^{2} } = \sqrt{(-41.5N)^{2} +(33.9N)^{2}} = 53.6 N (7)[/tex]
E)
Finally the angle regarding the x- axis of the resultant force vector, can be obtained using the definition of the tangent of an angle, as follows:[tex]\theta = arc tg \frac{33.9N}{(-41.5N)} = arc tg (-0.817) = -39. 2 \deg[/tex]
A. A piece of paper near a magnet
B. An aluminum nail near a magnet
C. An iron nail, not near a magnet
D. An iron nail near a magnet
Answer:
it’s c not d
Explanation:
took the test
Answer: D!!!
Explanation: jus got it wrong from the other answer.
Describe the motion of an object as it accelerates. IN YOUR OWN WORD!! ASAP
Answer:
The aceleration of an object is in the direction of the net force. If you push or pull an object in a particular direction, it accelerates in that direction. The aceleration has a magnitude directly proportional to the magnitude of the net force.
Explanation:
Hope this helps Plz mark brainliest
What do we call the material such as air that light travels through
Answer:
Transparent or Translucent
Explanation:
If two exactly the same cars are driving down a road, which one would have the most kinetic energy. The one that is moving faster, the one that is moving downhill, the one that is moving uphill, or the one that is moving slower.
Answer: the car that is moving downhill
Explanation:
A rocket will move upward as long as which condition applies?
A metal box of weigh 20 N rests on its 1 m by 0.6m side on floor. How
much is the pressure exerted by the metal box on the floor? Take g=
10 m/s
Answer:
P = 33.33 [Pa]
Explanation:
The pressure can be calculated by the relationship of the force over the area.
[tex]P =F/A[/tex]
where:
F = force = 20 [N]
A = area = 1 x 0.6 = 0.6 [m²]
Now replacing:
[tex]P=20/0.6\\P=33.33 [Pa][/tex]
A hare can run at a rate of 15 m/s, while a turbocharged tortoise can now crawl at a rate of 3 m/s, how much of a head-start (time-wise) does the tortoise need in order to tie the hare in a 250 meter race?
A.
16.7 seconds
B.
66.7 seconds
C.
83.3 seconds
D.
100 seconds
Answer:
t = 66.7 s
Explanation:
Given that,
Speed of a hare, v = 15 m/s
Speed of a turbocharged tortoise, v' = 3 m/s
The hare in a 250 meter race
Let the Hare takes time t. It can be calculated as follows :
[tex]t=\dfrac{250}{15}=16.67\ s[/tex]
Let a turbocharged tortoise takes t'. It can be calulated as follows :
[tex]t'=\dfrac{250}{3}= 83.33\ s[/tex]
To tie the race, required time is given by :
[tex]\Delta t = t'-t\\\\=83.33-16.67\\\\=66.66\ s\\\\\approx 66.7\ s[/tex]
Hence, the correct option is (b) i.e. 66.7 seconds.
Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?
Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]
Therefore, the drill's angular displacement during that time interval is 24.17 rad.
[2.21] Please help me find a) and b)
Answer:
A. 28.42 m/s
B. 41.21 m.
Explanation:
A. Determination of the initial velocity of the ball:
Time (t) to reach the maximum height = 2.9 s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) =?
Thus, we can obtain the initial velocity of the ball as follow:
v = u + gt
0 = u + (–9.8 × 2.9)
0 = u – 28.42
Collect like terms
u = 0 + 28.42
u = 28.42 m/s
Therefore, the initial velocity of the ball is 28.42 m/s.
B. Determination of the maximum height reached.
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = –9.8 m/s² (since the ball is going against gravity)
Initial velocity (u) = 28.42 m/s.
Maximum height (h) =?
Thus, we can obtain the maximum height reached by the ball as follow:
v² = u² + 2gh
0² = 28.42² + (2 × –9.8 × h)
0 = 807.6964 + (–19.6h)
0 = 807.6964 – 19.6h
Collect like terms
0 – 807.6964 = – 19.6h
– 807.6964 = – 19.6h
Divide both side by – 19.6
h = –807.6964 / –19.6
h = 41.21 m
Therefore, the maximum height reached by the ball is 41.21 m
3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterwards, the 3.00 kg block moves 1.71 m/s left. find the velocity of the 2.22 kg block afterwards
Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then
(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) v
where v is the velocity of the 2.22 kg block after collision. Solve for v :
6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) v
(2.22 kg) v = 2.70 kg•m/s
v = (2.70 kg•m/s) / (2.22 kg)
v ≈ 1.22 m/s
i.e. a velocity of about 1.22 m/s to the right.
An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life?
Answer:
The half-life is [tex] t_{1/2} = 1.005 h[/tex]
Explanation:
Using the decay equation we have:
[tex]A=A_{0}e^{-\lambda t}[/tex]
Where:
λ is the decay constantA(0) the initial activityA is the activity at time tWe know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that [tex]A = \frac{A_{0}}{2}[/tex]
[tex]\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}[/tex]
[tex]0.5=e^{-\lambda*1 h}[/tex]
Taking the natural logarithm on each side we have:
[tex]ln(0.5)=-\lambda[/tex]
[tex]\lambda=0.69 h^{-1}[/tex]
Now, the relationship between the decay constant λ and the half-life t(1/2) is:
[tex]\lambda = \frac{ln(2)}{t_{1/2}}[/tex]
[tex] t_{1/2} = \frac{ln(2)}{\lambda}[/tex]
[tex] t_{1/2} = \frac{ln(2)}{0.69}[/tex]
[tex] t_{1/2} = 1.005 h[/tex]
I hope it helps you!
A 30 N force toward the west is applied to an object. The object moves 50 m east during the time the force is applied. What is the change in kinetic energy of the object?
a) 1.0 J
b) 750 J
c) 1.7 J
d) -1500 J
Answer:
D.-1500Joules
Explanation:
The change in kinetic energy of the object s equivalent to the workdone by the body in the west direction (negative x direction)
Workdone = Force * Distance
Given
Force = 30N
Distance moved by the object = 30m
Required
Kinetic energy
Kinetic energy = 30 * 50
Kinetic energy = 1500Joules
Since the body moves in the negative direction, hence the kinetic energy will be -1500Joules
A vertical tube one meter long is open at the top. It is filled with 50 cm of water. If the velocity of sound is 344 m/s, what will the fundamental resonant frequency be (in Hz)?
Answer:
The fundamental resonance frequency is 172 Hz.
Explanation:
Given;
velocity of sound, v = 344 m/s
total length of tube, Lt = 1 m = 100 cm
height of water, hw = 50 cm
length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm
For a tube open at the top (closed pipe), the fundamental wavelength is given as;
Node to anti-node (N ---- A) : L = λ / 4
λ = 4L
λ = 4 (50 cm)
λ = 200 cm = 2 m
The fundamental resonance frequency is given by;
[tex]f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\[/tex]
Therefore, the fundamental resonance frequency is 172 Hz.
A car goes around a circular track at 30 m/s. If the radius of the curve is 90 m, what is the period of the car's revolution around the track?
Answer:
18.9s
Explanation:
Using the formula;
ω = v/r
Where;
ω = angular velocity (rad/s)
v = linear velocity (m/s)
r = radius of the circular track (m)
According to the given information, v = 30m/s, r = 90m
ω = v/r
ω = 30/90
ω = 3/9
ω = 0.3333 radians/seconds.
Since ω = 2π/T
Where;
π = 3.142
T = period (s)
ω = angular velocity
0.333 = 2 × 3.142/T
T = 2 × 3.142/0.333
T = 6.284/0.333
T = 18.87s
T = 18.9s
a mass of 2.00 kg rest on a rough horizontal table. The coefficient of static friction between the block and the table is 0.60. The block is attached to a hanging mass by a string that goes over a smooth pulley,as shown in the diagram. Determine the largest mass that can hang in this way without forcing the block to slide.
Answer:
1.2 kg
__________________________________________________________
We are given:
Mass of the block = 2 kg
Coefficient of Static Friction = 0.6
__________________________________________________________
Friction Force on the Block:
Finding the Normal Force:
We know that the normal force will be equal and opposite to the weight of the 2 kg block
So, Normal Force = mg
replacing the variables with the given values
Normal Force = (2)(9.8) [Taking g = 9.8]
Normal Force = 19.6 N
Friction force on the Block:
We know that:
Coefficient of Static Friction = Static Friction Force/Normal Force
replacing the variables
0.6 = Static Friction force / 19.6
Static Friction force = 0.6*19.6 N [Multiplying both sides by 19.6]
Static Friction force = 11.76 N
__________________________________________________________
Largest Mass that can Hang:
We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest
So, if the weight of the second block is less than the static friction force, it will hang
Weight of the second block ≤ 11.76
We know that weight = mg
mg ≤ 11.76
m(9.8) ≤ 11.76 [since g = 9.8]
m ≤ 1.2 kg [dividing both sides by 9.8]
From this, we can say that the maximum mass of the second block is 1.2 Kg
There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N. The bell has ____________
energy.
Answer:
The bell has a potential energy of 8550 [J]
Explanation:
Since the belt is 45 [m] above ground level, only potential energy is available. And this energy can be calculated by means of the following equation.
[tex]E_{p}= W*h\\E_{p} = 190*45\\E_{p}=8550[J][/tex]
A student throws a baseball upwards at an angle of 60 degrees to the horizontal. The initial
horizontal and vertical components are 12.5 m/s and 21.7 m/s, respectively.
60 degrees
Refer to the above information and diagram. What position will have the smallest magnitude
of vertical velocity?
Answer:
The initial horizontal and vertical components are 12.5 m/s and 21.7 m/s, respectively?
Explanation:
A container contains a number of marbles each having a mass 80 mg. If the total mass of the marbles in the container is 2 kg. find the numbers in the container
Answer:
There are 25,000 marbles in the container
Explanation:
There is a certain number of marbles in a container.
We know each marble has a mass of 80 mg and we also know the total mass of the marbles in the container is 2 Kg.
Since the data is given in different units of mass, we convert them into one unit in common.
Let's convert 2 Kg to milligrams. There are 1,000 grams in a kilogram, and there are 1,000 milligrams in a gram, thus there are one million milligrams in one kilogram, that is:
1 Kg = 1,000,000 mg
And 2 Kg = 2,000,000 mg
The number of marbles can be found by dividing the total mass by the individual mass:
No. of marbles = 2,000,000 / 80 = 25,000
There are 25,000 marbles in the container
What is the name for family labeled #4 (Yellow)?
#3
#5
#2
#
341 sud-
lasa 1
17:55
Alkaline Earth Metals
Metalloids
Transition Metals
Alkali Metals
Answer:
transition metals im sorry if this was too late
in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s about its vertical axis. the radius of the disk is 1 meter. What is the magnitude of the friction?
Answer:
25
Explanation:
n
Question 4
1 pts
A bus travels on an interstate highway at an average speed of 90 km/hrs. How far does it take to travel
in 30 mins? The distance equals speed times time, or d = st.
O 45 km
O 98 Km
O 56 km
O 432 Km
[tex]d = s \times t \\ d = 90 \times \frac{30}{60} \\ d = 90 \times \frac{1}{2 } \\ d = 45km[/tex]
3) A 10kg object rests on a frictionless surface when it is struck by a 300N force. At what rate will it accelerate?
3m/s/s
30m/s/s
0.3m/s/s
300m/s/s
Answer: 0.3m/s/s
(i'm really sorry if i'm wrong)
:(
Allure of the seas is one of the most expensive cruise ships around the world with a length of 362 meters(1,187 ft) and a height of 72 meters(236 ft) above water line. On her first day of operation she moves with a uniform acceleration of 83.5 km/hr2 from rest has gone 10 nautical miles. How many seconds she is in motion? Note: 1 nautical mile = 1.852 km (help 3 mins left
Answer:
about 2398 seconds
Explanation:
The relation between time, distance, and acceleration is ...
d = (1/2)at²
t = √(2d/a) = √(2·10·1.852 km/(83.5 km/h²)) ≈ √0.4436 h ≈ 0.6660 h
That is about ...
(0.6660 h)(3600 s/h) ≈ 2397.7 s
The cruise ship takes about 2397.7 seconds to cruise 10 nautical miles, accelerating all the way.
convert 100 Newton into dyne
Answer:10000000
Explanation:
a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop
Complete Question
a car traveling at 28.4 m/s undergoes a constant deceleration of 1.92 m/s2 when the breaks are applied. How many revolutions does each tire make before the car comes to a stop? Assume that the car does not skid and that each tire has a radius of 0.307 m. Answer in units of rev.
Answer:
The value is [tex]N = 109 \ rev[/tex]
Explanation:
From the question we are told that
The speed of the car is [tex]u = 28.4 \ m/s[/tex]
The constant deceleration experienced is [tex]a = 1.92 \ m/s^2[/tex]
The radius of the tire is [tex]r = 0.307 \ m[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2as[/tex]
Here v is the final velocity which is 0 m/s
So
[tex]0^2 = 28.4^2 + 2 * 1.92 * s[/tex]
=> [tex]s = 210.04 \ m[/tex]
Generally the circumference of the tire is mathematically represented as
[tex]C = 2 \pi r[/tex]
=> [tex]C = 2 * 3.142 * 0.307[/tex]
=> [tex]C = 1.929 \ m[/tex]
Generally the number of revolution is mathematically represented as
[tex]N = \frac{ s}{C}[/tex]
=> [tex]N = \frac{210.04}{1.929}[/tex]
=> [tex]N = 109 \ rev[/tex]
-I...................ok
Answer:
What?
Explanation:
Compute the specific heat capacity at constant volume of nitrogen (N2) gas. The molar mass of N2 is 28.0 g/mol.
Answer:
724.3J/Kg.K
Explanation:
CHECK THE COMPLETE QUESTION BELOW
Compute the specific heat capacity at constant volume of nitrogen (N2) gas.and compare with specific heat of liquid water. The molar mass of N2 is 28.0 g/mol.
The specific heat capacity can be computed by using expression below
c= CV/M
Where c= specific heat capacity
M= molar mass
CV= molar hear capacity
Nitrogen is a diatomic element, the Cv can be related to gas constant with 5/2R
Where R= 8.314J/mol.k
Molar mass= 28 ×10^-3Kg/mol
If we substitute to the expression, we have
c= (5R/2)/(M)
=5R/2 × 1/M
=(5×8.314) /(2×28 ×10^-3)
=724.3J/Kg.K
Hence, the specific heat capacity at constant volume of nitrogen (N2) gas is
724.3J/Kg.K
The specific heat of liquid water is about 4182 J/(K kg) which is among substance with high specific heat, therefore specific heat of Nitrogen gas is 724.3J/Kg.K which is low compare to that of liquid water.
It takes a truck 3.56 seconds to slow down from 112 km/h to 87.4 km/h. What is its average acceleration?
Answer:
1.92 m/s2
Explanation:
Determine the acceleration that results when a 12 N net force is applied to a 3 kg object.
Heya!!
For calculate aceleration, let's applicate second law of Newton:
[tex]\boxed{F=ma}[/tex]
⇒ Being:
→ F = Force = 12 N
→ m = Mass = 3 kg
→ a = aceleration = ?
Lets replace according formula and leave the "a" alone:
[tex]12\ N = 3\ kg * \textbf{a}[/tex]
[tex]\textbf{a} = 12\ N / 3\ kg[/tex]
[tex]\textbf{a} = 4\ m/s^{2}[/tex]
Result:
The aceleration of the object is of 4 m/s²
what is the force of an egg that is thrown at a brick wall if the egg has a mass of 0.3 kg and an acceleration of 50 m/s/s
Answer:
15N
Explanation:
F=ma so F=.3*50 therefore F=15N
The force of an egg that is thrown at a brick wall is equal to 15 N.
What is force?Force can be defined as the influence or effect that changes the state of the body of from motion to rest or vice versa. The S.I. unit of force is Newton (N) as well as force is a vector quantity. Force can change the direction or the speed of the moving object.
The force acting on an object can be calculated from the multiplication of the mass(m) and acceleration(a). The mathematical form of the second law of motion for force can be written as follows:
F = ma
Given, the mass of the egg, m = 0.3 Kg
The acceleration of the egg with which it is thrown on the wall, a = 50 m/s²
The force of an egg that is thrown at a brick wall can be calculated as:
F = ma = 50 ×0.3 = 15 N
Learn more about force, here:
brainly.com/question/13191643
#SPJ2