Answers
Answer:
C
Explanation:
sin of a right triangle is opposite divided by hypotenuse
in this case, that is 5/8.6 = 0.581395349
you can easily remember the trig functions with this acronym:
SOH-CAH-TOA
S: Sine
O: Opposite
H: Hypotenuse
C: Cosine
A: Adjacent
H: Hypotenuse
T: Tangent
O: Opposite
A: Adjacent
Explanation:
sin = Mp /0P
sin = 5/8,6
sin = 0,58
Related Questions
Using the American Engineering system of units (AES), a) Calculate the weight of a 170.5 lbm person on the surface of the earth, where the local acceleration due to gravity is 32.174 ft/s2 . Report your answer in pound-force, lbf. b) What would be the weight of a 170.5 lbm astronaut on the moon, where the local acceleration due to gravity is 5.32 ft/s2 . Report your answer in pound-force, lbf.
Answers
Answer:
a) the weight of the person is 170.5 lbf
b) weight of the astronaut on the moon is 28.2 lbf
Explanation:
Given the data in the question;
a)
we know that;
weight on the surface of the earth = m[tex]g_{earth[/tex]
given that m = 170.5 lbm and g = 32.174 ft/s²
we substitute
weight on the surface of the earth = 170.5 lbm × 32.174 ft/s²
= 5485.667 lbm-ft/s²
1 lbf = 32.174 lbm-ft/s²
so
weight on the surface of the earth = (5485.667 / 32.174) lbf
weight on the surface of the earth = 170.5 lbf
Therefore, the weight of the person is 170.5 lbf
b)
given that;
weight on the surface of the earth = m[tex]g_{moon[/tex]
m = 170.5 lbm and g = 5.32 ft/s²
weight on the surface of the earth = 170.5 lbm × 5.32 ft/s²
= 907.06 lbm-ft/s²
1 lbf = 32.174 lbm.ft/s²
weight on the surface of the earth = ( 907.06 / 32.174 ) lbf
weight on the surface of the earth = 28.2 lbf
Therefore, weight of the astronaut on the moon is 28.2 lbf
a force is applied to the right to drag a sled to the right across Loosely packed snow with a rightward acceleration (considered friction forces)
Answers
Answer:
1.67 m/s²
Explanation:
Total force is;
ΣF = F_f + F_a
From the free body diagram attached, we see that the frictional force F_f is acting in a negative direction.
Thus;
ΣF = -15 + 30
ΣF = 15 N
Now, to get the acceleration, we will use the formula;
ΣF = ma
We are given m = 9 kg
Thus;
15 = 9a
a = 15/9
a = 1.67 m/s²
Which subatomic particle identifies the element?
A.
protons
B.
electrons
C.
neutrons
D.
atomic mass
Answers
Answer:
A. Protons
Explanation:
Atomic number is an indicator of how many protons an element possesses in its nucleus. We use atomic number to identify an element. For example, carbon has 6 protons and an atomic number of 6.
If the hoop has speed vo at the bottom of
the pipe, what is its speed when it has rolled
halfway up the side of the pipe?
Answers
The speed of the hoop when it has rolled halfway up the side of the pipe is √(v₀² - gR).
Conservation of energyThe speed of the hoop when it has rolled halfway up the side of the pipe is calculated as follows;
K.E = P.E
- ¹/₂mv₀² + ¹/₂Iω² = (mgh₀ - mghf)
- ¹/₂mv₀² + ¹/₂Iω² = (0 - 0.5mgh) (hf = 0.5h) (half way up)
¹/₂Iω² = ¹/₂mv₀² - 0.5mgh
where;
I is moment of inertia of the hoop = mr²ω is angular speed = v/r¹/₂(mr²)(vf/r)² = ¹/₂mv₀² - 0.5mgh
¹/₂vf² = ¹/₂v₀² - ¹/₂gh
vf² = v₀² - gh
vf = √(v₀² - gh)
where;
h is the distance traveled half-way up the pipe = Rvf = √(v₀² - gR)
[tex]v_f = \sqrt{v_0^2 - gR}[/tex]
Learn more about moment of inertia of hoop here: https://brainly.com/question/14956994
A 20g bullet moving at 200m/s hits a bag of sand and comes to rest in 0.011s, calculate the momentum of the bullet just before hitting the bag
Answers
Answer:
momentum = mass * velocity ... kg•m/s
acceleration = v / t = 200 m/s / 11 ms ... m/s^2
force = mass * acceleration = .02 kg * (200 m/s / 11 ms) N
Explanation:
An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it rotates around its axis. Once the Rotor reaches its operating speed, the floor drops but the riders remain pinned against the wall of the cylinder. Suppose the cylinder makes 26.0 rev/min and has a radius of 3.70 m. 1) What is the coefficient of static friction between the wall of the cylinder and the backs of the riders
Answers
Answer:
μs = 0.36
Explanation:
While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.This acceleration is produced by the centripetal force.Now, this force is not a different type of force, is the net force acting on the riders in this direction.Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:[tex]F_{frmax} = \mu_{s}* F_{n} = m * g (1)[/tex]
where μs= coefficient of static friction (our unknown)As we have already said Fn = Fc.The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:[tex]F_{n} = m* \omega^{2} * r (2)[/tex]
Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:[tex]\mu_{s} = \frac{g}{\omega^{2} r} (3)[/tex]
Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:[tex]\omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)[/tex]
Replacing g, ω and r in (3):[tex]\mu_{s} = \frac{g}{\omega^{2} r} = \frac{9.8m/s2}{(2.72rad/sec)^{2} *3.7 m} = 0.36 (5)[/tex]4. A train starts its journey and accelerates at 5 ms 2. How long does it take for it to reach a velocity of 100 ms??
Answers
Answer:
20 seconds
Explanation:
A screw-jack used to lift a bus is a
A) first order lever
B) second order lever
C) pulley
D) screw
Answers
Answer:
c
Explanation:
A vertical piston cylinder assembly contains 10.0kg of a saturated liquid-vapor water mixture with initial quality of 0.85 The water receives energy by heat transfer until the temperature reaches 320*C. The piston has a mass of 204kg and area of 0.005m2. Atmospheric pressure of 100kPa acts on the top side of the piston. Local gravitational acceleration is 9.81m/s2 Calculate the amount of heat transfer between the water and the surroundings in kJ. Enter a numeric value only. 6735.66
Answers
Answer:
Explanation:
From the given information:
At state 1:
Initial Quality [tex]= x_1 = 0.85[/tex]
mass = 10.0 kg
At state 2:
Temperature [tex]T_2 = 320^0[/tex]
mass of the piston [tex]m_p = 204 \ kg[/tex]
area of the piston [tex]A_p = 0.00 5 \ m^2[/tex]
Atmospheric pressure [tex]P_{atm}= 100 \ kPa = 100 \times 10^3 \ Pa[/tex]
Gravitational acceleration = 9.81 m/s²
[tex]\mathbf{P= P_1=P_2}[/tex], This is because there exists no restriction to the movement of the piston and provided the process is frictionless. So, the process 1-2 is regarded as constant.
To calculate the applying force balance over the piston by using force balance in the vertical direction:
[tex]\mathbf{P_{AP} = P_{atmA_p} + m_pg}[/tex]
∴
(100 × 10³)×0.005 + 204 × 9.31 = P × 0.05
P = 500248 Pa
P = 500.25 kPa
At state 1:
[tex]\mathbf{P_1 = P = 500.25 \ kPa}[/tex]
[tex]x_1 = 0.85[/tex]
Hence, this is a saturated mixture of liquid and vapor
Using the steam tables at 500.25 kPa
[tex]V_f = 1.093 \times 10^{-3} \ m^3/kg \\ \\ V_g = 0.375 \ m^3/kg \\ \\ U_f = 639.72 \ kJ/kg \\ \\ U_g = 2560.72 \ kJ/kg[/tex]
∴
Specific volume at state 1 is given as:
[tex]V_1 = [ V_f +x_1(v_g -v_f) ] \ at \ 500.25 \ kPa \\ \\ V_1 = 0.319 \ m^3/kg[/tex]
volume at state 1 is given by:
[tex]V_1 = mV_1 = 10 \times 0.319 \\ \\ V_1 = 3.19 \ m^3[/tex]
Similarly, the specific internal energy is:
[tex]U_1 = [U_f +x_1 (U_o-Uf)] \ at \ 500.25 \ kPa[/tex]
[tex]U_1 = 639.72 +0.82 (2560.72 -639.72)[/tex]
[tex]U_1 = 2272.57 \ kJ/kg[/tex]
At state 2:
[tex]P = P_1 = P_2 = 500.25 \ kPa \\ \\ T_2 = 320^0 \ C[/tex]
Using steam tables at P = 500.25 kPa and T = 320° C
[tex]V_2 = 0.541 \ m^3/kg \\ \\ U_2 = 2835.08 \ kJ/kg[/tex]
∴
[tex]V_2 = mV-2 = 10 \times V_2 = 5.41 \ m^3[/tex]
[tex]\text{Now; Applying the 1st law of thermodynamics to the system}[/tex]
[tex]_1Q_2 -_1W_2 = \Delta V =m(u_2-u_1) \\ \\ where;\ _1W_2 = P(V_2-V_1) \\ \\ _1Q_2 -P(V_2-V_1) = m(u_2-u_1) \\ \\ _1Q_2 - 500.25(5.91 -3.19) = 10( 2835.08 -2272.57) \\ \\ \mathbf{ _1Q_2 = 6735.66 \ kJ}[/tex]
You start biking at the top of a steep hill. As you travel downhill, you apply
your brakes to control your speed. What are the energy transformations
taking place in this system?
A. Kinetic energy to mechanical energy to chemical energy
B. Potential energy to kinetic energy to heat energy
C. Thermal energy to mechanical energy to potential energy
D. Electric energy to kinetic energy to chemical energy
Answers
An electron, moving toward the west, enters a uniform magnetic field. Because of this field the electron curves upward. The direction of the magnetic field is An electron, moving toward the west, enters a uniform magnetic field. Because of this field the electron curves upward. The direction of the magnetic field is downward. towards the west. upward. towards the south. towards the north.
Answers
Answer:
towards the north.
Explanation:
The uniform magnetic field has strength and direction in all points. The upwards motion of the field or he electronic curves will show a northern direction. Hence, the field can be created by taking the opposite magnetic in the two directions. The straight line of motion of the field will indicate the flow in the north direction. The magnetic field lines will remain parallel and stay uniform to poles.when is an object considered to be in motion
Answers
Answer:
An object is considered to be in motion only when its position changes over time with reference to a point which will be known as orgin
What is the acceleration of a 7 kg object if a force of 63 N is applied?
Answers
Answer:
9m/s^2
Explanation:
F=ma
a=F/m
a=63/7
a=9m/s^2
He-Ne Laser device emits photons of wave length 632.8 nm by rate 4.5 x 1020 photon/s, so
the power of the laser beam =
a. 3.14 W
b. 141.3 W
c. 314.1 w
d. 431.4 W
Answers
Answer: Option b.
Explanation:
We know:
Wavelength = 632.8 nm
Fluence = 4.5*10^20 photon/s
The energy of a single photon of wavelength λ is:
E = (h*c)/λ
where:
h = 6.6*10^(-34) J*s
c = 3*10^8 m/s
And we should rewrite the wave length in meters, so:
λ = 632.8 nm = 632.8*10^(-9) m
replacing these in the energy equation, we get:
E = (6.6*10^(-34)J*s)*(3*10^8 m/s)/(632.8*10^(-9) m) = 3.13*10^(-19) J
So each one of the 4.5x10^20 photon that flow each second have this energy, then the power is:
P = (3.13*10^(-19) J)*(4.5*10^20 /s) = 140.85 J/s
and 1 W = 140.85 J/S
Then the power is:
P = 140.85 W
Then the correct answer is the option b, where the units are a little bit different than mine because I used really simplified values for c and h.
A child balancing while standing on a chair is demonstrating which type of energy?
A. Potential energy,
B. Nuclear energy,
C. Electrostatic energy,
Answers
Answer:
Potential Energy
Explanation:
HELP - What are some good topics to add to a Pipeline presentation? So yes, a presentation about pipelines, like for example in the US. Have any ideas??
Answers
Netwtons 1st law of motion
Answers
Answer:
Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. This postulate is known as the law of inertia.
Explanation:
I need help!!!!!!!!!!!pleaseeeeeee
Answers
And the second one is the the 1 and 2
Help please :)
Infrared waves are transmitted by ....... ?
Answers
Suppose that a uniform rope of length L resting on a frictionless horizontal surface, is accelerated along the direction of its length by means of a force F, pulling it at one end. A mass M is accelerated by the rope. Assuming the mass of the rope to be m and the acceleration is a. Stated in terms of the product ma, what is the tension in the rope at the position 0.3 L from the end where the force F is applied if the mass M is 1.5 times the mass of the rope m?
Answers
Answer:
2.2 ma
Explanation:
Given :
Length of the rope = L
Mass of the rope = m
Mass of the object pulled by the rope = M
M = 1.5 m
So, L [tex]$\rightarrow$[/tex] m
For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]
∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]
= 0.3 m
And for remaining 0.7 L = [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]
= 0.7 m
By Newtons law of motion,
F - T = ( 0.3 m) a .........(1)
T = ( M + 0.7 m) a
T = ( 1.5 m + 0.7 m) a
T = ( 2.2 m ) a ..............(2)
So from equation (1) and (2), we have
Tension on the rope
T = 2.2 ma
As a sports car travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the sports car by the air and the road. If the power developed by the engine is 0.824 hp, calculate the total friction force acting on the sports car (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W.
Answers
Answer:
20.49 N
Explanation:
From the question given above, the following data were obtained:
Power (P) = 0.824 hp
Velocity (v) = 30 m/s
1 h = 746 W
Force (F) =?
Next, we shall convert 0.824 hp to Watts (W). This can be obtained as follow:
1 h = 746 W
Therefore,
0.824 hp = 0.824 hp × 746 W / 1 h
0.824 hp = 614.704 W
Finally, we shall determine the force as follow:
Power (P) = 614.704 W
Velocity (v) = 30 m/s
Force (F) =?
P = F × v
614.704 = F × 30
Divide both side by 30
F = 614.704 / 30
F = 20.49 N
3. If a spring extends by 3 cm when a 4 N weight is suspended from it, find the extension
when the weight is changed to
(a) 8 N
(b) 10 N
(c) 14 N
Answers
10N - 7.5cm
14N - 10.5cm
What is the wavelength of this wave
Answers
[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Waves.
Wavelength is the distance between two consecutive crest or trough.
hence, here the distance is 10cm
So the wavelength is 10cm
===> 10 cm
I'm asking for a quick favor. I'm trying to understand an equation that has to do with Projectile motion, Bernoulli's principle, and Magnus Effect. Basically focused on understanding air resistance on a projectile. I would like to discuss this privately rather than have it on this public forum. I'll give you 100 of my points if you help.
Answers
Explanation:
Projectile motion, Bernoulli's principle, and Magnus Effect.
Sure I would be happy to discuss projectile motion!
I'll do it if you mark brainliest :) I need the points thanks
The amplitude of a wave
determines the volume of a
sound.
True
O False
Answers
3. A woman drove her car from home to her daughter's school. The odometre on her dashboard says she travelled 4.5 km to do this. She then immediately drove back home, using a different route, which was 5.5 km long. The whole journey took 30 minutes.
a. What distance did she travel?
b. What was her displacement?
C. What was her average speed during the journey?
Answers
Answer:
Look Below -->
Explanation:
a. She traveled 10 km, add 4.5 km + 5.5 km = 10 km (Distance is the total units travelled, so just add them all up :) )
b. Her displacement is 0 km because she went back home. (Displacement is the difference between the end and starting points)
c. 3 km/hr (30 minutes / 10 km)
If you increase the frequency of a sound wave four times, what will happen to its speed?
А
The speed will increase four times.
B.
The speed will decrease four times,
C. The speed will remain the same.
D.
The speed will increase twice.
E.
The speed will decrease twice.
Answers
PLEASE HELP 15 POINTS AND BRAINIEST!!!
Reporting fake answers
Answers
what is the acceleration of an object that has a velocity of 25.0m/s and is moving in a circle of a radius of 10.0 m?
Answers
Answer:
[tex]a=62.5\ m/s^2[/tex]
Explanation:
Given that,
The velocity of the object, v = 25 m/s
Radius of the circle, r = 10 m
We need to find the acceleration of the object. The centripetal acceleration act on it. It is given by :
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(25)^2}{10}\\\\a=62.5\ m/s^2[/tex]
So, the acceleration of the object is equal to [tex]62.5\ m/s^2[/tex].
In what order were the following energy sources
discovered by humans