Answer:
is point 2
Explanation:
There is a good cause for concern that technology will create social isolation
If an object is placed between the focal point and twice the focal length of a convex lens, which type of image will be produced?
A.
real, upright, and magnified
B.
virtual, inverted, and smaller
C.
virtual, upright, and magnified
D.
real, inverted, and magnified
E.
real, upright, and smaller
Answer: I think its D
Explanation: Hope this was helpful...
A potter's wheel is a uniform disk of mass 4.50 kg and radius 0.650 m and can spin freely around a vertical axis through its center. With the wheel spinning at an angular speed of 4.70 rad/s, a small piece of clay of mass 0.870 kg is dropped at the outer edge of the wheel and sticks to it. Find the final angular speed of the wheel clay. Treat the piece of clay as a point particle. Group of answer choices
Answer:
3.39 rad / s.
Explanation:
Given data:
mass of disk = 4.50 Kg
radius of wheel = 0.650 m
mass of the clay = 0.870 kg
The moment of inertial of the wheel = I = 4.5 kg x ( 0.65 m )2 / 2 = 0.95 kg . m2.
Now, applying the principle of angular momentum conservation :
Iω_i = ( I + mr2 )ω_f.
where ω_i = initial angular speed= 4.70 rad/s, ω_f = final angular speed
Hence, ω_f = Iω_i / ( I + mr2 )
= ( 0.95 kg . m2 x 4.7 rad / s ) / [ 0.95 kg . m2 + 0.87 kg x ( 0.65 m )2 ]
= 3.39 rad / s.
Hence, correct answer is : 3.39 rad / s.
2.3 The motion of an object is accelerated
when its speed:
a
decreases
b remains constant
increases.
Answer:
The motion of an object is accelerated when its speed increases.
When the temperature of a certain solid, rectangular object increases by AT, the length of one
side of the object increases by 0.010% = 1.0 x 10-4 of the original length. The increase in volume
of the object due to this temperature increase is
Explanation:
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Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is |v with arrow| = 230 km/s and the orbital period of each is 15.5 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 1030 kg.)
Answer:
[tex]1.554\times 10^{32}\ \text{kg}[/tex]
Explanation:
M = Mass of each star
T = Time period = 15.5 days
v = Orbital velocity = 230 km/s
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
Radius of orbit is given by
[tex]R=\dfrac{vT}{2\pi}[/tex]
We have the relation
[tex]\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}[/tex]
The mass of each star is [tex]1.554\times 10^{32}\ \text{kg}[/tex]
A farmer plants the same crop in a field year after year. Every year there are months during which the field is left without any plants. The farmer notices a decline in soil
quality during these months. What causes this decrease in soil quality?
Erosion
Drought
Desertification
Consumption of nutrients by the crops
Answer:
Drought
Explanation:
three small balls each of mass 13.3g are suspended separately from a common point by silk threads, each 1.17m long. The balls are identically charged hang at the corners of equilateral triangle 15.3 cm on a side. Find the charge on each ball
The charge on each ball is 6.1 ×10⁻⁸ C.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity.
An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. A spring balance can be used to calculate the Force. The Newton is the SI unit of force. Force is a vector quantity and it has both magnitude and direction.
Let the balls have equal charge of q. Then, electrostatic force acting on each particle = √3 q²/r² dyne = √3 q²/15² dyne =0.0076 q².
Weight of each ball = 13.3 ×980 dyne = 13034 dyne.
As the electrostatic force and the weight acts perpendicular to each other.
Hence, 0.1/√3 = 0.0076 q²/13034
q = 6.1 ×10⁻⁸ C
Hence, the charge on each ball is 6.1 ×10⁻⁸ C.
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When scientists use one of their five senses to gather information, they are
A.
drawing a conclusion.
B.
making an inference.
C.
predicting a relationship
D.
making an observation.
Answer:
making an observation
Explanation:
HELP ITS DUE IN 4 MINUTES
Answer:
igneous = melted rocks formed by cooled magma
sedimentary is brken rocks, kayers with fossil...
metamophic is rocks formed by pressure and heat
What is the force of gravity between two 40.0kg masses that are separated by 3.00m?
Answer:
[tex]f = g \times \frac{m1 \times m2}{ {d}^{2} } [/tex]
[tex]f = 6.67 \times {10}^{ - 11} \times \frac{40 \times 40}{9} [/tex]
F=1.2x 10^-8Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.40 cm. If the potential difference across the plates was 23.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.
Answer:
E = 1.64 x 10⁶ V/m
Explanation:
The electric field in the region between the plates can be given by the following formula:
[tex]E = \frac{\Delta V}{d}[/tex]
where,
E = Electric Field = ?
ΔV = Poetential Difference across the plates = 23 KV = 23000 V
d = distance between plates = 1.4 cm = 0.014 m
Therefore, using these values in the equation, we get:
[tex]E = \frac{23000\ V}{0.014\ m}[/tex]
E = 1.64 x 10⁶ V/m
1. Describe the components of the reflex arc
I need help please .
Answer:
option 5
Explanation:
because all u do is have to add them up
1. A person kicks a rock off a cliff horizontally with a speed of 20 m/s. It takes 7.0 seconds to hit the
ground, find:
a. height of the cliff
b. final vertical velocity
C. range
D.speed and angle of impact
This problem involved half projectile.
initial velocity, vo = 20 m/s
time of flight, t = 7 s
(a) Simply use the formula to get the height, h:
h = vo*t - (1/2)gt^2
(b) To get the final vertical velocity or terminal velocity (vf), use the formula:
(vf)^2 - (vo)^2 = 2gh
(c) Use the formula find the horizontal distance traveled, R:
R = vo * cos(θ) * t
But since the angle involved with respect to horizontal is zero, and cos(0) = 1, we have
R = vo * t
Hope this helps~ `u`
Jai
You are a member of an alpine rescue team and must get a box of supplies, with mass 3 kg , up an incline of constant slope angle 30.0, so that it reaches a stranded skier who is a vertical distance 4 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 0.05. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s2 .What is the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier
Answer:
v = 9.04 m / s
Explanation:
For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.
W = Em_f - Em₀ (1)
Starting point. Lower slope
Em₀ = K = ½ m v²
highest point. Where is the skier at a height h
Em_f = U = m g h
The work of rubbing
W = -fr x
the negative sign is because the friction force opposes the movement.
Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular
let's use trigonometry to break down the weight
cos θ = W_y / W
sin θ = Wₓ / W
W_y = W cos θ
Wₓ = W sin θ
Y axis
N - Wₓ = 0
N = mg sin θ
X axis
fr = m a
the friction force has the expression
fr = μ N
fr = μ mg sin θ
we look for the job
W = - μ mg sin θ x
where x is the distance along the slope
we substitute in 1
-μ mg sin θ x = mg h - ½ m v²
let's use trigonometry to find the distance x
tan 30 = h / x
x = h / tan 30
we substitute
- [tex]\mu \ mg \ sin \theta \ \frac{h}{tan 30} \ x[/tex] = m gh - ½ m v²
we use
tan 30 = sin30 / cos30
v² = 2g h + 2 μ g h cos 30
v = [tex]\sqrt{ 2gh \ (1+ cos 30}[/tex]
let's calculate
v = [tex]\sqrt{ 2 \ 9.8 \ 4 \ (1 + 0.05 \ cos \ 30)}[/tex]
v = 9.04 m / s
SCIENCE
The growth of algae in ocean water is limited by their need for
a.
warm ocean currents.
b.
carbon dioxide and sunlight.
c.
dissolved oxygen.
d.
low salinity.
Answer:
c is trueeeeeeeeeeeeeee
The growth of algae in ocean water is limited by their need for dissolved oxygen. Thus, the correct answer is option C.
What is algae?The term "algae" refers to a large and diverse group of photosynthetic eukaryotic organisms. It is a polyphyletic grouping of species from several distinct clades. Most are aquatic and autotrophic, lacking many of the distinct cell and tissue types found in land plants, such as stomata, xylem, and phloem.
Algae, like all organisms, normally grow in balance with their ecosystems, with the amount of nutrients in the water limiting their growth. Deep ocean water contains fewer algae because algae require sunlight and carbon dioxide to thrive.
Therefore, due to need for dissolved oxygen the growth of algae in ocean water is limited.
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1. If airbags reduce the impact force from an accident why has there been questions over their safety?
2. Are airbags the safest option to prevent serious injury or death from a car accident?
Answer:Air bags can leave you in even more injury, From the impact they give
You could end up with a broken nose,arm
concussion
26. Which member of the carbon family is a non-metal?
Please help
Explanation:
Check out the Periodic Table attached here. The Carbon family is the vertical group starting with Carbon, Group 14. Nonmetals are in green, which element there is a nonmetal?
Surface tension is often calculated using a machine that lifts a wire ring from the surface of a liquid. In this case the ring and liquid have some cohesive forces and attract rather than repel. In order to lift a ring of radius 2.75 cm off of the surface of a pool of blood plasma, a vertical force of 2.00*10-2 N greater than the weight of the ring is required. Consider the situation just before the ring breaks contact with the blood plasma where the blood plasma makes a contact angle of approximately zero degrees along the circumference of the ring and is stretched down vertically on both sides of the ring.
Required:
Calculate the surface tension of blood plasma from this information.
Answer:
0.116 N/m
Explanation:
Since the net force acting on the ring must be greater than 2.00 × 10⁻² N, and the surface tension T = F/L where F = net force = 2.00 × 10⁻² N and L = circumference of ring = 2πr where r = radius of ring = 2.75 cm = 2.75 × 10⁻² m.
So, T = F/L
= F/2πr
= 2.00 × 10⁻² N ÷ 2π(2.75 × 10⁻² m)
= 1/2.75π N/m
= 1/8.64 N/m
= 0.116 N/m
A vessel having a capacity of 0.05 m³ contains a mixture of saturated water an saturated steam at a temperature 245°C the mass of the liquid present is 10 kg. find the following: i- The pressure. ii- The mass. iii- The specific volume. iv- The specific enthalpy. v- The specific internal energy.
A graduated beaker with 375 mL of water is sitting on a scale which measures the weight of the glass and water to be 7.60 N. When a rock is put into the glass, the volume level of the water changes to 450 mL and the scale reading changes to 9.22 N. What is the specific gravity of the rock
Answer:
Volume of water displaced = 450 - 375 = 75 ml
Vr = volume of rock = 75 ml
Wr = 9.22 - 7.60 = 1.62 N weight of 75 ml of rock
Density of rock = 1.62 N / 75 ml = .0216 N / ml
Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml
Density of rock / density of water = .0216 / .0098 = 2.20
The specific gravity of the rock in the given water volume is 0.2.
The given parameters;
initial volume of the water, = 375 mlweight of the water, = 7.6 Nfinal volume of water = 450 mlchange in scale reading = 9.22 NThe specific gravity of the rock is calculated as follows;
[tex]S.G = \frac{weight \ in \ air}{Weight \ in \ water} \\\\S.G = \frac{450 - 375}{375} \\\\S.G = 0.2[/tex]
Thus, the specific gravity of the rock in the given water volume is 0.2.
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Please help me!!!!!!!!!
Hi there! :)
[tex]\large\boxed{17.32 m/s}[/tex]
Use the following equation in solving for kinetic energy:
KE = 1/2mv² where:
KE = kinetic energy (J)
m = mass (kg)
v = velocity (m/s)
Plug in the given values:
12,000 = 40v²
Divide both sides by 40:
12,000 / 40 = v²
300 = v²
Take the square root of both sides:
√300 = v
v ≈ 17.32 m/s
What kind of weather would MOST LIKELY lead to hurricane formation?
Answer:
The recipe for a hurricane is a combination of warm, humid wind over tropical waters. The temperature of tropical waters must be at least 80 degrees F for up to 165 feet below the ocean’s surface. As this warm water meets the wind that blows west from Africa across the ocean, it causes the water to vaporize. The water vapor then rises into the atmosphere, where it cools and liquefies.
Explanation:
I hope this helps
A hoop is rotating at 12 rad/s on a horizontal plane about an a vertical axis that passes through its
edge when a second body with mass one quarter of the disc's is placed safely on it just opposite of the
pivot. What is the percentage change of the kinetic energy of the hoop?
A. 20%
B. 80%
C. 36%
D. 64%
Answer:
the correct answer is C
Explanation:
For this exercise we can use the conservation of angular momentum, let's define a system formed by the disk and the other body, so that the torques during the collision have been internal and the moment is conserved.
Initial instant. Before placing the second body
L₀ = I₁ w₁
Final moment. Right after placing the body
L_f = (I₁ + I₂) w
the moment of inertia of the ring with respect to an axis passing through its center is tabulated
I_{cm} = m r²
we use the parallel axes theorem to find the moment of inertia about an axis passing through one end
I₁ = I_{cm} + m d²
where d is the distance from the center of mass to the new axis
d = r / 2
we substitute
I₁ = mr² + m (r/2)²
I₁ = 5/4 m r²
The second body does not specify a specific shape, so we can assume it to be punctual
I₂ = m’ (2r)²
I₂ = 4 m’ r²
in the exercise indicate that
m ’= m / 4
I₂ = m r²
how the moment of inertia is conserved
L₀ = L_f
I₁ w₁ = (I₁ + I₂) w
5/4 m r² w₁ = (5/4 m r² + m r²) w
5/4 m r² w₁ =9/4 m r² w
w = 5/9 w₁
we calculate
w = 5/9 12
w = 6.67 rad / s
having the initial and final angular velocities we can find the kinetic energy of the hoop
K₀ = ½ I₁ w₁²
K_f = ½ (I₁) w2
the energy ratio is
[tex]\frac{K_f}{K_o}[/tex] = [tex]\frac{w^2}{w_1^2}[/tex]
\frac{K_f}{K_o} = (6.67 / 12)²
\frac{K_f}{K_o} = 0.309
[tex]\frac{K_f}{K_o} %[/tex] = 31%
the correct answer is C
A projectile was fired horizontally from a cliff 20m above the ground. If
the horizontal range of the projectile is 40m, calculate the initial velocity
of the projectile.
The initial velocity of the projectile is 19.8m/s
Explanation:
First, find time.
From our kinematics equations:
delta y = Vi•t + (1/2)at^2
rearrange,
t = sqrt[(2•delta y)/a]
t = sqrt[(2•20m)/9.8m/s^2]
t = 2.02s
Next, plug time into new kinematics equation to solve for the Vi in the x direction (horizontal)
delta x = Vi•t + (1/2)at^2
delta x = Vi•t
Rearrange:
Vi = delta x/t
Vi = 40m/2.02s
Vix = 19.8m/s
A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 520 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tensions in the rope to the left and to the right of the mountain climber.
Answer:
The answer is "892.90 N"
Explanation:
Following are the solution to these question:
Calculating the vertical force of the summation that is equal to zero:
[tex]\to TL \cos 65 + TR \cos 80 -520 = 0\\\\\to 0.4226\ TL + 0.1736\ TR = 520\\[/tex]
Calculating the sum of horizontal forces that is equal to zero:
[tex]\to TL\sin 65 - TR \sin 80 = 0 \\\\\to 0.9063TL - 0.9846TR = 0\\\\\to TL = (\frac{0.9846}{0.9063})\ TR \ \ = 1.0866\ TR\\\\\to 0.4226(1.0866) \ TR +0.1736\ TR =520 \ N\\\\\to 0.6328 \ TR = 520 \\\\\to TR = 821.74 \ N \\\\\to TL = 1.0866 \times 821.74 = 892.90\ N[/tex]
A bus is moving at a speed of 150km/hr. Begins to slow at a constant rate of 3.0m/s each second. Find how far it goes before stopping
Answer:
Distance = 13.9 meters
Explanation:
Given the following data;
Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s
Decelerating speed = 3m/s
To find the distance travelled with this speed;
Distance = maximum speed/decelerating speed
Distance = 41.67/3
Distance = 13.9 meters
Therefore, the bus would travel a distance of 13.9 meters before stopping.
Please help !!!!!!!!!!!!!!!
Answer:
9v
Explanation:
please someone help me !!
Answer:
D
Explanation:
because if the solvent is more than the solvent then we can't resolve it.
so our product will be suspended
Anyone know how to do this???
Answer:
World War 1 was caused by entangled alliances, nationalism, imperialism, and major
advancements in military technology. Does the Treaty of Versaille address those issues?
Explain your answer using facts. (5 points)