Answer:
a) [tex] v_{0y} = 9.72 m/s [/tex]
b) [tex] v_{0x} = 20.85 m/s [/tex]
c) [tex] t = 2.01 s [/tex]
d) [tex] h_{max} = 4.82 m [/tex]
Explanation:
a) The initial vertical velocity is given by:
[tex] v_{0y} = v*sin(\theta) [/tex]
Where:
θ: 25°
v: is the magnitude of the speed = 23 m/s
[tex] v_{0y} = 23 m/s*sin(25) = 9.72 m/s [/tex]
b) The initial horizontal velocity can be calculated as follows:
[tex] v_{0x} = v*cos(\theta) = 23 m/s*cos(25) = 20.85 m/s [/tex]
c) The flight time can be calculated using the following equation:
[tex] v_{0x} = \frac{x}{t} [/tex]
Where:
x: is the total distance = 42 m
[tex] t = \frac{x}{v_{0x}} = \frac{42 m}{20.85 m/s} = 2.01 s [/tex]
d) The maximum height is given by:
[tex] v_{fy}^{2} = v_{0y}^{2} - 2gh_{max} [/tex]
Where:
[tex] v_{fy}[/tex]: is the final vertical velocity =0 (at the maximum heigth)
g: is the gravity = 9.81 m/s²
[tex]h_{max} = \frac{v_{0y}^{2}}{2g} = \frac{(9.72 m/s)^{2}}{2*9.81 m/s^{2}} = 4.82 m[/tex]
I hope it helps you!
Which of the following best describes the relationship between the speed of light and the type of medium through which it travels?
A.
Light travels fastest in solids
B.
Light travels fastest in liquids
C.
The less dense the medium, the slower a light wave travel
D.
The more dense the medium, the slower a light wave will travel
Answer:
D
Explanation:
When light makes contact with any object it naturally slows down, and the more dense the object the slower it travels.
Which action would decrease the strength of attraction between a magnet and an iron object
A. Moving the magnet farther away from the iron object
B. Moving the iron object closer to the magnet
C. Decreasing the number of electrons on the magnet
D. Increasing the number of electrons on the magnet
Answer:
a. moving the magnet farther away and an iron object.
Explanation:
Answer:A
Explanation:
A p e x
What type of numbers in a measurement are always significant?
Answer:
Non-zero digits
Explanation:
umm (ꏿ﹏ꏿ;)
Answer:
There's three rules on determining how many important quantities are in a number: Non-zero digits are always predominant. Any zeros between two significant digits are always major. The final zero or dangling zeros in the decimal segment are really only important.
I hope this answered your question, have a nice day!
What are the 6 basic machines, and what are they and what can the be, and look like.
Answer:
Pulleys, levers, inclined planes, wedges, screws, and wheel and axle.
Explanation:
asheed and Sofia are riding a merry-go-round that is spinning steadily (uniform circular motion). Sofia is twice as far from the axis as is Rasheed. Sophia's acceleration is _________ that of Rasheed. *
Answer:
The same as
Explanation:
First of all, they are in a merry go round, which is a circular one. And thus, we would be dealing with circular motions.
The angular velocity of a merry go round, in this instance, is given by
w = Δ/Δt
The formula has nothing whatsoever to do with distance, and as such, both Rasheed and Sofia would have the same angular velocity.
Transferring this further, the angular acceleration is given as
α = Δw/Δt
Remember I said the velocity has nothing to do with distance, well, so does the acceleration as we can see from the formula stated. And therefore, both Rasheed and Sofia would have the same angular acceleration
A car starts from rest and reaches a velocity of 30 m/s in 10 s. If the car
has a mass of 1500 kg, what net force is exerted upon the car?
Answer:
f= ma
Explanation:
v-u/ t
30-0/10
3m/s
maas 1500
1500*3
4500
As the rate of radioactive decay becomes smaller half life become...
What is the momentum of 100 kg running at 4m/s north?
a. A remote-control car with a constant velocity drives off the top of a wall that is 10.0 m high and lands 4.60 m from the base of the wall. Draw a diagram of the problem. Label the known and unknown quantities. b. What is the car's speed before it drives off the top of the wall? c. What is the car's vertical speed just before impact? d. How far from the base of the wall would the remote-control car have landed if it had been subject to the gravitational field on the Moon (1.7 N/kg) ?
Answer:
a. Please see the attached graph and drawing combined created with Microsoft Excel
b. Approximately 3.2217 m/s
c. Approximately 14.0 m/s
d. Approximately 11.05 meters
Explanation:
a. The given parameters are;
The height of the wall, h = 10.0 m
The distance from the base of the wall the car lands = 4.60 m
The time, t, it takes the car to land is given by the equation for free fall as follows;
h = 1/2·g·t²
Where;
g = The acceleration due to gravity = 9.81 m/s²
From the equation for free fall, we have;
h/(1/2·g) = t²
∴ t = √(h/(1/2·g) ) = √(10/(1/2·9.81) ) ≈ 1.4278
The time it takes the car to land, t ≈ 1.4278 seconds
b. The horizontal speed of the car = Horizontal distance/(Time) = 4.6/1.4278 ≈ 3.2217 m/s
The horizontal speed of the car before it drives off the wall ≈ 3.2217 m/s
c. The car's vertical speed just before impact is given by the following equation;
v = u + gt
Where;
u = The initial vertical speed = 0
t = The time it takes before impact ≈ 1.4278 seconds
∴ v = 9.81 × 1.4278 ≈ 14.0 m/s
The car's vertical speed just before impact, v ≈ 14.0 m/s
d. Whereby the car is subject to the gravitational field of the moon, we have;
Gravitational force per kilogram = 1.7 N/kg
∴ Gravitational acceleration = 1.7 m/s².
The time it takes the car to land whilst subject to the gravitational field of the moon is therefore;
[tex]t_{Moon}[/tex] = √(h/(1/2·g)) = √(10/(1/2 × 1.7)) ≈ 3.43
[tex]t_{Moon}[/tex] ≈ 3.43 seconds
The horizontal distance covered, at the car's horizontal speed in the time of free fall ≈ 3.2217 m/s × 3.43 seconds ≈ 11.05 meters
The horizontal distance covered, at the car's horizontal speed in the time of free fall = The distance the car will land from the base of the wall ≈ 11.05 meters
The distance the car will land from the base of the wall ≈ 11.05 meters.
1500 POINTS TO WHOEVER ANSWERS THIS FIRST!!!!!!!!!!
1. You drop a bowling ball 15m from the top of the science building. What is its
velocity as it impacts the ground?
2. A cheetah accelerates from 0 to 13m/s in 3 seconds. How far does it travel
during that time?
3. Your foolish friend shoots a bullet straight in the air at a muzzle velocity of
380 m/s. How high does the bullet go (neglecting air resistance)?
4. Your foolish friend shoots a bullet straight in the air at a muzzle velocity of
380 m/s. What is the bullets velocity when it falls back down and hits your
friend?
Answer: Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.
Explanation:
Need help ASAP please help
Answer:
.....
Explanation:
Which factors affect the rate of deposition? Check all that apply.
color of the sediment
density of the sediment
size of the sediment
texture of the sediment
velocity of the agent of erosion
Answer:
density of the sediment, size of the sediment, and velocity of the agent of erosion.
Explanation:
Have a nice day!
Answer: b., c., e.
Explanation:
Bicyclist travels at an average velocity of 11.2 km/h [W]. How far will the bicyclist travel in 175 minutes?
Answer:
32.67km
Explanation:
~= round off
11.2km/h~3.11m/s
175minutes=10500s
10500*3.11~32, 666.67m
=32.66667km
~ 32.67km
What is the average emf between the ends of the wings of a plane flying at a speed of 500 km/hr when the vertical component?
Answer:
This question is incomplete
Explanation:
This question is incomplete because there are some parameters that are missing. However, when calculating the average emf and speed/velocity is provided, then
average emf = Brvl
where Br is the vertical component of the earth's magnetic field (in weber per square meter, Wb/m²)
v is velocity (in m/s)
l is the length (in meter) of the wing span
Why is struggling an important part of the process of learning?
Answer:
Students need a safe environment to take risks and struggle. It’s uncomfortable to struggle, but struggling—falling down and getting back up—is an important facet to learning. Productive struggle is not about being in pain or becoming frustrated.
Please mark branliest!
Explanation: it shows that your really trying your best and when you mess up its telling you that your actually learning from your mistakes.
At 96° F the saturation density of air is 0.04 kg/m3. If the weather report says the relative humidity is 90 % when the temperature is 96°F, the humidity is
Answer:
The value is [tex]H = 0.036 \ kg/m^3[/tex]
Explanation:
From the question we are told that
The saturation density of air is [tex]\rho = 0.04 \ kg/m^3[/tex]
The temperature for this saturation density to occur [tex]T_s = 96^o F[/tex]
The relative humidity at this temperature is [tex]R_h = 90\% = 0.90[/tex]
Generally the relative humidity is mathematically represented as
[tex]R_H = \frac{H}{\rho}[/tex]
Here H represents humidity
=> [tex]0.90 = \frac{H}{0.04}[/tex]
=> [tex]H = 0.90 * 0.04[/tex]
=> [tex]H = 0.036 \ kg/m^3[/tex]
The graph below shows the velocity of a car as it attempts to set a speed
record.
Velocity vs. Time
1400
1300
1200
1100
1000
4 45
3
1 (s)
At what point is the car the fastest?
A. t = 1.0 s
B. t = 4.2 s
C. t = 3.0 s
D. t = 4.5 s
From the graph, it is clear that, the velocity is at a time of 1 s is highest. The velocity at 1 second corresponds to 1250 km/hr. Then it decreases with time.
What is velocity - time graph ?The velocity - time graph shows the change in velocity with respect to time. The velocity is placed in y -axis and time is given in x - axis. The slope of the curve in velocity - time graph gives the acceleration of the object.
Similarly, the position of the object in meter after a t seconds can be determined from the velocity - time graph. It is the rate of change in velocity of the object.
From the graph, it is clear that, the curve has its peak at 1 second. After that the peak descends down. Hence, the maximum velocity of the car is at a time of 1 second at which the velocity is 1250 km/hr.
Find more on velocity - time graph :
https://brainly.com/question/28357012
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Harnessing tidal power would involve:_______.a) letting in-flowing and/or out-flowing tide flow through turbines in a dam.b) huge floats mechanically hooked to generators.c) electrodes in the water.d) bioluminescence of marine organisms.e) all of the above
Answer:
a) letting in-flowing and/or out-flowing tide flow through turbines in a dam.
Explanation:
Tidal power makes use of the energy from the tidal force and wave action in order to generate electricity. it is a predictable source of energy.
Tidal Barrages
The system allows tides to enter, seawater flows via the dam and is trapped into the basin when the tides subside and system’s gates close.
When the tides start to move out, the gates in the dam are opened up this consist of turbines and water begins to flow out, hitting the turbines this eventually turn to produce energy.
Power is produced when the tidal range, which is the difference between low and high tide, has to be more than 5 meters.
Which of the following is NOT a correct statement?*
As the volume of a contained gas increases, the pressure in the container will
decrease.
As the temperature decreases, the kinetic energy of the particles decreases.
As pressure increases with a constant temperature, the volume decreases.
O As temperature increases, the kinetic energy of the particles decreases.
AnswerAmontons's law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. ... If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure 1).:
Explanation:
During the slowing down process, there is a time when the angular speed is 0.5 rev/s. The fan arm is 0.8 m long. What is the speed of a point on the end of the fan arm at this time
Answer:
2.51 m/s
Explanation:
Given that,
Angular speed of a fan, [tex]\omega=0.5\ rev/s = 3.14\ rad/s[/tex]
Length of the fan arm, r = 0.8 m
We need to find the speed of a point on the end of the fan arm at this time. Let v is the speed. It is given in terms of angular speed is given by :
[tex]v=r\omega\\\\v=0.8\times 3.14\\\\v=2.51\ m/s[/tex]
So, the required speed is 2.51 m/s.
what can you infer from the statement. velocity is an object is zero
HELP PLEASE!!!! :( Does anyone have Lab: Conservation of Linear Momentum for online Physics class?? WILL GIVE YOU 80 PTS
Answer:
this is it
Explanation:
the control and experiemental groups i didnt know
which property of potential energy distinguishes it from kinetic energy
Answer:
Shape and position
Explanation:
Hope this helps! :)
You tie the loose end of a 0.1 kg yo-yo string to your finger and then release the yo-yo so that it spins down toward the ground (the yo-yo is released from rest and the end of the string tied to your finger remains motionless). After the yo-yo falls a distance of 0.9 m, it has a translational speed of 4 m/s and an angular speed of 180 rad/s. What is the moment of inertia of the yo-yo
Answer:
The answer is "[tex]5.06 \times 10^{-6} \ kg \ m^2[/tex]"
Explanation:
[tex]\to E_1=0..............(i)\\\\\to E_2= \frac{mV^2}{2} +\frac{Iw^2}{2} - mgh.............(ii)\\\\ \Delta E=0\\\\\to mgh= \frac{mV^2}{2} +\frac{Iw^2}{2} \\\\ \to 2 \ mgh= mV^2 +Iw^2\\\\ \to 2 \ mgh- mV^2 =Iw^2\\\\ \to m(2gh- V^2) =Iw^2\\\\ \to I= \frac{m(2gh- V^2)}{w^2}[/tex]
[tex]= 5.06 \times 10^{-6} \ kg \ m^2[/tex]
How much tension must a rope withstand if it is used to accelerate a 1200-kg car vertically upwards at 0.8m/s2?
Answer:
T = 12732 [N]
Explanation:
To solve this problem we must use newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
By means of the attached free body diagram, we can better understand the solution to this problem.
As we see in the diagram the tension Force T lifts the car upwards so the movement is upwards. Therefore we have two forces one upward (positive) of the force T and the other negative downward due to the weight of the vehicle.
As the movement is up the acceleration is also up (with a positive sign).
ΣF = m*a
where:
F = forces [N] (units of Newtons)
m = mass = 1200 [kg]
a = acceleration = 0.8 [m/s²]
T - 1200*9.81 = 1200*(0.8)
T = 12732 [N]
Answer:
12,720
Explanation:
Maybe I’ll make one more question. What type of electricity is made when you spin something really fast?
Answer:
i didnt know ther was differnt types
Explanation:
Which form of energy has the most potential globally
a) Solar energy
b) Bio energy
c) Wind energy
d) Geothermal energy
Answer:
i think the answer is solar,
Explanation:
because it is a natural resource im sorry if im wrong
If a second planet were of the same radius R and made of the same material but had a hollow center of radius 0.50 R , what would be the acceleration of gravity at its surface
Answer:
[tex]g_2=\dfrac{7}{8}g[/tex]
Explanation:
G = Gravitational constant
M = Mass of planet
R = Radius of planet
Acceleration due to gravity on first planet
[tex]g=\dfrac{GM}{R}[/tex]
Assuming that the planets have the same mass density [tex]\rho[/tex]
Density of first planet
[tex]\rho=\dfrac{M}{V}\\\Rightarrow \rho=\dfrac{M}{\dfrac{4}{3}\pi R^3}\\\Rightarrow M=\rho \dfrac{4}{3}\pi R^3[/tex]
Density of second planet
[tex]\rho=\dfrac{M_2}{V_2}=\dfrac{M_2}{\dfrac{4}{3}\pi R^3-\dfrac{4}{3}\pi (0.5R)^3}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(1-\dfrac{1}{2^3})}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(1-\dfrac{1}{8})}\\\Rightarrow \rho=\dfrac{M_2}{\dfrac{4}{3}\pi R^3(\dfrac{7}{8})}\\\Rightarrow M_2=\rho\dfrac{4}{3}\pi R^3(\dfrac{7}{8})\\\Rightarrow M_2=M\dfrac{7}{8}[/tex]
Acceleration due to gravity on the second planet
[tex]g_2=\dfrac{GM_2}{R}\\\Rightarrow g_2=\dfrac{GM\dfrac{7}{8}}{R}\\\Rightarrow g_2=\dfrac{7}{8}g[/tex]
The acceleration due to gravity of the planet would be [tex]\dfrac{7}{8}[/tex] times the acceleration due to gravity on the first planet.
describe the motion of the fireball in the air
Explanation:
doesn't get the question and the answer to
Answer:
(this is the sample answer:After the launch, the fireball moves upward, than it falls back down to EarthEartEarthEarthEarthEarth
Explanation:
Which of the following is how the substance in a longitudinal wave flows?
A. Up and Down
B. At Random
C. Perpendicular to the direction of the wave
D. In the direction of the wave
Please help I need this turned in tonight.
Answer:
The answer is D. In the direction of the wave
Explanation:
I took the test, hope this helps! :)