The correct choice is
A) I. Sound traveling faster through the air in the Sahara desert than in Antarctica.
Explanation:
Sound is a mechanical wave, which is a periodic disturbance of the space that propagates through vibrations of the particles in the medium filling the space. In particular, sound waves are longitudinal waves, in which the direction of oscillation of the particles in the medium is parallel to the direction of propagation of the wave.
Let's analyze now each statement:
I. Sound traveling faster through the air in the Sahara desert than in Antarctica. --> TRUE. In fact, the higher the temperature of the air, the faster the sound propagates. In particular, there is a linear relationship between speed of sound and temperature of the air:
[tex]v=331.3 + 0.6T(^{\circ}C)[/tex]
where [tex]T(^{\circ}C)[/tex] is the temperature of the air in Celsius degrees.
II. An astronaut hearing the loud explosion of a ship from a mile away in space. --> FALSE. This is not possible: in fact, sound is a mechanical wave, so it propagates only through a medium, and it cannot propagate in a vacuum (space is essentially vacuum, so sound cannot propagate in space).
III. Speeds and frequencies of sounds change as they pass through different mediums. --> FALSE. When a wave passes through different mediums, only the speed and the wavelength of the wave change, while the frequency remains the same.
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Answer: I. Sound traveling faster through the air in the Sahara desert than in Antarctica.
Explanation:
Frequency stays the same as sound passes through different mediums. An explosion in space is not going to be heard if you are in the vacuum of space yourself.
A car is moving with a constant velocity of 25 m/s. Which of the following is true?
a) the force from the engine is greater than all the forces of friction.
b) the net force on the car is zero.
c) the inertia is changing.
d) the forces of friction are proportional to the acceleration of the car.
e) All of the above.
b) the net force on the car is zero.
Explanation:
Let's analyze each option one by one:
a) the force from the engine is greater than all the forces of friction. --> FALSE. In fact, the car is moving at constant velocity: this means that its acceleration is zero,
a = 0
and so Newton's second law becomes
[tex]\sum F = ma = 0[/tex]
where [tex]\sum F[/tex] is the net force on the car and m is its mass. This means that the net force on the car is zero: so, the force from the engine cannot be greater than all the forces of friction, otherwise the net force cannot be zero.
b) the net force on the car is zero. --> TRUE, for what we said at point A)
c) the inertia is changing. --> FALSE. The inertia of an object just depend on the mass and the velocity of the object: as neither the mass nor the velocity are changing in this problem, then the inertia of the car is not changing.
d) the forces of friction are proportional to the acceleration of the car. --> FALSE. Generally, the force of friction acting on an object moving on a flat surface is
[tex]F_f = \mu mg[/tex]
where [tex]\mu[/tex] is the coefficient of friction, m is the mass, and g the acceleration of gravity. Therefore, the force of friction does not depend on the acceleration of the car.
e) All of the above. --> FALSE
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if you push a 4-kg mass...
Answer:
B
Explanation:
F = ma , a = F/m
a1 = F/10 and a2 = F/4
Since Force is constant, a2 will we greater than a1
A nucleus in a transition from an excited state emits a gamma-ray photon with an energy of 2.5 MeV. (a)
What is the photon frequency? (b) What is the photon wavelength? (c) How does the wavelength compare
with typical nuclear radii (of the order 10−15 m)?
a) The frequency of the photon is [tex]7.16\cdot 10^{20}Hz[/tex]
b) The wavelength of the photon is [tex]4.19\cdot 10^{-13} m[/tex]
c) The wavelength of the photon is about 100 times larger than the nuclear radius
Explanation:
a)
The energy of a photon is given by
[tex]E=hf[/tex] (1)
where:
[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant
f is the frequency of the photon
The photon in this problem has an energy of
[tex]E=2.5 MeV = 2.5\cdot 10^6 eV[/tex]
And keeping in mind that
[tex]1eV = 1.6\cdot 10^{-19} J[/tex]
we can convert to Joules:
[tex]E=(2.5\cdot 10^6)(1.9\cdot 10^{-19})=4.75\cdot 10^{-13} J[/tex]
And now we can use eq.(1) to find the frequency of the photon:
[tex]f=\frac{E}{h}=\frac{4.75\cdot 10^{-13}}{6.63\cdot 10^{-34}}=7.16\cdot 10^{20}Hz[/tex]
b)
The wavelength of a photon is related to its frequency by the equation
[tex]c=f\lambda[/tex]
where
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the photon in this problem,
[tex]f=7.16\cdot 10^{20}Hz[/tex]
Re-arranging the equation, we find its wavelength:
[tex]\lambda=\frac{c}{f}=\frac{3\cdot 10^8}{7.16\cdot 10^{20}}=4.19\cdot 10^{-13} m[/tex]
c)
The size of the nuclear radius is approximately
[tex]d \sim 10^{-15} m[/tex]
While we see that the wavelength of this photon is
[tex]\lambda=4.19\cdot 10^{-13} m[/tex]
Therefore, the ratio between the wavelength of the photon and the nuclear radius is
[tex]\frac{\lambda}{d}=\frac{\sim 10^{-13}}{\sim \cdot 10^{-15}}=100[/tex]
So, the wavelength of the photon is approximately a factor 100 times larger than the nuclear radius.
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An idea that is supported by some evidence, but is not yet the basis for a field of study is called a
An idea that is supported by some evidence, but is not yet the basis for a field of study is called a working theories.
Explanation:
The combined form of one’s skills, the work experiences from testing to reveal some truth about the issues or being involved one’s work in an experiment usually called as working theory.
These type of theories are more elaborate, interconnected, make some experiences with the solution to sort out the issues and seems to be useful in event monitoring. These can be supported with certain evidences but still not acting as the basement for the field study.
Final answer:
A hypothesis based on evidence that requires further investigation and testing before becoming a field of study.
Explanation:
A hypothesis is a suggested explanation based on evidence that can be tested through observation or experimentation. This idea is supported by some evidence but is not yet the basis for a field of study. Multiple strands of evidence are needed to potentially develop a verified theory.
What variable is represented on the y-axis?
Question 1 options:
Position
Velocity
Distance
Time
Question 2 (Mandatory) (1 point)
What variable is represented on the x-axis?
Question 2 options:
Distance
Time
Position
Velocity
Question 3 (Mandatory) (1 point)
What variable is the dependent variable?
Question 3 options:
Velocity
Position
Distance
Time
Question 4 (Mandatory) (1 point)
What variable is the independent variable?
Question 4 options:
Position
Time
Velocity
Distance
Question 5 (Mandatory) (1 point)
What is the starting position of the object in the graph?
Question 5 options:
The object starts at position 0 at time 0.
The object starts south of position 0 at time 0.
The object starts slightly north of position 0 at time 0.
The object starts far north of position 0 at time 0
Question 6 (Mandatory) (1 point)
What is the direction of motion of the object in the graph?
Question 6 options:
The object moves in a northward direction away from the starting position.
The object moves in a positive direction away from position 0.
The object moves in a southward direction toward position 0.
The object moves in a northeast direction away from the starting position.
Question 7 (Mandatory) (1 point)
What is the type of motion and the relative speed of the object in the graph?
Question 7 options:
At first, the object is not moving. Then, the object begins to gain speed.
First the object begins slowing down at a slow rate. Then, the object begins slowing down at a fast rate.
First the object is moving at a slow and constant speed. Then, the object begins moving at a fast and constant speed.
First, the object is speeding up at a slow rate. Then, the object beings speeding up at a fast rate.
Question 8 (Mandatory) (1 point)
What is the starting position of the object in the graph?
Question 8 options:
The object starts far north of position 0 at time 0.
The object starts south of position 0 at time 0.
The object starts at position 0 at time 0.
The object starts slightly north of position 0 at time 0.
Question 9 (Mandatory) (1 point)
What is the direction of motion of the object in the graph?
Question 9 options:
At first, the object is not moving, then the object begins moving in a southward direction toward position 0. Finally, the object is not moving, again.
At first, the object is moving toward the right, then the object begins moving in a northward direction away from position 0. Finally, the object begins moving toward the right, again.
At first, the object is not moving, then the object begins moving in a negative direction toward position 0. Finally, the object is not moving again.
At first, the object is moving in an eastward direction, then the object begins moving southward toward position 0. Finally, the object begins moving in an eastward direction, again.
Question 10 (Mandatory) (1 point)
What is the type of motion and relative speed of the object in the graph?
Question 10 options:
At first, the object is moving at a slow and constant speed. Then, the object begins moving at a fast and constant speed. Finally, the object moves at a slow and constant speed, again.
At first, the object is moving at a constant speed. Then, the object begins gaining speed. Finally, the object is moving at a constant speed, again.
At first, the object is not moving. Then, the object begins moving at a fast and constant. Finally, the object is not moving again.
At first, the object is moving at a constant speed. Then, the object begins slowing down. Finally, the object begins moving at a constant speed, again.
Question 11 (Mandatory) (4 points)
Compare and contrast a scalar quantity and a vector quantity. Match each description to the correct term.
Question 11 options:
Speed
Has magnitude and direction
Velocity
Has magnitude only.
1.
Scalar Quantity
2.
Vector Quantity
Question 12 (Mandatory) (6 points)
Compare and contract speed and velocity. Match each description to the correct term.
Question 12 options:
The rate at which an object changes its position with respect to time
Equation: Change in position / elapsed time
The slope of a Position vs. Time graph
Equation: Distance traveled / elapsed time
A measure of how fast an object travels
The slope of a Distance vs. Time graph
1.
Speed
2.
Velocity
Question 13 (Mandatory) (2 points)
An object travels with a speed of 30 meters in 10 seconds. If average speed is calculated using the equation, "distance traveled / elapsed time," what is the speed of the object?
Question 13 options:
3 speed
3 seconds
3 meters per second
3 meters
Distance/ Time which means Distance is on horizontal and time is on vertical
Which following is true off free falling object of different mass
A.they have the same acceleration but different g value
B.they have the same g value but different acceleration
C.they experience the same force of gravity
D.the rate of acceleration is the same for all objects
Answer:
Answer: C
explanation:
They could be same or different
ie:
|-5|,|-5| = 5,5
|-5|,|5| = 5,5
Answer:
C
Explanation:
9.8 m/s/s as well
how does a pith ball electroscope work
Answer:
look at the explanation
Explanation:
This pith-ball electroscope is used to detect the presence of a static electricity charge. The two lightweight “pith” balls suspended from the strings are attracted to objects with a static electric charge. The pith balls can also be charged by touching them to an object with a static electric charge.
A 0.614mole sample of ideal gas at 12degree occupies a volume of 4.32.what is the pressure of the gas
Answer:
336.9520 atm
Explanation:
The Gas Equation is as follows;-
Pressure×Volume=Number of Moles × Universal Gas Constant ×Temperature(in Kelvin)
Given Parameters
Number of moles-0.614 mol
Temperature 12°C or 12+273.15 ie 285.15°F
Volume-4.32 L
Universal Gas Constant-8.314 J/mol·K
Pressure -?(in atm)
Plugging in all the values in the Gas Equation:-
Pressure=[tex]\frac{0.614 × 8.314× 285.15}{4.32} atm[/tex]
Pressure=336.9520 atm
An electromagnet can be made by wrapping wire around which object?
Answer:
iron
Explanation:
Increasing the number of coils of wire wrapped around the nail increases the strength of the electromagnet, as measured by the number of paper clips the magnet can pick up
iron
Explanation:
Increasing the number of coils of wire wrapped around the nail increases the strength of the electromagnet, as measured by the number of paper clips the magnet can pick up
The driver of a car applies the brakes for 4.0 s.As a result an average force of 7500N is exerted in the car to slow it down.What is the change in momentum of the car.
Answer:
The change in momentum of the car is 30,000 kg m/s
Explanation:
Given,
The force exerted on the car to slow down, F = 7500 N
The time period of force, t = 4 s
The rate of change of momentum of the object is equal to the force acting on it.
Therefore,
(mv - mu) / t = F
Where v and u are the final and initial velocity of the car. The change in momentum of the car,
mv - mu = F x t
= 7500 x 4
= 30,000 kg m/s
Hence the change in momentum of the car is 30,000 kg m/s
The current theory of the structure of the
Earth, called plate tectonics, tells us that the
continents are in constant motion.
Assume that the North American continent
can be represented by a slab of rock 5100 kmon a side and 30 km deep and that the rock
has an average mass density of 2850 kg/m3
.
The continent is moving at the rate of about
3.8 cm/year.
What is the mass of the continent?
Answer in units of kg.
What is the kinetic energy of the continent?
Answer in units of J.
A jogger (of mass 80 kg) has the same kinetic
energy as that of the continent.
What would his speed be?
Answer in units of m/s.
1) Mass of the continent: [tex]2.2\cdot 10^{21} kg[/tex]
2) Kinetic energy: 1118 J
3) Speed of the jogger: 5.3 m/s
Explanation:
1)
First of all, we calculate the volume of the continent. It is a slab of side
[tex]L=5100 km = 5.1\cdot 10^6 m[/tex]
and thickness
[tex]t=30 km = 3.0\cdot 10^4 m[/tex]
So its volume is
[tex]V=tL^2=(3.0\cdot 10^4)(5.1\cdot 10^6)^2=7.8\cdot 10^{17} m^3[/tex]
The density of the slab is
[tex]\rho = 2850 kg/m^3[/tex]
Therefore, we can calculate the mass using the relationship
[tex]\rho = \frac{m}{V}[/tex]
where m is the mass. And solving for m,
[tex]m=\rho V=(2850)(7.8\cdot 10^{17})=2.2\cdot 10^{21} kg[/tex]
2)
The kinetic energy of the continent is given by
[tex]K=\frac{1}{2}mv^2[/tex]
where
[tex]m=2.2\cdot 10^{21} kg[/tex] is its mass
v = 3.8 cm/year is its speed
We have to convert the speed into m/s. Keeping in mind that
1 cm = 0.01 m
[tex]1 year = 365\cdot 24\cdot 60 \cdot 60 =3.15\cdot 10^7 s[/tex]
We find
[tex]v=3.18 \frac{cm}{y} \cdot \frac{0.01}{365\cdot 24 \cdot 60 \cdot 60}=1.0\cdot 10^{-9} m/s[/tex]
So now we can find the kinetic energy:
[tex]K=\frac{1}{2}(2.2\cdot 10^{21})(1.0\cdot 10^{-9})^2=1118 J[/tex]
3)
The kinetic energy of the jogger is given by
[tex]K=\frac{1}{2}m'v'^2[/tex]
where
m' = 80 kg is the mass of the jogger
v' is the speed of the jogger
Here we want the jogger to have the same kinetic energy of the continent, so
[tex]K=1118 J[/tex]
And by re-arranging the equation, we can find what speed the jogger must have:
[tex]v'=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1118)}{80}}=5.3 m/s[/tex]
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A 17.5 kg block is dragged over a rough, horizontal surface by a constant force of 88.2 N
acting at an angle of 26.9*
above the horizontal. The block is displaced 44 m, and the
coefficient of kinetic friction is 0.168.Find the work done by the 88.2 N force.
The acceleration of gravity is 9.8 m/s
2.
Answer in units of J
Find the magnitude of the work done by the
force of friction.
Answer in units of J.
Answer:
Part 1.: Work done by the force is 3461 J
Part 2: Work done by the force of friction is 972.8 J
Explanation:
part 1.
Since the displacement is horizontal, the only component of the 88.2 N force applied on the block is the horizontal component (which carries the cos(26.9) (the cosine of the angle with the horizontal). Then the component of the force acting on the 44 m displacement is: 88.2 * cos(26.9) . Then the work done by this component can be found by multiplying this quantity times the displacement (44 m), which gives:
88.2 * cos (26.9) * 44 = 3460.8878 J
which we can round to the nearest whole number as: 3461 J
Part 2.
The force of kinetic friction is the product of the normal force exerted by the surface on the block, times the coefficient of kinetic friction. Now the surface reacts with the normal force opposing the net vertical force applied by the block on the surface. Notice that in this case, the normal force doesn't just oppose the weight (m * g) of the object, since there is a vertical component of the external 88.2 N force ( the one that uses the sin(26.9) ) that is acting opposite to the gravitational force. This pointing upwards force component is going to subtract from the weight, thus making the normal force smaller than what it would be if just the object's weight was acting vertically. In magnitude, this is the normal force calculation:
Normal = Weight - 88.2 * sin(26.9)
Normal = [17.5 * 9.8 - 88.2 * sin(26.9) ] N
Normal = 131.6 N
At this point, we can calculate the value of the force of kinetic friction (product between the normal force and the coefficient of kinetic friction):
Kinetic friction force: 0.168 * 131.6 N = 22.1088 N
Now the product of this force times the displacement will give us the magnitude of the work done by the force of friction:
22.1088 N * 44 m = 972.7872 J
that we can round to 972.8 J
It is important to notice that the work done by the force of friction is negative due to the fact that the force acts opposite to the displacement of the block.
a soccerball is kicked straight out from a hill at 15 m/s and lands 42m away. how tall is the hill
Answer:
The height of the hill is, h = 38.42 m
Explanation:
Given,
The horizontal velocity of the soccer ball, Vx = 15 m/s
The range of the soccer ball, s = 42 m
The projectile projected from a height is given by the formula
S = Vx [Vy + √(Vy² + 2gh)] / g
Therefore,
h = S²g/2Vx² (Since Vy = 0)
Substituting the values
h = 42² x 9.8/ (2 x 15²)
= 38.42 m
Hence, the height of the hill is, h = 38.42 m
A particle is moving in a straight line.at t second, it acceleration is (4-kt), where k is a constant. when t=6, the acceleration of a particle is zero determine its velocity as a function of time
Answer:
v = 4t - t²/3
Explanation:
4 - k*6 = 0
k = 4/6 = 2/3
dv/dt = 4- 2/3 t
integral of dv = integral of (4-2/3 t) dt
v = 4t - t²/3
Plants are producers. What is the role of plants in an ecosystem?
A.
Plants help animals make their own food.
B.
Plants are food for some consumers.
C.
Plants use oxygen to make their own food.
D.
Plants break down dead animals.
PLSSSSS HELP ME QUICK!
Answer:
(C) Plants use oxygen to make their own food
Answer:
B.
Explanation:
We can use process of elimination here. A is not correct. Animals do not make their own food. C is not correct. Plants use carbon dioxide, not oxygen to power photosynthesis. D is not correct. Decomposers break down dead animals. Therefore, B is correct. Plants are food for some consumers.
A racing car’s velocity is increased from 44 m/s to 66 m/s in 11 seconds. What is the acceleration? and what is the displacement?
The acceleration of the racing car is 2 m/s² and the displacement over the 11 seconds is 605 m.
Explanation:The given physics problem involves calculating the acceleration and the displacement of a racing car that has increased its velocity from 44 m/s to 66 m/s in a time interval of 11 seconds.
To find the acceleration, we use the formula for average acceleration, which is the change in velocity divided by the time taken. The change in velocity (Δv) is 66 m/s - 44 m/s = 22 m/s. So, the acceleration (a) is 22 m/s divided by 11 s, which equals 2 m/s².
For the displacement, we can use the kinematic equation s = ut + 0.5at², where u is the initial velocity, a is the acceleration, and t is the time. With u = 44 m/s, a = 2 m/s², and t = 11 s, we get the displacement s = 44 m/s * 11 s + 0.5 * 2 m/s² * (11 s)², resulting in 484 m + 121 m, or a total displacement of 605 m.
The acceleration of the racing car is [tex]\( 2 \text{ m/s}^2} \)[/tex] and the displacement of the racing car during this time is [tex]\( 605 \text{ m}} \)[/tex].
To find the acceleration and displacement of the racing car, we will use the following kinematic equations:
1. Acceleration [tex](\(a\))[/tex]:
[tex]\[ a = \frac{\Delta v}{\Delta t} \][/tex]
2. Displacement [tex](\(s\))[/tex]:
[tex]\[ s = v_i t + \frac{1}{2} a t^2 \][/tex]
Step 1: Calculate the Acceleration
Given:
Initial velocity [tex](\(v_i\))[/tex]: 44 m/s
Final velocity [tex](\(v_f\))[/tex]: 66 m/s
Time interval [tex](\(\Delta t\))[/tex]: 11 s
The change in velocity [tex](\(\Delta v\))[/tex] is:
[tex]\[ \Delta v = v_f - v_i = 66 \text{ m/s} - 44 \text{ m/s} = 22 \text{ m/s} \][/tex]
Using the acceleration formula:
[tex]\[ a = \frac{\Delta v}{\Delta t} = \frac{22 \text{ m/s}}{11 \text{ s}} = 2 \text{ m/s}^2 \][/tex]
Step 2: Calculate the Displacement
Using the kinematic equation for displacement:
[tex]\[ s = v_i t + \frac{1}{2} a t^2 \][/tex]
Given:
Initial velocity [tex](\(v_i\))[/tex]: 44 m/s
Time [tex](\(t\))[/tex]: 11 s
Acceleration [tex](\(a\))[/tex]: [tex]2 m/s^2[/tex]
Substitute the values into the equation:
[tex]\[ s = 44 \text{ m/s} \times 11 \text{ s} + \frac{1}{2} \times 2 \text{ m/s}^2 \times (11 \text{ s})^2 \][/tex]
[tex]\[ s = 484 \text{ m} + \frac{1}{2} \times 2 \text{ m/s}^2 \times 121 \text{ s}^2 \][/tex]
[tex]\[ s = 484 \text{ m} + 1 \times 121 \text{ m} \][/tex]
[tex]\[ s = 484 \text{ m} + 121 \text{ m} \][/tex]
[tex]\[ s = 605 \text{ m} \][/tex]
The temperature of the liquid in a container decreases as the liquid evaporates. Use kinetic theory to explain why.
Answer:
Temperature decreases because the number of collision of the molecules decreases as they escape or evaporate. Molecules are in constant motion. Increase in temperature leads to increase in average kinetic energy of the molecules.
The kinetic theory of matter states that all matter is made up of tiny particles that are constantly moving. The faster the particles are moving, the higher the temperature of the matter.
Why is The temperature of the liquid in a container decreases as the liquid evaporates?When a liquid evaporates, the fastest-moving particles are the ones that are most likely to escape from the surface of the liquid. This leaves behind the slower-moving particles, which means that the average temperature of the liquid decreases.
Another way to think about it is that evaporation is a cooling process because it removes the fastest-moving particles from the liquid. These particles have the most kinetic energy, so their removal lowers the average kinetic energy of the remaining particles and thus lowers the temperature of the liquid.
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If the value of g=9.8;R=6400;G=6.67x10^-11then what is the value of M.
FORMULA =g=GM/R^2
Or gR=Gm/R
Answer:
The mass of the earth is, M = 5.98 x 10²⁴ kg
Explanation:
Given data,
The acceleration due to gravity, g = 9.8 m/s²
The radius of Earth, R = 6400 km
=6.4 x 10⁶ m
The universal gravitational constant, G = 6.67 x 10 ⁻¹¹ Nm²kg⁻²
Using the equation,
GM = gR²
M = gR² / G
Substituting the values,
M = 9.8 x (6.4 x 10⁶)² / 6.67 x 10 ⁻¹¹
= 5.98 x 10²⁴ kg
Hence, the mass of the earth is, M = 5.98 x 10²⁴ kg
A mass of 5kg accelerates at 3m/s/s, how much force was put on it?
Answer:
15N
Explanation:
According to Newton's Second Law of Motion
F = m*a
mass = m = 5Kg
acceleration = a = 3m/s^2
=> F = 5kg * 3m/s^2
=> F = 15 N
F = m a
The force is (5kg)x(3 m/s^2)= 15 Newtons, and it's still there.
If the force stops, the acceleration stops.
You launch a water balloon from the ground with a speed of 8.3 m/s at an angle of 27°. a. What is the horizontal component of the velocity? (1 point) b. What is the vertical component of the velocity? (1 point) c. How long does it take the water balloon to reach its highest point? (1 point) d. What is the maximum height of the water balloon? (1 point) e. What is the total amount of time that the water balloon is in the air? (1 point) f. How far does the water balloon land from where you launched it? (1 point)
a) The horizontal component of the velocity is 7.4 m/s
b) The vertical component of the velocity is 3.8 m/s
c) The balloon reaches the highest point after 0.39 s
d) The maximum height is 0.74 m
e) The total time of flight is 0.78 s
f) The range of the balloon is 5.77 m
Explanation:
a)
The motion of the balloon is the motion of a projectile, which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
The horizontal component of the velocity (which is constant) is given by
[tex]v_x = u cos \theta[/tex]
where
u = 8.3 m/s is the initial velocity of the balloon
[tex]\theta=27^{\circ}[/tex] is the angle of projection
Substituting,
[tex]v_x = (8.3)(cos 27^{\circ})=7.4 m/s[/tex]
b)
The vertical component of the initial velocity of a projectile is given by
[tex]u_y = u sin \theta[/tex]
where
u is the initial velocity
[tex]\theta[/tex] is the angle of projection
Here we have
u = 8.3 m/s
[tex]\theta=27^{\circ}[/tex]
Substituting,
[tex]u_y = (8.3)(sin 27^{\circ})=3.8 m/s[/tex]
c)
The vertical component of the velocity of the balloon follows the suvat equation
[tex]v_y = u_y - gt[/tex]
where
[tex]v_y[/tex] is the vertical velocity at time t
[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
The balloon reaches the maximum height when the vertical velocity becomes zero:
[tex]v_y = 0[/tex]
So we get:
[tex]0=u_y -gt\\t=\frac{u_y}{g}=\frac{3.8}{9.8}=0.39 s[/tex]
d)
The maximum height of the balloon can be calculated using the suvat equation:
[tex]s=u_y t - \frac{1}{2}gt^2[/tex]
where
[tex]u_y = 3.8 m/s[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
t = 0.39 s is the time at which the highest point is reached
Substituting,
[tex]s=(3.8)(0.39)-\frac{1}{2}(9.8)(0.39)^2=0.74 m[/tex]
e)
The total time of flight of a projectile is twice the time needed to reach the maximum height, and it is given by
[tex]t=\frac{2u_y}{g}[/tex]
where
[tex]u_y[/tex] is the initial vertical velocity
[tex]g[/tex] is the acceleration of gravity
Here we have
[tex]u_y = 3.8 m/s[/tex]
[tex]g=9.8 m/s^2[/tex]
Substituting,
[tex]t=\frac{2(3.8)}{9.8}=0.78 s[/tex]
f)
The range of a projectile is the horizontal distance covered by the projectile, so it can be found by multiplying its horizontal velocity (which is constant) by the time of flight:
[tex]d=v_x t[/tex]
where
[tex]v_x[/tex] is the horizontal velocity
t is the time of flight
Here we have
[tex]v_x = 7.4 m/s[/tex]
[tex]t = 0.78 s[/tex]
Substituting,
[tex]d=(7.4)(0.78)=5.77 m[/tex]
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All of the following would be questions that could be scientifically investigated except:
Answer:
"What is the best advice a parent can give a child?"
Explanation:
The other answers have one clear response. Ex: Orange popsicles melt faster than grape popsicles. That would be a fact.
But parental advice can vary, depending on your opinion. I may say that all parents must teach their children not to talk to strangers, while someone else may say that parents should advice their kids to treat everyone fairly. Nothing can be proven as the only appropriate response.
Hope this helps!
Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speec
up to 45 m/s. If it takes a train 8 seconds to reach 45 m/s, what is the acceleration of the train?
m/s2
Answer:
The acceleration of the train is 5 m/s².
Explanation:
Given:
let the initial velocity of a train = 5 m/s and
final velocity of a train = 45 m/s
time taken = 8 s
To find:
acceleration: ?
Solution:
We define acceleration as change in velocity per unit time that is the difference between the final velocity and initial velocity divided by time.
[tex]Acceleration = \frac{\textrm{final velocity} - \textrm{initial velocity}}{time} \\[/tex]
On substituting the above values we get the required acceleration
[tex]Acceleration = \frac{45 - 5}{8}\\ =\frac{40}{8}\\ =5\ m/s^{2}[/tex]
Therefore,the acceleration of the train is 5 m/s².
2) What are the directions of the velocity and acceleration of an object in
uniform circular motion?
how does a metal leaf electroscope work
Answer: An electroscope is a device that detects static electricity by using thin metal or plastic leaves, which separate when charged. ... Electrical charges move to the metal and down to the foil leaves, which then repel each other. Since each leaf has the same charge (positive or negative), they repel each other.
Answer:
An object with a suspected static electric charge is brought near the metal plate or ball of the electroscope. Electrical charges move to the metal and down to the foil leaves, which then repel each other. Since each leaf has the same charge (positive or negative), they repel each other.
Find the velocity of a turtle who swims 100 meters toward the shore in 25 seconds
Answer:
The speed of a turtle is 4m/s
Explanation:
speed = distance/time
distance = 100m
time = 25s
s = 100/25
= 4m/s
Answer: 20m/s
Explanation:
Data;
Distance (s) = 100m
Time (t) = 25s
Velocity (v) = s/t
v = 100/25
v = 20m/s
A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the horizontal. The magnitude of the friction force acting on the block is:
The force of friction is 25 N
Explanation:
For this problem, we can apply Newton's second law of motion along the horizontal direction:
[tex]\sum F_x = ma_x[/tex]
where
[tex]\sum F_x[/tex] is the net force in the horizontal direction
m is the mass of the block
[tex]a_x[/tex] is the horizontal acceleration
Here the block is moving at constant speed, so its acceleration is zero, therefore:
[tex]a=0 \rightarrow \sum F_x = 0[/tex] (1)
The net force in the horizontal direction can be written as:
[tex]\sum F_x = Fcos \theta -F_f[/tex] (2)
where
[tex]Fcos\theta[/tex] is the horizontal component of the pulling force, with F = 50 N being the magnitude and [tex]\theta=60^{\circ}[/tex] being the direction, acting forward[tex]F_f[/tex] is the force of friction, acting backwardCombining (1) and (2), we find the magnitude of the force of friction:
[tex]Fcos \theta -F_f = 0\\F_f = F cos \theta =(50)(cos 60^{\circ})=25 N[/tex]
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How does energy change (transforms) as the mass is dropping?
Answer:
Mass has total mechanical energy, which is the sum of kinetic and potential energy. as the mass is dropping, potential energy is converted into kinetic energy so mechanical energy is preserved If there is no friction. If there is friction, some of the mechanical energy is lost as heat energy so it changes.
Explanation:
100 points, will mark Brainliest, please show work (As I'd like to learn how to solve this on my own in the future :^)
"With other strict requirements, an astronaut cannot be taller than 6 foot, 3 inches or have a mass greater than 95 kilograms. Determine the weight of the most massive astronaut on the moon where the gravity is one-sixth that of the Earth."
Answer choices:
120
140
160
180
Thanks in advance!
Answer:
180
Explanation:
The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second. The Moon's surface gravity is weaker because it is far less massive than Earth. A body's surface gravity is proportional to its mass, but inversely proportional to the square of its radius.
Answer:
160kg
Explanation:
Given m= 95kg g= 1.6345m/s²
Unknown Fg= ?
Formula Fg=mg
Fg=(95)(1.6345)
Fg= 155.2775
Fg=160kg
If the box weighs 1,500 N, how much work dose the force of gravity do on the box
Answer:
The work done by the force of gravity, W = 0 J
Explanation:
Given data,
The weight of the box, w = 1500 N
Therefore the gravitational force acting on the box is , F = 1500 N
The displacement of the box S = 0
The work done is defined as the product of force acting on the body to the displacement it caused.
The formula for work done is,
W = F x S
= 1500 N x 0
= 0
Hence. the work done by the force of gravity is, W = 0 J
The work done by the force of gravity on a box weighing 1500 N depends upon the distance the box is moved. If we assume a free fall situation, the work done would be calculated using the formula: W = 1500N * h, where h is the height from which the box falls.
Explanation:The calculation of the work that the force of gravity does on an object comes down to two important factors: the force of gravity acting on the object, represented as weight, and the distance the object is moved. In this case, the box weighs 1500 N, which is the force of gravity acting on it. If we assume that the box is in free fall, i.e., it's falling to the ground under gravity, we can calculate the work done by the force of gravity.
To calculate this, we need to know the height from which the box is falling (let's denote that as h). The work done, according to the work-energy principle, is equal to the force times the distance covered and the cosine of the angle between force and distance vectors. Considering the angle between the gravity force and the displacement is 0 degrees in a free fall, the work done by gravity is W = 1500N * h * cos(0) = 1500N * h. Here, cos(0) is equal to 1.
As you can see, the work done by gravity relies heavily on the distance/height textbox falls. If the box doesn't move, then no work is done by gravity even though the force of gravity exists, because work is dependent on displacement.
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Robert lifts a 150-newton bucket of water 0.5 meters in 3 seconds. What is the amount of work done on the
bucket?
a. 25 joules
b. 25 watts
c. 75 joules
d. 75 watts
Answer:
c. 75 joules
Explanation:
Work done(joules)=Force(newton)×Distance(meters)
Given: force=150-newton, Distance=0.5 meters
∴Work done=150×0.5=75 joules.
The amount of work done on the bucket will be 75 joules. Option C is correct.
What is work done?Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive and negative.it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.
The given data in the problem is;
F is the force lift = 150-
d is the displacement = 0.5 meters
t is the time = 3 second
W is the amount of work done
The amount of work done on the bucket is found as;
[tex]\rm W = F \times d \\\\ W = 150 \times 0.5 \\\\ W=75 \ Joule[/tex]
Hence the amount of work done on the bucket will be 75 joules. Option C is correct.
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