Answer:
t_pass = 2.34 m
t_stop = 4.68 s
Thus, for the car passing at constant speed the pedestrian will have to wait less.
Explanation:
If the car is moving with constant speed, then the time taken by it will be given as:
[tex]t_{pass} = \frac{D}{v}[/tex]
where,
t_pass = time taken = ?
D = Distance covered = 23 m
v = constant speed = (22 mi/h)(1609.34 m/1 mi)(1 h/3600 s) = 9.84 m/s
Therefore,
[tex]t_{pass} = \frac{23\ m}{9.84\ m/s} \\[/tex]
t_pass = 2.34 m
Now, for the time to stop the car, we will use third equation of motion to get the acceleration first:
[tex]2as = v_{f}^{2} - v_{i}^2\\a = \frac{v_{f}^{2} - v_{i}^2}{2D}\\\\a = \frac{(0\ m/s)^{2}-(9.84\ m/s)^2}{(2)(23\ m)}\\\\a = -2.1\ m/s^2[/tex]
Now, for the passing time we use first equation of motion:
[tex]v_{f} = v_{i} + at_{stop}\\t_{stop} = \frac{v_{f}-v_{i}}{a}\\\\t_{stop} = \frac{0\ m/s - 9.84\ m/s}{-2.1\ m/s^2}[/tex]
t_stop = 4.68 s
Constant velocity is the velocity which covers the same distance for each interval of the time.
The time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.
What is constant velocity?Constant velocity is the velocity which covers the same distance for each interval of the time.
It can be given as,
[tex]v=\dfrac{x}{t}[/tex]
As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.
Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.
Thus amount of time, [tex]t_{pass}[/tex] is,
[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]
As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.
Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.
Thus amount of time, [tex]t_{pass}[/tex] is,
[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]
According to the third equation of the motion acceleration can be given as,
[tex]v^2-u^2=2ax\\a=\dfrac{v^2-u^2}{2x}\\a=\dfrac{0^2-9.84^2}{2\times 23}\\a=-2.1 \rm \; m/s^2[/tex]
Now, use the first equation of motion, to get the required time,
[tex]v=u+at\\0=9.84+(-2.1)t\\t=4.68\rm \; s[/tex]
Therefore, the time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.
For more details about equation of motion, refer to the link:
https://brainly.com/question/8898885
How fast is a ball going when it hits the ground after being dropped from a
height of 16 m? The acceleration of gravity is 9.8 m/s2
A. 22.5 m/s
B. 28.1 m/s
O C. 17.7 m/s
D. 19.3 m/s
Hi there!
[tex]\large\boxed{\text{C. 17.7m/s}}[/tex]
Use the following kinematic equation to solve:
vf² = vi² + 2(ad)
Since the initial velocity is 0 m/s because it started at rest, we can eliminate this part of the equation:
vf² = 2ad
Plug in the given acceleration and distance:
vf² = 2(9.8)(16)
vf ≈ 17.7. The correct answer is C.
The amplitude of a wave
determines the volume of a
sound.
True
O False
what is tangential velocity
Answer:
Tangential velocity is the linear speed of any object moving along a circular path
Explanation:
Answer:
Tangential velocity is the linear speed of any object moving along a circular path. A point on the outside edge of a turntable moves a greater distance in one complete rotation than a point near to the center.
Explanation:
i hope this helps and can i get brainliest pls?
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.
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
I need help!!!!!!!!!!!pleaseeeeeee
What is the wavelength of this wave
[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
What is the acceleration of a 7 kg object if a force of 63 N is applied?
Answer:
9m/s^2
Explanation:
F=ma
a=F/m
a=63/7
a=9m/s^2
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)
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
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?
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].
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??
Answer:
20 seconds
Explanation:
Which subatomic particle identifies the element?
A.
protons
B.
electrons
C.
neutrons
D.
atomic mass
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.
A screw-jack used to lift a bus is a
A) first order lever
B) second order lever
C) pulley
D) screw
Answer:
c
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
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]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.
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.
In what order were the following energy sources
discovered by humans
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.
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
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?
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
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
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?
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)
In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.105 T magnetic field.
Part (a) What electric field strength, in volts per meter, is needed to select a speed of 3.8 × 106 m/s?
Part (b) What is the voltage, in kilovolts, between the plates if they are separated by 0.75 cm?
Answer:
a).[tex]$3.99 \times 10^5 \ v/m$[/tex]
b). 2.9925 kV
Explanation:
Given :
For mass spectrometer
The magnetic field = B
B = 0.105 T
a). Given speed, v = [tex]$3.8 \times 10^6 \ m/s$[/tex]
We known
[tex]$\frac{E}{B}=v$[/tex]
∴ [tex]$E= 3.8 \times 10^6 \times 0.105$[/tex]
[tex]$=3.99 \times 10^5 \ v/m$[/tex]
b). Now spectrometer, d = 0.75 cm
[tex]$d=0.75 \times 10^{-2} \ m$[/tex]
We known
[tex]$E=\frac{V}{d}$[/tex]
[tex]$V = E\times d$[/tex]
[tex]$V = 3.99 \times 10^5 \times 0.75 \times 10^{-2}$[/tex]
[tex]$V = 2.9925 \times 10^3 \ V$[/tex]
= 2.9925 kV
A child balancing while standing on a chair is demonstrating which type of energy?
A. Potential energy,
B. Nuclear energy,
C. Electrostatic energy,
Answer:
Potential Energy
Explanation:
PLEASE HELP 15 POINTS AND BRAINIEST!!!
Reporting fake answers
Time Dilation. A group of scientists discover a new, rare isotope and are able to store a small amount of it. They determine that the isotope is unstable and half of their sample will decay in 13.0 monthsmonths . The scientists need a new laboratory to properly conduct measurements on the isotope. If laboratory is built before half the sample decays there will still be enough of the isotope available for experiments. However, it will take 21.0 monthsmonths to build the new lab. Fortunately, the scientists can quickly start construction and they have access to a spaceship that can travel at speeds approaching the speed of light c = 3.00
This question is incomplete, the complete question is;
Time Dilation. A group of scientists discover a new, rare isotope and are able to store a small amount of it. They determine that the isotope is unstable and half of their sample will decay in 13.0 months . The scientists need a new laboratory to properly conduct measurements on the isotope. If laboratory is built before half the sample decays there will still be enough of the isotope available for experiments. However, it will take 21.0 months to build the new lab. Fortunately, the scientists can quickly start construction and they have access to a spaceship that can travel at speeds approaching the speed of light c = 3.00 x 10⁸ m/s.
If they place their sample of the isotope on the spaceship, at what speed must it travel in order for the new laboratory to be completed on Earth by the time half of the isotope on the spaceship decays? Assume that it took one month for the scientists to actually start construction and launch the spaceship so half the sample will remain in 12.0 months but it will still take 21.0 months to build the new lab.
Answer:
the required speed is 2.4618 x 10⁸ m/s
Explanation:
Given the data in the question;
Time taken to complete a lab = 21 months
it took them one one to actually start so
Time remaining so that sample only decays by 1/2 = (13 - 1) = 12 months
we need to find at what speed the spaceship should travel so that time on earth ( = 12 months) become time on spaceship ( = 21 months)
we make use of time dilation equation;
t' = t/√( 1 - v²/c² )
where t is time in rest frame = 12 and t' is time in moving frame = 21
so we substitute
21 = 12/√( 1 - v²/c² )
21√( 1 - v²/c² ) = 12
√( 1 - v²/c² ) = 12/21
we square both side
( 1 - v²/c² ) = ( 12/21 )²
( 1 - v²/c² ) = 0.3265
v²/c² = 1 - 0.3265
v²/c² = 0.6735
v² = 0.6735 × c²
v = √0.6735 × √c²
v = 0.8206 × c
given that speed of light c = 3 x 10⁸ m/s
v = 0.8206 × 3 x 10⁸ m/s
v = 2.4618 x 10⁸ m/s
Therefore, the required speed is 2.4618 x 10⁸ m/s
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.
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?
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
when is an object considered to be in motion
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
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
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]
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.
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 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
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: