As there is only one force acting on the ball i.e. gravitational force, the acceleration will be constant and downward. Also because ball moves in the direction of the acceleration, the velocity increases.
Acceleration - constant; velocity - increasing.
What is Acceleration?
Acceleration is the rate of change of velocity of an object over time. It can be positive, negative, or zero. Positive acceleration is when an object speeds up, negative acceleration is when an object slows down, and zero acceleration is when an object's velocity remains constant.
What is velocity?
Velocity is a measure of the speed and direction of an object's motion. It is typically measured in meters per second (m/s) or kilometers per hour (km/h). Velocity can also be expressed in other units, such as miles per hour (mph) or feet per second (ft/s). Velocity is related to a object's acceleration, as an object's acceleration is the rate of change in its velocity over time.
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Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 16.7 kg and an initial velocity of = 7.26 m/s, due east. Object B, however, has a mass of mB = 29.3 kg and an initial velocity of = 4.39 m/s, due north. Find the (a) magnitude and (b) direction of the total momentum of the two-object system after the collision.
Answer:
a) v = 3,843 m / s, b) 46.7º North- East
Explanation:
Moment is a vector quantity, so one of the best ways to solve this problem is to solve each component separately.
The system is formed by the two vehicles so that the moment is preserved during the crash
Direction to the East
initial instant. Before the crash
p₀ = mₐ vₐ₀
final insttne. After the crash
p_f = (mₐ + m_b) vₓ
p₀ = p_f
mₐ vₐ₀ = (mₐ + m_b) vₓ
vₓ = [tex]\frac{m_a}{m_a + m_b} \ v_{ao}[/tex]
let's calculate
vₓ = [tex]\frac{16.7}{16.7 + 29.3} \ 7.26[/tex]
vₓ = 2,636 m / s
direction north
initial p₀ = m_b v_{bo}
final p_f = (mₐ + m_b) v_y
p₀ = p_f
m_b v_{bo} = (mₐ + m_b) v_y
v_y = [tex]\frac{m_b}{m_a+m_b} \ v_{bo}[/tex]
let's calculate
v_y = [tex]\frac{29.3}{16.7 + 29.3} \ 4.39[/tex]
v_y = 2.796 m / s
the final speed of the two two vehicles is
v = (2,636 i ^ + 2,796 j ^) m / s
a) the magnitude of the velocity
let's use the Pythagorean theorem
v = [tex]\sqrt{v_x^2 + v_y^2}[/tex]
v = [tex]\sqrt{2.636^2 + 2.796^2}[/tex]
v = 3,843 m / s
b) let's use trigonometry to find the direction
tan θ = v_y / vₓ
θ = tan⁻¹ v_y / vₓ
θ = tan⁻¹ (2,796 / 2,636)
θ = 46.7º
This direction is 46.7º North East
The rollercoaster is near the
bottom of the hill after the first big
drop. Which best describes the
potential and kinetic energy?
A. It has mostly kinetic energy.
B. It has mostly potential energy.
C. The kinetic energy is decreasing.
D. The potential energy is about 50%
and increasing.
Answer:
A. It has mostly kinetic energy
Explanation:
Kinetic energy refers to movement. Potential energy refers to height. In this case, the big drop just got over. So, when the coaster is at the bottom, it has more kinetic than potential energy . Potential energy is still present but kinetic is more at the bottom.
On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total weight of the balloon, including its load and the hot air in it, is 20,000 N. a. Show that the weight of the displaced air is 20,000 N. b. Show that the volume of the displaced air is 1700 m3 .
Explanation:
Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.
Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.
Weight of the air displaced = density of air × volume
The density of air at 1 atm pressure and 20º C is 1.2 kg/m³
the volume V = 20,000/(1.2×9.8) = 1700 m³
TIME REMAINING
15:56:15
A plant root is an example of
Type here to search
Answer:
h
Explanation:
A cylindrical diving bell is open at the bottom and closed at the top, and is 5m tall. The bell is open to atmospheric air until it is placed in the water, but the bell remains upright (open end facing down, closed end facing up). The pressure of the air inside the bell will naturally increase by 105 Pa for every 10m of depth the bell descends within the water. Assume the temperature of the air remains constant for this process, and that the air can be approximated as an ideal gas
Required:
a. If the bell is lowered 40 meters below the surface, how many meters of air space are left inside the bell?
b. Explain why water doesn't completely flood the bell as it enters the water V (m).
Answer:
a) y = 0.35 m, b) hydrostatic balance
Explanation:
a) For this fluid mechanics exercise, let's use that the pressure at a given level is the same, let's set a level on the bell shape.
The pressure inside is
P_interior = P₀ + ρ g h ’
The pressure outside
P_exterior = Pₐ + ρ g h
as the point is at the same level the pressures are equal
P_interior = P_exterior
P₀ + ρ g h ’= Pₐ + ρ g h
h ’= (Pₐ- P₀) + ρ g h
To calculate P₀ they indicate that the pressure increases 10⁵ Pa for every 10 m, we use a direct rule of proportions or rule of three
P₀ = 10⁵ (40 + h ’) / 10 = 4 10⁵ + h’ 10⁴
the positive sign is because the water inside the hood also increases the air pressure.
we substitute
(4 10⁵ + h’ 10⁴) + ρ g h’ = Pₐ + ρ g h
h’ (ρ g + 10⁴) = Pₐ - 4 10⁵ + ρ h h
h’ (1000 9.8 + 10⁴) = (1 10⁵ -4 10⁵) + 1000 9.8 40
h' (1.98 10⁴) = -3 105 + 3.92 10⁵
h’ = [tex]- \frac{0.92 \ 10^5 }{1.98 \ 10^4 }[/tex]
h ’= -4.65 m
as the hood is only 5 m high, the free air space is
Y = 5 - 4.65
y = 0.35 m
it is very little free space
B) The pressure outside and inside the hood is the same, the water rises inside the hood until the pressures equalize and at this point the force is equal and in the opposite direction, which is why the system is in hydrostatic balance.
A 1.8 kg book has been dropped from the top of the football stadium. Its speed is 4.8 m/s when it is 2.9 m above the ground. How high is the stadium?
Answer:
the height of the stadium is 4 m
Explanation:
The computation of the height of the stadium is shown below:
but before that total mechanic energy should be determined
E = PE + KE
where
PE = mgh
and, KE = 1 ÷2 mv^2
Now
E = mgh + 1 ÷2 mv^2
= (1.8) (9.8) (2.9) + 1 ÷ 2 (1.8) (4.8)^2
= 71.9J
= 72J
Now the height of the stadium is
TE = mgh
72 = (1.8) × (9.8) × h
So, h = 4 m
Hence, the height of the stadium is 4 m
an object is acted by force of 22 newtons to the right and a force of 13 newtons to left
Answer:
Explanation:
I'm going to guess that you want the net force.
These two forces are acting in opposite directions. Therefore the forces are subtracted in effect.
F = F1 - F2
F1 = 22N to the right
F2 = 13N to the left
F = 22 - 13 = 9 N to the right.
Specifying the direction is very important. Forces do have directions and you must specify what that is. Otherwise, the question should be marked incorrect.
g A velocity selector consists of crossed electric and magnetic fields. The electric field has a magnitude of 480 N/C and is in the negative z direction. What should the magnetic field (magnitude and direction) be to select a proton moving in the negative x direction with a velocity of 3.50 cross times 10 to the power of 5 m/s to go un-deflected
Answer:
B = 1.37 mT
Explanation:
Given that,
The magnitude of the electric field, E = 480 N/C
The speed of the proton, [tex]v=3.50 \times 10^5\ m/s[/tex]
We need to find the magnitude of the magnetic field. In a velocity selector, the electric field is balanced by the magnetic field. So,
[tex]qE=qvB[/tex]
Where
B is the magnetic field
[tex]B=\dfrac{E}{v}\\\\B=\dfrac{480}{3.5\times 10^5}\\\\B=1.37\times 10^{-3}\ T\\\\or\\\\B =1.37\ mT[/tex]
So, the magnetic field is equal to 1.37 mT.
A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are µs = 0.50 and µk = 0.40. For P = 3.6 lb,
Determine:
(a) Whether slipping occurs between the belt and either cylinder,
(b) The angular acceleration of each cylinder.
Slipping doesn't occur between the belt and cylinder B because the force of static friction is greater than force exerted on cylinder B.
Given the following data:
Mass A = 5 lb to kg = 2.27 kg.
Mass B = 20 lb to kg = 9.02 kg.
Force = 3.6 lb to N = 16.02 Newton.
How to calculate angular acceleration.In order to calculate the angular acceleration of each cylinder, we would take moment about the two cylinders.
For cylinder A:
[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_A\alpha _A = F_A (0.1)\\\\(\frac{m_Ar^2}{2}) \alpha _A = F_A (0.1)\\\\(\frac{2.27 \times 0.1^2}{2}) \alpha _A = F_A (0.1)\\\\0.1F_A=0.01135\alpha _A\\\\F_A=\frac{0.01135\alpha _A}{0.1} \\\\F_A= 0.1135\alpha _A[/tex]
For cylinder B:
[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_B\alpha _B = F_B (0.2)\\\\(\frac{m_Br^2}{2}) \frac{\alpha _A}{2} = F_B (0.2)\\\\(\frac{9.02 \times 0.1^2}{4}) \alpha _A = F_B (0.2)\\\\0.1F_B=0.02255\alpha _A\\\\F_B=\frac{0.02255\alpha _A}{0.2} \\\\F_B= 0.1128\alpha _A[/tex]
For the belt, we have
[tex]\sum F_A =0\\\\P-F_B-F_A=0\\\\16.02-0.1128\alpha _A-0.1135\alpha _A=0\\\\16.02=0.2263\alpha _A\\\\\alpha _A=\frac{16.02}{0.2263} \\\\\alpha _A=70.79 \;rad/s^2[/tex]
Also, we would determine the angular acceleration of cylinder B:
[tex]0.1\alpha _A=0.2\alpha _B\\\\0.1 \times 70.79 = 0.2\alpha _B\\\\7.079= 0.2\alpha _B\\\\\alpha _B=\frac{7.079}{0.2} \\\\\alpha _B=35.40\;rad/s^2[/tex]
Next, we would calculate the forces acting on the cylinders:
[tex]F_A = 0.1135\alpha _A\\\\F_A = 0.1135 \times 70.79\\\\F_A = 8.04 \;Newton[/tex]
[tex]F_B = P-F_A\\\\F_B = 16.02 - 8.04\\\\F_B = 7.98\;Newton[/tex]
Next, we would determine the force of static friction:
[tex]F_s = \mu_s N = \mu_s m_B g\\\\F_s = 0.50 \times 9.02 \times 9.8\\\\F_s=44.198\;Newton[/tex]
From the above calculation, we can deduce that the force of static friction is greater than force exerted on cylinder B. Therefore, slipping doesn't occur between the belt and cylinder B.
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Use the scenario to answer the question.
An astronomer discovers a new galaxy using a telescope. The astronomer wants to investigate how the galaxy is moving relative to the Milky Way galaxy.
In one or two sentences, make a hypothesis about the movement of the galaxy and explain at least one way to test the hypothesis.
Answer:
The galaxies outside of our own are moving away from us, and the ones that are farthest away are moving the fastest. This means that no matter what galaxy you happen to be in, all the other galaxies are moving away from you
Explanation:
The hypothesis about the movement of the galaxy is that galaxies are moving far from each other continuously.
What is the milky way galaxy?The milky way galaxy is a galaxy that contains over a hundred billion stars and it also includes our solar system. Its name describes its appearance when viewed from the earth. All the individual stars in the whole sky are a portion of the Milky Way Galaxy, the term "Milky Way" is because of the band of light.
The astronomer has discovered a new galaxy which means our universe is continuously expanding. This is because the universe encloses everything that exists.
Galaxies are moving in space and since the universe space is continuously expanding so the galaxies continuously move from each other. The farther the galaxy is from the milky way which is an observable part, the faster will be moving the galaxy and the closer the galaxy is to the milky way, the slower will be movement of the galaxy.
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Humans impact the Earth in good AND bad ways.
A) True
B) False
Answer:
True
Explanation:
yes we can see that we are helping animals but we create pollution which is very bad
Describe the formation of the land, the atmosphere, and the oceans of earth
Which of the following is true of the
thermocline layer of the ocean?
A. rapidly decreases in temperature
B. warmest and least dense of the ocean layers
C. is the bottom layer of the ocean
D. is the top layer of the ocean
Answer:
d is trueeeeeeeeeeeeeeeee
Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 60.0 N is exerted and the astronaut's acceleration is measured to be 0.870 m/s2. (a) Calculate her mass (in kg). kg (b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut's acceleration. Propose a method in which recoil of the vehicle is avoided.
Answer:
a) m = 69.0 kg
b) release some gas in the opposite direction to the astronaut's movement
Explanation:
a) Let's use Newton's second law
F = m a
m = F / a
m = 60.0 / 0.870
m = 69.0 kg
b) when we exert a force on the astronaut it acquires a momentum po, as the astronaut system plus spacecraft is isolated, the momentum is conserved
p₀ = p_f
m v = M v '
v ’= [tex]\frac{m}{M} \ v[/tex]
so we see that the ship is moving backwards, but since the mass of the ship is much greater than the mass of the astronaut, the speed of the ship is very small.
One method to avoid this effect is to release some gas in the opposite direction to the astronaut's movement so that the initial momentum of the astronaut plus the gas is zero and therefore no movement is created in the spacecraft.
a disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular acceleration?
b. how long did it take the disk to reach this velocity?
Answer:
a) α = 1.875 [tex]\frac{rad}{s^{2} }[/tex]
b) t = 8 s
Explanation:
Given:
ω1 = 0 [tex]\frac{rad}{s}[/tex]
ω2 = 15 [tex]\frac{rad}{s}[/tex]
theta (angular displacement) = 60 rad
*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*
a) α = ?
(ω2)^2 = (ω1)^2 + 2α(theta)
[tex]15^{2}[/tex] = [tex]0^{2}[/tex] + 2(α)(60)
225 = 120α
α = 1.875 [tex]\frac{rad}{s^{2} }[/tex]
b)
α = (ω2-ω1)/t
t = (ω2-ω1)/α = (15-0)/1.875 = 8
t = 8 s
A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick to it.
Required:
What is the turntable's angular speed, in rpm, just after this event?
Answer:
The turntable's angular speed after the event is 28.687 revolutions per minute.
Explanation:
The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:
[tex]I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}[/tex] (1)
Where:
[tex]I_{T}[/tex] - Moment of inertia of the turntable, in kilogram-square meters.
[tex]r[/tex] - Distance of the block regarding the center of the turntable, in meters.
[tex]m[/tex] - Mass of the object, in kilograms.
[tex]\omega_{o}[/tex] - Initial angular speed of the turntable, in radians per second.
[tex]\omega_{f}[/tex] - Final angular speed of the turntable-objects system, in radians per second.
In addition, the momentum of inertia of the turntable is determined by following formula:
[tex]I_{T} = \frac{1}{2}\cdot M\cdot r^{2}[/tex] (2)
Where [tex]M[/tex] is the mass of the turntable, in kilograms.
If we know that [tex]\omega_{o} \approx 7.330\,\frac{rad}{s}[/tex], [tex]M = 1.5\,kg[/tex], [tex]m = 0.54\,kg[/tex] and [tex]r = 0.1\,m[/tex], then the angular speed of the turntable after the event is:
[tex]I_{T} = \frac{1}{2}\cdot M\cdot r^{2}[/tex]
[tex]I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}[/tex]
[tex]I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}[/tex]
[tex]\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}[/tex]
[tex]\omega_{T} = 3.004\,\frac{rad}{s}[/tex] ([tex]28.687\,\frac{rev}{min}[/tex])
The turntable's angular speed after the event is 28.687 revolutions per minute.
Which of the following is NOT true about
newspaper weather maps?
A. They report the temperature of the area in degrees
Fahrenheit.
B. They let you know how much precipitation to expect.
C. They provide more detail than weather service maps.
D. They tell you about the warm and cold air fronts.
Select all the correct answers.
Which statements about our solar system are false?
Our solar system is made up of the Sun and other objects that orbit the Sun.
The Sun is the only star in our solar system.
Dwarf planets have several other bodies in their path orbiting the Sun just as they do.
The Kuiper Belt is between Uranus and Neptune.
A large number of irregularly shaped comets are located in a vast ring between the orbits of Mars and Jupiter.
D and E are false.
A, B, and C are true.
Answer:
- Our solar system is made up of the Sun and other objects that orbit the Sun.
- The Sun is the only star in our solar system.
- Dwarf planets have several other bodies in their path orbiting the Sun just as they do.
Explanation:
plato
A rod that is 96.0 cm long is made of glass that has an index of refraction equal to 1.60. The rod has its ends ground to convex spherical surfaces that have radii equal to 8.00 cm and 19.1 cm. An object is in air on the long axis of the rod 19.9 cm from the end that has the 19.1-cm radius.
(a) Find the image distance due to refraction at the 19.1-cm radius surface.
(b) Find the position of the final image due to refraction at both surfaces.
(c) Is the final image real or virtual?
Please help me someone !
Answer:
The object is moving at constant speed.
Explanation:
The spaces between the dots are equal.
What is the equation for frequency?
a. number of cycles +unit of time
b. number of cycles - unit of time
c. number of cycles ×unit of time
d. number of cycles/ unit of time
Answer:
d
Explanation:
does altitude has an effect on weight, yes or no
Answer: yes
Explanation:
Weight is the gravitational force experienced on a body. If you move up to higher altitudes, the distance between you and earth increases. ... Yes, weight drops as you go up in altitude (because of diminishing gravity), though your mass remains the same. However, the effect is not huge.
It is almost as if each outer planet is a solar system in its own right.
True or False
stainless steel, tell us about its properties and what should be taken into account when using it?
A student hangs a block from a light string that is attached to a massive pulley of unknown radius R, as shown in the figure. The student allows the block to fall from rest to the floor. Which two of the following sets of data that could be measured or determined should the student use together to determine the final angular velocity of the pulley just before the block hits the floor? Select two answers. Justify your selections.
Answer:
The mass of the block, the distance of the block above the floor, and the time it takes the block to reach the floor, because these quantities can be used to determine the acceleration of the block.
The radius and the mass of the pulley, because these quantities can be used together to determine the rotational inertia of the pulley.
Explanation:
If the motion starts from rest, the initial angular velocity will be zero and the final angular velocity can be determined with the product of angular acceleration and time of motion of the pulley.
Angular velocity is defined as the change in the angular displacement per change in time of motion. This can be expressed mathematically as follows;
[tex]\omega = \frac{\Delta \theta}{\Delta t} = vr[/tex]
where;
Ф is the angular displacementt is the time of the motionv is the linear velocityr is the radius of the circular path.In a circular motion that starts from rest and ends with final velocity, the equation is given as;
[tex]\omega_f =\omega_i + \alpha t[/tex]
Where;
[tex]\omega_f[/tex] is the final angular velocity[tex]\omega_i[/tex] is the initial angular velocity[tex]\alpha[/tex] is the angular accelerationThus, if the motion starts from rest, the initial angular velocity will be zero and the final angular velocity can be determined with the product of angular acceleration and time of motion of the pulley.
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Three 5 Ohm resistors are connected in series to a 10 Volt power supply. What is the current through each resistor?
Answer74.3
:
Explanation:
An electromagnetic wave has a frequency of 5.0 x 1014 Hz. What is the
wavelength of the wave? Use the equation 2 = and 3.0 x 108 m/s for the
speed of light.
A. 1.7 x 10-8 m
О
B. 6.0 x 1022 m
O C. 6.0 x 10-7 m
O D. 1.7 x 105 m
Answer:
c
Explanation:
wavelength = speed of light/ frequency
= (3x 10^8 m/s)/(5.0 x 10^14 Hz)
= 6.0 x 10^-7 m
Which of the following is an example of heat transfer by conduction?
A. Heat is transferred to the air above a candle flame.
B. Heat is transferred to the soil on a sunny day.
c. Heat is transferred to your hand from a warm cup.
D. Heat is transferred to the air from a warm lightbulb.
Answer:
option c
Heat is transferred to your hand from a warm cup
conduction is the process of transferring of heat from one material to another when they are in contact
hope it helps
1.
is the rate that velocity changes
O Acceleration
O
Time
O
Distance
O
Mass
Answer:
a) acceleration
Explanation:
Acceleration is, by definition, the change of an object's velocity.
A 41.6-kg person, running horizontally with a velocity of +4.21 m/s, jumps onto a 14.6-kg sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?
Answer:
a) v = 3.116 m / s, b) μ = 1.65 10⁻²
Explanation:
a) to find the velocity of the set, let's define a system formed by the person and the sled, so that the forces during the collision are internal and the moment is conserved
initial instant. Before the crash
p₀ = M v₀
final instant. After the crash
p_f = (M + m) v
the moment is preserved
M v₀ = (M + m) v
v = [tex]\frac{M}{M+m} \ v_o[/tex]
let's calculate
v = [tex]\frac{41.6}{41.6 + 14.6} \ 4.21[/tex]
v = 3.116 m / s
b) for this part let's use the relationship between work and kinetic energy
W = ΔK
as the body has its final kinetic energy is zero
the work of the friction forces is
W = - fr x
the negative sign is because the friction forces always oppose the movement
let's write Newton's second law
Y axis
N - W_sled -W_person = 0
N = mg + M g
N = (m + M) g
X axis
fr = ma
the friction force has the expression
fr = μ N
fr = μ g (m + M)
we substitute
- μg (m + M) x = 0- ½ (m + M) v²
μ = [tex]\frac{1}{2} \ \frac{v^2 }{g \ x }[/tex]
let's calculate
μ = [tex]\frac{1}{2} \ \frac{3.116^2}{9.8 \ 30.0}[/tex]
μ = 0.0165
μ = 1.65 10⁻²