Answer: Calendar
Explanation: I'm pretty sure it is calendar.
As tall as a tree, as small as a bug, As fast as an eagle, as slow as a slug. Many times in front or behind but never on top, always aligned. What am i
Most generation of electricity involves turning turbines to spin coils of metal. However some do not. Give a type of energy that does not involve spinning coils of metal. How does it generate electricity?
Answer:
Photovoltaic Energy
Explanation:
There are other forms of energy that do not involve spinning of metals to generate electricity, one of many such is the Solar Photovoltaic Energy.
Photovoltaic is a form of energy in which sunlight Is directly into electricity. This type of energy is often used to power devices such as small wrist watches, rechargeable lamps, and others. They are connected together to panels that are joined together in arrays to power people or heavy power plants. Photovoltaic power plants are happening to be one of the fastest growing sources of electricity generation around the world.
A tennis rackets delivers 500 N on impact with a tennis ball. What is the force of the tennis ball on the racket?
A: this cannot be determined without knowing the mass of the tennis ball
B: the force on the tennis racket is more than 500 N
C: the force on the tennis racket is 500 N
D: the force on the tennis racket is less than 500 N
Answer:
the a.
Explanation:
because if you compute you can't know the distance
Integrated Concepts A lightning bolt strikes a tree, moving 20.0 C of charge through a potential difference of 1.00×102 MV . (a) What energy was dissipated? (b) What mass of water could be raised from 15ºC to the boiling point and then boiled by this energy? (c) Discuss the damage that could be caused to the tree by the expansion of the boiling steam.
Answer:
a) 2*10^9 J
b) 764.8 kg
Explanation:
Given that
Energy of charge, q = 20 C
Potential difference, ΔV = 1*10^2 MV = 1*10^2 * 10^6 V = 1*10^8 V
a)
To find the energy dissipated, we use the formula
ΔU = qΔV
ΔU = 20 * 1*10^8
ΔU = 2*10^9 J
b)
Change in temperature, ΔT = 100 - 15°
ΔT = 85° C
Change in energy, ΔU = 2*10^9 J
Specific heat of water, C = 4180 j./Kg.K
Latent heat of vaporization, L(v) = 2.26*10^6 J/Kg
Q1 = mcΔT
Q2 = mL(v)
Net energy needed, U = Q1 + Q2
U = mcΔT + mL(v)
U = m (cΔT + L(v))
m = U /[cΔT + L(v)]
Being that we have all the values, we then substitute
m = 2*10^9 / [(4180 * 85) + 2.26*10^6]
m = 2*10^9 / (3.553*10^5 + 2.26*10^6]
m = 2*10^9 / 2.615*10^6
m = 764.8 kg
c)
Having 765 kg of steam at the temperature would have extreme effect on the tree, damaging it permanently. Possibly even blowing it to pieces
The density of ice is 917 kg/m^3, and the density of sea water is 1025 kg/m^3. A swimming polar bear climbs onto a piece of floating ice that has a volume of 6.78 m^3. What is the weight of the heaviest bear that the ice can support without sinking completely beneath the water?
Answer:
w_bear = 7175.95 N , m = 732.24 kg
Explanation:
Let's analyze the situation a bit, the ice block is in equilibrium with the thrust given by Archimedes' law that is directed towards the bottom, the weight of the ice and the weight of the bear.
B -W_ice - w_bear = 0
w_bear = B - W_ice
The thrust is given by
B = ρ g V
B = 1025 9.8 6.78
B = 68 105.1 N
Note that we use the total volume of the block since the problem indicates that it is submerged.
The weight of the ice is
W_ice = m g
the density is
ρ_ice = m_ice / V
m_ice = rho_ice V
we substitute
W_ice = ρ_ice g V
W_ice = 917 9.8 6.78
W_ice = 60929.15 N
we substitute in the first equation
w_bear = 68105.1 - 60929.15
w_bear = 7175.95 N
the mass of this bear is
w_bear = m g
m = w_bear / g
m = 7175.95 / 9.8
m = 732.24 kg
Is a bike being pedaled potential or kinetic?
Answer:
Since there is no motion yet, there is no kinetic energy. This means that the work needed to get up the hill is the potential energy at the top. Once you have coasted down the hill and are at the bottom, all of the energy is kinetic.
Explanation:
.
A rock is thrown with an initial vertical velocity 50 m/s at an angle of 40 degrees.
a. What is the horizontal component of the velocity?
b. What is the vertical component of the velocity?
c. What is the hang timel?
d. What is the peak height?
e. What is the range?
Answer:
[tex]38.3\ \text{m/s}[/tex]
[tex]32.14\ \text{m/s}[/tex]
6.55 seconds
[tex]52.65\ \text{m}[/tex]
[tex]254.84\ \text{m}[/tex]
Explanation:
u = Initial velocity of rock = 50 m/s
[tex]\theta[/tex] = Angle of throw = [tex]40^{\circ}[/tex]
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Horizontal component is given by
[tex]u_x=u\cos\theta\\\Rightarrow u_x=50\times \cos40^{\circ}\\\Rightarrow u_x=38.3\ \text{m/s}[/tex]
The horizontal component of the velocity is [tex]38.3\ \text{m/s}[/tex]
Vertical component is given by
[tex]u_y=u\sin\theta\\\Rightarrow u_y=50\times \sin40^{\circ}\\\Rightarrow u_y=32.14\ \text{m/s}[/tex]
The horizontal component of the velocity is [tex]32.14\ \text{m/s}[/tex]
Time of flight is given by
[tex]t=\dfrac{2u\sin\theta}{g}\\\Rightarrow t=\dfrac{2\times 50\sin40^{\circ}}{9.81}\\\Rightarrow t=6.55\ \text{s}[/tex]
The hang time of the rock is 6.55 seconds
Maximum height is given by
[tex]h=\dfrac{u^2\sin^2\theta}{2g}\\\Rightarrow h=\dfrac{50^2\sin^240^{\circ}}{2\times 9.81}\\\Rightarrow h=52.65\ \text{m}[/tex]
Maximum height is [tex]52.65\ \text{m}[/tex]
Range is given by
[tex]d=\dfrac{u^2\sin2\theta}{g}\\\Rightarrow d=\dfrac{50^2\sin(2\times40)^{\circ}}{9.81}\\\Rightarrow d=254.84\ \text{m}[/tex]
The range is [tex]254.84\ \text{m}[/tex]
The timeline below shows some major discoveries in biology.
Which of the following theories did these discoveries lead to?
A.
the theory of evolution
B.
the cell theory
C.
the germ theory
D.
the atomic theory
Each piece has 15 of 4 meters of fabric.
Answer:
The cell theory
Explanation:
Answer:
The cell theory
Explanation:
Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same direction.
Answer:
The magnitude of the force between the two parallel wires is 0.0111 N.
Explanation:
Given;
length of the two parallel wires, L = 42 m
distance between the two wires, r = 0.03 m
current in both wires, I₁, I₂ = 6.3 A
Therefore, the magnitude of the repulsive force between the two parallel wires is given by;
[tex]F = \frac{\mu_0 I_1I_2l}{2\pi r}\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^{-7} \ T.m/A \\\\F = \frac{(4\pi *10^{-7})(6.3)^2(42)}{2\pi (0.03)}\\\\F = 0.0111 \ N[/tex]
Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.
Select the correct responses to the questions from the drop-down menus. Which type of wave vibrates both side to side and up and down? P waveS wavesurface wave Which type of wave is the first to arrive during an earthquake? P waveS wavesurface wave Which type of wave may make the ground roll? P waveS wavesurface wave
The S wave vibrates both side to side and up and down whereas P wave is the first to arrive during an earthquake
Seismic waves during earthquake:During the eathquake, two types of seismic waves are generated , P wave and S wave.
The S waves produces vibrations to the ground in a shearing force perpendicular to the direction of propagation of the wave, moving the ground up and down or from side to side. S waves are called secondary waves because arrive after P waves.
The P waves is analogous to longitudinal vibrations, producing vibrations to the ground to compress and expand in the direction of propagation f the wave. They are called primary waves because they are the first type of wave to arrive at seismic recording stations.
Learn more about seismic waves:
https://brainly.com/question/13062627?referrer=searchResults
Answer:
i got it right so chillllll :)
Explanation:
Select the correct responses to the questions from the drop-down menus.
Which type of wave vibrates both side to side and up and down?
✔ S wave
Which type of wave is the first to arrive during an earthquake?
✔ P wave
Which type of wave may make the ground roll?
✔ surface wave
7.
As the distance from the Sun increases the time to orbit the Sun
a. Increases
b. Decreases
c. Stays the same
As the distance from the Sun increases the time to orbit the Sun increases. (a)
Examples:
-- Nearest planet to the sun: Mercury. Time to orbit the sun: 88 days
-- 3rd planet from the sun: Earth. Time to orbit the sun: 1 year
-- 5th planet from the sun: Jupiter. Time to orbit the sun: 12 years
-- 9th closest object to the sun: Pluto. Time to orbit the sun: 248 years
Another pair of examples:
-- Object in a near Earth orbit: International Space Station
Time to orbit the Earth: 90 minutes
-- Object in a far Earth orbit: the Moon
Time to orbit the Earth: 27.3 days
The reason for all of this is: Two things about orbits.
1). The larger the orbit is, the farther the object has to travel around it.
2). The farther out the object is, the slower it travels in its orbit.
This is simply the way gravity works.
A 0.1 kg tennis ball is brought from a speed of 15 m/s to a speed of 45 m/s when hit by a tennis racket. If this done over 0.05 seconds, what is the power generated by the tennis racket
Kinetic energy is the energy of _____.
Speed, Mass, Work, Motion
Answer: Motion
Explanation:
Little Billie has a mass of 25 kg. The acceleration of gravity on the moon is one-sixth of the value on Earth. What is Little Billie's weight on the moon? _____________ What is little Billie's mass on the moon? _____________
Answer:
Mass is constant everywhere,
But weight is different,
If earth g = 10 then moon's is 1.6666667
Now billie's weight in moon is 41.6667
Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Remember: cal=m*deltaT
(WRITE THE WORK!/Give explanation.)
I’ll give BRAINLIEST.
Answer:
12552 J or 3000 calories
Explanation:
Q = m × c × ∆T
Where;
Q = amount of heat energy (J)
m = mass of water (g)
c = specific heat capacity (4.184 J/g°C)
∆T = change in temperature
For 50mL of water, there are 50g, hence, m = 50g, c = 4.184 J/g°C, initial temperature = 0°C, final temperature = 60°C.
Q = m × c × ∆T
Q = 50 × 4.184 × (60 - 0)
Q = 209.2 × 60
Q = 12552 J
Hence, the amount of heat energy used to heat the water is 12552 J or 3000 calories
Consider an underwater spherical air bubble (you can assume the index of refraction of air is 1 and water is 1.33) with radius R. This bubble can act as a lens. What is its focal length?
Answer:
-2R
Explanation:
The focal length can be calculated using this formula
1/f = (n1/n2-1)(1/r1-1/r2)
1/f = (1/1.33-1)(1/r + 1/r)
= -0.133/1.33(2/R)
F = -2R
The focal length is this
if you double the magnitude of a vector, does it follow that the magnitude of the components double? Give example.
Answer:
The magnitude of the vector clearly doubles if each of its components is doubled.
Explanation:
what are two ways to advance basketball down the Court
having a plan and listening to your coach
A rock falls from a vertical cliff that is 4.0 m tall and experiences no significant air resistance as it falls. At what speed will it's gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy in m/s?
Answer:
v = 6.3 m/s
Explanation:
Since no significant air resistance exists, total mechanical energy must be kept constant at any time.At the top of the cliff, all the energy is gravitational potential energy, as follows:[tex]E_{i} = K_{i} + U_{i} = 0 + U_{i} (1)[/tex]
If we choose the ground level as our zero reference level for the gravitational potential energy, Ui is simply:Ui = m*g*h (1)At any height, the sum of the kinetic and the gravitational potential energy must be equal to (1).We know from the question, that at the point of interest, both types of energies must be equal each other, so we can write the following expression from (1):[tex]m*g* h = 2*\frac{1}{2}*m*v^{2} (2)[/tex]
Dividing both sides by m, simplifying, and solving for v, we get:[tex]v = \sqrt{g*h} =\sqrt{9.8m/s2*4.0m} = 6.26 m/s (3)[/tex]
v = 6.3 m/s (with two significative figures)help asap
How does the scientist know that the substance goes through a change of state during the time period labeled B on the graph?
a.the temperature remains the same
b.the temperature increases gradually
c.the particles start slowing down
d.the particles start moving faster
Answer:
The answer is A. The temperature remains the same.
A golf ball starts with a speed of 2 m/s and slows at a constant rate of 0.5 m/s2, what is its velocity after 2 s?
Answer:
3m/s
Explanation:
Given parameters:
Initial speed = 2m/s
Acceleration = 0.5m/s²
Time = 2s
Unknown:
Final speed = ?
Solution:
To solve this problem, we apply the right motion equation;
V = U + at
V is the final speed
U is the initial speed
a is the acceleration
t is the time
V = 2 + (2 x 0.5)
V = 2 + 1
V = 3m/s
A 5 kg ball is moving at a velocity of +2 m/s when it
speeds up to +5 m/s in 7 seconds.
Calculate the acceleration.
Heya!!
For calculate aceleration, lets applicate formula:
[tex]\boxed{a=\frac{V-V_o}{t} }[/tex]
Δ Being Δ
V = Final Velocity = 5 m/s
Vo = Initial Velocity = 2 m/s
a = Aceleration = ? m/s²
t = Time = 7 s
⇒ Let's replace according the formula:
[tex]\boxed{a=\frac{5\ m/s - 2 \ m/s}{7\ s} }[/tex]
⇒ Resolving
[tex]\boxed{ a=0,428\ m/s^{2}}[/tex]
Result:
The aceleration of the object is 0,428 m/s²
Good Luck!!
Answer:
Ksasdasd
Explanation:
A basketball is shot with an initial velocity of 16 m/s at an angle of 55° to the
horizontal. What is the approximate horizontal distance that the ball travels in
1.5 s?
Answer:
approximately 13.77 m
Explanation:
We use the kinematic equations for the horizontal motion which is a uniform non-accelerated motion, and with initial velocity equal to 16 * cos(55). This is described by the equation:
[tex]x=16 * cos(55^o) * t = 9.177 * t[/tex]
then at t = 1.5 seconds, the covered distance becomes:
x = 9.177 * (1.5) = 13.7655 m
Round the answer to the number of decimals that the problem asks.
An engineer is designing a runway for an airport. Of the planes that will use the airport, the
lowest acceleration rate is likely to be 3 m/s2
. The takeoff speed for this plane will be 70 m/s.
Assuming this minimum acceleration, what is the minimum allowed length for the runway?
Answer:
816m
Explanation:
70^2m/s=2(3m/ss)(x)
x=816
Which best explains why an MRI is useful for diagnosing cancer but ineffective in treating it?
The frequency of the radio waves used is so high that the waves cannot be used to target specific cells.
The wavelength of the radio waves used is so short that the waves cannot be used to target specific cells.
The amount of energy carried by radio waves is enough to provide images but not enough to kill cells.
The amount of energy carried by radio waves is so high that it provides images but can also damage cells.
Answer:The amount of energy carried by radio waves is enough to provide images but not enough to kill cells.
Explanation:
The reason of MRI should be useful is that the energy amount that should be carried out via the radio waves.
What is Magnetic resonance imaging (MRI) ?It refers to the medical imaging technique that should be applied in the radiology to create the pictures with respect to the anatomy and the process of physiological with related to the body.
Also, it is applied for strong the magnetic field, gradient related to the magnetic field, also the radios waves for produce image in the body organs.
Learn more about wave here: https://brainly.com/question/12734041
PLZ HELP I BEG YOU :(
I DON'T KNOW WHAT THE ANSWERS ARE
Answer:
9. qulitative(spell check)
10. false
11. true
Explanation:
What is energy?
A. A change that appears in an object when forced is applied.
B. The property of a body that gives it mess.
C. The amount of heat produced by the body.
D. The ability of an object to undergo change.
E. The ability of a body to move.
Answer:
A
Explanation:
Answer: a or e
Explanation:
Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of the dive, the speed of the airplane is 150 m/s. What is the apparent weight of the 70-kg pilot at that point?
Answer:
3311N
Explanation:
r = radius = 600m
V = speed = 150m/s
Mass = weight = 70kg
The weight of pilot when calculated due to circular motion
W = tv
Fv = mv²/r
Fv = 70x150²/600
Fv = 79x22500/600
= 15750000/600
= 2625N
Real Weight of the pilot = m x g
= 70 x 9.8
= 686N
The apparent Weight is calculated by
Mv²/r + mg
= 2625N + 686N
= 3311 N
Therefore the apparent Weight is 3311N
A fish is able to jump vertically out of the water with a speed of 4.45 m/s. What is the speed of the fish as it passes a point 0.6 m above the water?
How much time, after leaving the water does it take for the fish to pass a point 0.6 m above the water while it is on its way down?
Answer:
Explanation:
given
initial velocity u = 4.45m/s
Height = 0.6m
g = 9.8m/s²
Required
final velocity v
Using the equation of motion;
v² = u²-2gH (upward motion of the fish makes g to be negative)
v² = 4.45²-2(9.8)(0.6)
v² = 19.8025-11.76
v² = 8.0425
v = 2.84 m/s
Hence the speed of the fish as it passes a point 0.6 m above the water is 2.84m/s
To get the time, we will use the formula
v = u - gt
2.84 = 4.45 - 9.8t
2.84-4.45 = -9.8t
-1.61 = -9.8t
t = 1.61/9.8
t = 0.164secs
Hence the time taken is 0.164secs
an explanation can be consistent but may not be the best explanation for a phenomenon
True
False
Answer:
true
Explanation: