Answer:
S = 16.3125m
Explanation:
Given the following data;
Acceleration, a = 14.5m/s²
Time, t = 2.25secs
Since the bottle rocket starts from rest, its initial velocity is 0m/s.
To find the distance S, we would use the second equation of motion.
S = ut + ½at²
Substituting into the equation, we have
S = 0(2.25) + ½*14.5*2.25
S = 0 + 7.25*2.25
S = 16.3125m
Therefore, the bottle rocket covered a distance of 16.3125 meters.
Which is the best explanation for why Toms technique works ?
An Egyptian pyramid contains approximately 1.95 million stone blocks. The average weight of each block is 2.55 tons. What is the weight of the pyramid in pounds?
Answer:
More than 2,300,000 limestone and granite blocks were pushed, pulled, and dragged into place on the Great Pyramid. The average weight of a block is about 2.3 metric tons (2.5 tons).
Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly walk 30.0 m west to a bench where you sit and watch the sunrise. It takes you 27.0 s to walk from your house to the windmill and then 47.0 s to walk from the windmill to the bench. For the entire trip from the front door to the bench, what are your :
a. average velocity
b. average speed
Answer:
Explanation:
Total displacement for entire trip = final position - initial position
= 50 m - 30 m = 20 m
Total time = 27 + 47 = 74 s
Average velocity = Total displacement / total time
= 20 / 74 = .27 m /s
Total distance covered in entire trip = 50 + 30 = 80 m
Total time = 74 s
Average speed = Total distance covered / total time
= 80 / 74 = 1.08 m /s .
How do pulleys help move objects?
Pulleys are powerful simple machines. They can change the direction of a force, which can make it much easier for us to move something. If we want to lift an object that weighs 10 kilograms one meter high, we can lift it straight up or we can use a pulley, so we can pull down on one end to lift the object up.
Answer:
Pulleys are powerful simple machines. They can change the direction of power, which can make it much easier for us to move something. If we want to lift an object that weighs 10 kilograms one meter high, we can lift it straight up or use a pulley, so we can pull one end down and lift the object.
Explanation:
calculate the average speed of talias car during the trip
Answer:
We're no strangers to love
You know the rules and so do I
A full commitment's what I'm thinking of
You wouldn't get this from any other guy
I just wanna tell you how I'm feeling
Gotta make you understand
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
We've known each other for so long
Your heart's been aching but you're too shy to say it
Inside we both know what's been going on
We know the game and we're gonna play it
And if you ask me how I'm feeling
Don't tell me you're too blind to see
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
No, I'm never gonna give you up
No, I'm never gonna let you down
No, I'll never run around and hurt you
Never, ever desert you
We've known each other for so long
Your heart's been aching but
Never gonna give you up
Never gonna let you down
Never gonna run around and desert you
Never gonna make you cry
Never gonna say goodbye
Never gonna tell a lie and hurt you
No, I'm never gonna give you up
No, I'm never gonna let you down
No, I'll never run around and hurt you
I'll never, ever desert you
Explanation:
RICK ROLLED
How many miles per day can you walk at a MODERATE Intensity level and your heart rate is 170?
Answer:
Not enough detail as it is very defendant on the person and a bunch of factors in health, but overall your heart rate shouldn't reach 170 as an adult walking at a moderate intensity level, that would be closer to extreme intensity.
Explanation:
The main idea of this paragraph is:
a. Boots is my dog.
b. Boots can sit.
c. Boots is very smart.
d. Most dogs can't dance.
28. Which of the following correctly shows the order of highest amount of friction to the lowest amount of
friction?
a. Static, Rolling, Sliding
b. Sliding, Rolling, Static
c. Rolling, Static, Sliding
d. Static, Sliding, Rolling
Answer:
[tex]\mathrm{d.\:Static,\: Sliding,\:Rolling}[/tex]
Explanation:
Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.
During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.
Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.
A protein molecule in an electrophoresis gel has a negative charge. The exact charge depends on the pHpH of the solution, but 30 excess electrons is typical. What is the magnitude of the electric forceon a protein with this charge in a 1500 N/C electric field?
Answer:
The magnitude of the force = 7.2 × 10⁻¹⁵ C
Explanation:
The total quantization of charge q on an electron = n × e
where;
n = 30
e = 1.6 × 10⁻¹⁸ C
q = 30 × 1.6 × 10⁻¹⁸ C
q = 4.8 × 10⁻¹⁸ C
Now, the magnitude of the force is determined by using the formula:
F = qE
F = ( 4.8 × 10⁻¹⁸ C) ( 1500 N/C)
F = 7.2 × 10⁻¹⁵ C
A safety plug is designed to melt when the pressure inside a metal tank becomes too high. A gas
at 51.0 atm and a temperature of 23.0°C is contained in the tank, but the plug melts when the
pressure reaches 75.0 atm. What temperature did the gas reach?
Waves in the ocean are tearing apart the shoreline. Which of the following two Earth Systems are interacting with each other.
Answer:
the Indian Ocean on 26 December 2004. This event claimed 227,898 dead and missing from 14 countries. The difference in mortality rates between these tsunamis reflects, in part, the benefits of understanding how tsunami waves are generated and move, and educating citizens to make scientifically
sound and potentially life-saving decisions.
A tsunami is a series of rapidly propagating, shallow-water ocean waves that develops when a submarine earthquake, landslide, or volcanic eruption displaces a large volume of water. Powerful earthquakes, with magnitudes of 9 or greater, caused both the 2004 and 2011 tsunamis. The earthquakes resulted from the movement of large tectonic plates. The 11 March 2011 earthquake occurred at 32 km (20
mi.) deep in Earth’s crust about 130 km (81 mi.) east of the city of Sendai. This location is on the boundary between two tectonic plates—the Pacific plate to the east and North American plate to the west. This
boundary fractured, releasing energy that was transmitted through the rocks and elevated portions of the
ocean floor. This drastic movement transmitted energy to the overlying ocean water, which generated
tsunami waves that radiated outward. The waves washed over the nearby coastlines and were felt around
the globe within hours (Figure 1.1).
Explanation:
Answer:
I believe Geosphere (lithosphere) and Hydrosphere
Explanation:
I hope it's right if not please notify me.
Artificial satellites in space can help you find locations on
Earth. True or false?
In the equation for the gravitational force between two objects, which quantity must be squared?
•mi
•m2
•G
•d
Answer:
d
Explanation:
The quantity that must be squared in the equation of gravitational force is distance d.
According to the universal gravitational law, the square of the distance between two objects is inversely proportional to the force of gravity.
Therefore, the quantity to be squared is dThe formula is given as:
Fg = [tex]\frac{G m_{1} m_{2} }{d^{2} }[/tex]
So d is the quantity that must be squared
if an electric is not grounded, it is best to reach out and touch it to provide the ground
Answer:
No. Touching a live electric current is never a good idea.
Answer:
false you would electrocute yourself
Explanation:
!!!!!!!!!!! LOGICAL !!!!!!!!!
An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-quency will the car and passenger vibrate onthe springs? Answer in units of Hz. The acceleration of gravity is 9.81 m/s^2.
Answer:
The required frequency = 0.442 Hz
Explanation:
Frequency [tex]f = ( \dfrac{1}{2 \pi}) \omega[/tex]
where;
[tex]\omega = \sqrt{\dfrac{k}{m} }[/tex]
Then;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )[/tex]
However;
[tex]k = \dfrac{F}{x}[/tex] and;
mass [tex]m = m_{car } + m_{person}[/tex]
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )[/tex]
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )[/tex]
where;
[tex]F = m_{person}g[/tex]
Then;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )[/tex]
replacing the values;
[tex]f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )[/tex]
[tex]\mathbf{f = 0.442 \ Hz}[/tex]
Can someone please help meee .
Answer:
32 amu is the right choice because both protons and neutrons have a mass of 1 amu. Electrons have no mass so go with the last choice
Some giant ocean waves have a wavelength of 25 m and travel at 6.5 m/s with a frequency of 0.26 HZ. What is the period of such a wave ?
Answer:
3.85s
Explanation:
Given parameters:
Wavelength = 25m
Velocity = 6.5m/s
Frequency = 0.26Hz
Unknown:
Period of the wave = ?
Solution:
The period of a wave is the inverse of the frequency of the wave.
Period = [tex]\frac{1}{frequency}[/tex]
Period = [tex]\frac{1}{0.26}[/tex] = 3.85s
A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.
Part A
What is the speed of the rock just before it hits the street?
Express your answer with the appropriate units.
Part B
How much time elapses from when the rock is thrown until it hits the street?
Express your answer with the appropriate units.
Answer:
A) v = 28.3 m/s
B) t = 4.64 s
Explanation:
A)
Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:[tex]v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h (1)[/tex]
Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:[tex]\Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)[/tex]
So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:Δh = 26.0 m + 14. 8 m = 40.8 m (3)Replacing now in (1), we can solve for vf, as follows:[tex]v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)[/tex]
B)
In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =02) Time elapsed from this point until it hits the street, with vo=0.For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:[tex]v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)[/tex]
Replacing by the givens in (5) and solving for Δt, we get:[tex]\Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)[/tex]
For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:[tex]\Delta h = \frac{1}{2} * g * t^{2} (7)[/tex]
Replacing by the givens and solving for t in (7), we get:[tex]t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)[/tex]
So, total time is just the sum of (6) and (8):t = 2.9 s + 1.74 s = 4.64 sHeather drives her Super-Beetle around a turn on a circular track which has a radius of 200 m. The Super-Beetle has a mass of 1500 kg and the coefficient of static friction between the road and tires is 0.6.
a. What is the force of static friction the road can apply batore the car starts to selon (use Ft= uFn).
b. What is the maximum speed the car can travel before it would start to slide?
Answer:
a) The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.
Explanation:
a) Let suppose that the car is on a horizontal ground and travels at constant speed. The vehicle experiments a centripetal acceleration due to friction, which can be seen in the Free Body Diagram (please see image attached for further details). By Newton's Laws, we construct the following equations of equilibrium:
[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (1)
[tex]\Sigma F_{y} = N -m\cdot g = 0[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]N[/tex] - Normal force from ground to the car, measured in newtons.
[tex]v[/tex] - Maximum speed of the car, measured in meters per second.
[tex]R[/tex] - Radius of the circular track, measured in meters.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
By applying (2) in (1):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex] (3)
The force of static friction the road can apply in the car ([tex]f[/tex]), measured in newtons, is: ([tex]\mu_{s} = 0.6[/tex], [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
[tex]f = \mu_{s}\cdot m \cdot g[/tex]
[tex]f = (0.6)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]f = 8826.3\,N[/tex]
The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) Then, we calculate the maximum speed of the car by (3):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex]
[tex]\mu_{s}\cdot g = \frac{v^{2}}{R}[/tex]
[tex]v = \sqrt{\mu_{s}\cdot g\cdot R}[/tex]
If we know that [tex]\mu_{s} = 0.6[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 200\,m[/tex], then the maximum speed of the car can travel before it would start to slide is:
[tex]v =\sqrt{(0.6)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (200\,m)}[/tex]
[tex]v \approx 34.305\,\frac{m}{s}[/tex]
The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.
In the laboratory, a ball is dropped onto a force-sensing platform several times, each time hitting a different surface (foam, feathers, clay, etc.). The momentum of the ball changes by the same amount in each trial; in each trial, the average scale reading is F, and the time of collision t are measured. What quantities would need to be graphed to exhibit a straight-line relationship
Answer:
Graphing the momentum against the change in moment yields a linear relationship.
Explanation:
This is an impulse experiment,
I = ∫ F .dt
where the force and time of the collision are measured, therefore if we assume an average force the integral reduces to
I = F t
Furthermore, the momentum is equal to the change in moment of the ball, this change in moment can be found using the energy relations measuring the height of the ball and calculating its speed, in the two intervals for the descent and for the exit, possibly the heights are different so the moment change is different from zero.
Starting point. Higher
Em₀ = U = mgh
Lower end point, just before hitting the scale
[tex]Em_{f}[/tex] = K = ½ m v²
in the path in the air there is no friction
Em₀ = Em_{f}
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
this height is different for the descent and ascent of the ball, so we have two moments
Δp = [tex]p_{f}[/tex] - p₀
Δp = m (v_{f} -v₀)
therefore we have the relationship
I = Δp
Graphing the momentum against the change in moment yields a linear relationship.
Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.70 m/s2 , how much time passes after the police car is passed by a speeder and before the police car overtakes the speeder (assumed moving at constant speed)
Answer:
t= 16.75 s
Explanation:
We will solve this exercise using the kinematic expressions
corridor that goes at constant speed, suppose that its speed is v₁ = 20 m/s, it does not appear in the statement, we start counting the time when it passes the policeman.
x₁ = v₁ t
The policeman starts from rest, so his initial velocity is zero and he has an acceleration a = 2.70 m /s², to use the same time counter we take into account that the policeman left at = 1.00 s after passing the corridor
x₂ = v₀ (t-t₀) + ½ a (t-t₀)²
x₂ = ½ a (t-1)²
at the point where the two meet, the position must be the same
x₁ = x₂
v₁ t = ½ a (t-1)²
(t-1)² = [tex]\frac{2 v_1 t}{a}[/tex]
t² - 2t + 1 - \frac{2 v_1 t}{a} +1 = 0
t² - 2(1 + [tex]\frac{v_1}{a}[/tex]) t +1
let's we solve the second degree equation
t² - 2 ( 1 + [tex]\frac{20}{2.7}[/tex]) t + 1=0
t² - 16.81 t +1=0
t = [ 16.81 ± [tex]\sqrt{ 16.81^2 - 4 )}[/tex] ] /2
t = [16.81 ± 16.695]/2
t₁= 16.75 s
t2= 0.06 s
Time t₂ is less than the reaction time of humans, so the correct answer is the first time
t= 16.75 s
A 50kg refrigerator is being moved across a kitchen floor with an applied force of
300N. There is a known friction force of 50N acting against the motion of the
refrigerator.
What was the acceleration of the refrigerator? [ Select ]
m/s2
Answer:
5m/s²
Explanation:
Given parameters:
Mass of refrigerator = 50kg
Applied force = 300N
Frictional force = 50N
Unknown:
Acceleration of the refrigerator = ?
Solution:
To solve this problem:
Net force = m x a
m is the mass
a is the acceleration
Net force = Applied force - Frictional force
300 - 50 = 50 x a
250 = 50 x a
a = 5m/s²
A ball bearing is projected vertically upwards from the ground with a velocity of 15ms. Calculate the time taken by the ball to return to the ground (g=10ms^2)
Answer:
t = 3 [s]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}-g*t[/tex]
where:
Vf = final velocity [m/s]
Vo = initial velocity = 15 [m/s]
g = gravity acceleration = 10 [m/s²]
t = time [s]
Now replacing we have:
[tex]0 = 15 -10*t\\10*t=15\\t= 1.5[s][/tex]
Note: In the equation above the gravity acceleration is negative, because the movement of the ball bearing is pointing againts the gravity acceleration.
The time calculated is only when the ball bearing reaches the highest elevation, and it will take the same time for descending, therefore the total time is:
t = 1.5 + 1.5 = 3 [s]
describe measurement in our daily life
3. As the mass of an object increases, the force of gravity
Answer:
As the mass of an object increases, the force of gravity increases as well.
Explanation:
Objects with more mass have more gravity. They work together.
find the volume of an object with a density of 3.2 g/mL and a mass of 12 g.
what is the formula of moment of force
These three bulbs are powered by the battery. What will happen if the middle light burns out? A. The two other bulbs will go out. B. The two other bulbs will stay on. C. Just the bulb closest to the battery will stay on. D. Just the bulb farthest from the battery will stay on.
Answer:
C
Explanation:
Before there is a transfer of charges between objects, they are uncharged. What does this mean? (pls answer by 7:35)
Answer: This means that the objects didn't hold or have any charges before it was charged by something or someone
Explanation:
Like a dead phone it dosn't have any "charge", but after you put the charger in the wall you can plug the phone in and get it charged It's an on going cycle of energy being turned into other energy.
Answer:
The answer is C
Explanation:
The positive and negative charges are equal.
Do it in order.
from smallest to largest
Answer:
The earth, The sun, the solar system and the milky way.