Answer: Are these free point?
Explanation:
How much did the pressure drop in the storm's center from November 9, 1200z, until November 11, 0000z
The pressure dropped by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.
Using the attached map below, in the morning of November 9, the pressure situated at the storm's center = 1000 MB isobar located at the center.
Meanwhile, on November 11, 0000z the pressure situated at the storm's center = 976 MB isobar
The difference in this pressure is regarded as the pressure drop in the storm's center and it is determined as follows;
= 1000 MB - 976 MB
= 24 MB
Therefore, we can conclude that the pressure drop by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.
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What is speed of sound in air?
Answer:
speed of sound, speed at which sound waves propagate through different materials. In particular, for dry air at a temperature of 0 °C (32 °F), the modern value for the speed of sound is 331.29 metres (1,086.9 feet) per second
help me in the Infer one
Answer:
Overcurrents
Circuit breakers ("fuses" back in the day) are supposed to trigger when you are requesting past it more power (thus more current, since the tension is fixed and power is the product of voltsge times current). So whoever made that part of the grid did not foresee that many appliances.
Another option could be a short circuit somehwere but I think it is unlikely since the breaker would not trigger only when all three loads are powered on.
Determine the unbalanced force necessary to accelerate a 2.60 kg object at a rate of 14.0 m/s².
Answer:
Explanation:
F = ma
F = 2.60(14.0)
F = 36.4 N
Objects 1 and 2 attract each other with a gravitational force of 34 units. If the distance separating objects 1 and 2 is changed to one-third the original value, then the new gravitational force will be ____ units.
Answer:
F12 = G M1 M2 / R12^2
F12' = G M1 M2 / R12'^2
F12' / F12 = R12'^2 / R12^2 = (1/3)^2
F12' = 1/9 F12
The new force is 1/9 the of the old force
A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 4.7 cm from its equilibrium position?
A)The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.
Select all that apply.
1. force constant k
2. total energy E
3. mass m
4. maximum velocity vmax
5. amplitude A
6. potential energy U at x
7. kinetic energy K at x
8. position x from equilibrium
B)What is the kinetic energy of the object on the spring when the spring is compressed 4.7 cm from its equilibrium position?
C)What is the potential energy U of the toy when the spring is compressed 4.7 cm from its equilibrium position?
Hi there!
Part A:
The only quantities explicitly given to us are:
3. mass (m)
4. Maximum velocity (vmax)
5. Amplitude (A)
8. Position x from equilibrium
Part B:
To solve, we must begin by calculating the force constant, 'k'.
We can use the following relationship:
[tex]v = \sqrt{\frac{k}{m}(A^2-x^2)[/tex]
We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:
[tex]\frac{v^2}{(A^2-x^2)} = \frac{k}{m}[/tex]
[tex]\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44[/tex]
Now, we can find the velocity at 4.7cm (0.047m) using the equation:
[tex]v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s[/tex]
Plug this value into the equation for kinetic energy:
[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}[/tex]
Part C:
The potential energy of a spring is given as:
[tex]U = \frac{1}{2}kx^2[/tex]
Find 'k' using the derived quantity above:
[tex]\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m[/tex]
Now, calculate potential energy:
[tex]U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}[/tex]
A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s, how high is the cliff, if it took 12 s for the ball to reach the ground?
I need the Formula
Hi there!
We can use the equation:
d = v₀t + 1/2at²
Where:
v₀ = initial velocity downward
a = acceleration due to gravity
t = time
Plug in given values:
d = 4(12) + 1/2(9.8)(12²)
d = 48 + 705.6 = 753.6 m
What is the voltage if the current is 4 A and the resistance is 10 Ω?
Answer:
40 volts
Explanation:
Use the equation [tex]V=IR[/tex]
[tex]V=(4)(10)[/tex]
[tex]V=40[/tex]
A cognitive bias is a kind of logical fallacy.
True or false?
Answer:
True
Explanation:
It definetly is because, if we are biased it sets us up to fail on some things.
6. How are the temperature of the universe and Cosmic Microwave Background (CMB) related?
A. Astronomers use the temperature of CMB as the warmest temperature in the universe
B. Astronomers calculate the temperature of the universe based on the coldest part of the CMB
C. Astronomers consider the temperature of the universe to be the temperature of CMB
D. Astronomers never consider the temperature of CMB when looking at the temperature of the universe
Answer:
I think the answer is (A)...
Hope this helps!
A 1.1 kg ball drops vertically onto a floor, hitting with a speed of 23 m/s. It rebounds with an initial speed of 5.0 m/s. (a) What impulse acts on the ball during the contact
Hi there!
We know that:
I = Δp = m(vf - vi)
Plug in the given values. Remember to take into account direction ⇒ let the rebound velocity be positive and initial be negative.
I = 1.1(5 - (-23)) = 30.8 Ns
can anyone explain how to do it for me? i don't understand...
Answer:
15[m].
Explanation:
1) the required distance is AD, for more info see the attached picture.
2) [tex]AD=\sqrt{AA_1^2+A_1D^2} =\sqrt{81+144}=15[m].[/tex]
An object accelerates from rest, with a constant acceleration of 6.4 m/s2, what will its velocity be after 7s?
I also need the Formula
Hi there!
The formula for velocity given acceleration:
v = at
Plug in given values:
v = 6.4(7) = 44.8 m/s
She had a quiet voice what is the adjective in this sentence
Answer:
Quiet, is the adjective in this sentence
Explanation:
She had a quiet voice. This statement uses the adjective quiet. Her voice can be characterized as quiet.
What is adjectives?
A word used to characterize a person, place, or thing is called an adjective. Adjectives are used to modify nouns or pronouns.
There are many adjectives that describe quiet, including muted, peaceful, reticent, silent, soft, gentle, mild, placid, private, secluded, sedate, serene, timid, smooth, calm, tranquil, modest, restrained, sober, and subdued.
Hence, the significance of the adjectives is aforementioned.
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Tectonic plate movement is the reason why northern California has a very different landscape than southern California. Two different tectonic plates, each moving in different directions, border the western side of the North American Plate. Use the map to identify the two tectonic plates that border the North American Plate to the west.
Answer:
Remember, NORTH ^, EAST >, SOUTH v, WEST <
Explanation:
It doesn't have to be a super complex answer. All you have to do is look to the left (west) of the North American plate. What are the 2 plates that you see? The Pacific and the Juan de Fuca, yeah? To the South, there is the Cocos amongst a few others.
I am not going to share the answer for sure as I haven't completed the test yet but that's how I'm solving it. You should write the answer in your own words anyways. Hope this helps! Have a good day :)
Answer:
The Juan de Fuca Plate and the Pacific Plate both border the west side of the North American Plate.
Explanation:
Edmentum
Need help with dot product
[tex]\textbf{A}\cdot\textbf{B} = 11.5[/tex]
Explanation:
The dot product between two vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B}[/tex] is defined as
[tex]\textbf{A}\cdot\textbf{B} = AB\cos{\theta}[/tex]
where A and B are the magnitudes of the vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B},[/tex] respectively and [tex]\theta[/tex] is the angle between the two. Since A = 3, B = 5 and [tex]\theta = 40°,[/tex] the dot product [tex]\textbf{A}\cdot\textbf{B}[/tex] is
[tex]\textbf{A}\cdot\textbf{B} = (3)(5)(0.766) = 11.5[/tex]
Accelerations are produced by
A. Masses
B.accelerations
C. Velocities
D.unbalanced, net forces
can someone please help no.1,2,3
Answer:
12345678910i don't know
Amy uses 20 N of force to push a lawnmower 10 meters. How much work does she do?
Answer:
200J
Explanation:
work done = force x distance
= 20 x 10
= 200J
A tennis player strikes the tennis ball with an initial velocity of 44.7 m/s horizontally. The ball is initially 1.28 m above the ground and 12.9 m from the 0.914 m tall net. Does the tennis ball make it over the net?
Hi there!
We can begin by finding the total time taken for the ball to reach the net using the equation:
dₓ = vₓt
12.9 = 44.7t
12.9/44.7 = t = 0.289 s
Now, we can use the following equation to solve for displacement in the Y direction:
d = y₀ + vit + 1/2at²
There is no initial vertical velocity, so:
d = y₀ + 1/2at²
Plug in known values:
d = 1.28 + 1/2(-9.8)(0.289²)
d = 0.87m
Thus, since 0.87 m < 0.914 m, the tennis ball does NOT make it over the net.
What state of matter is jelly?
Answer:
Hey mate.....
Explanation:
This is ur answer.....
SolidsJelly is a colloid where colloidal particles are solids which are dispersed in the liquid. Jelly is a fluid which can not be considered as a particular one kind of state. It is in the category of 'gel' which is a colloidal form.
Hope it helps!
Brainliest pls!
Follow me! ;)
True or False: The basketball should be dribbled below the waist.
if fire needs oxygen to burn, where does the sun get oxygen if there is no oxygen in space?
The sun is not a burning fire.
It's much much much hotter than that.
The sun's energy is the result of continuous nuclear fusion in it's core. We know how to do that on Earth, but the only thing we've been able to use it for so far is hydrogen bombs and other thermonuclear weapons.
Which is a similar theme in both "The Stag at the Pool" and "Hans in Luck"? O A. The things we think are valuable sometimes turn out not to be valuable after all. 0 B. It is important to know whom we can trust and whom we cannot O C. By working hard, we will end up with possessions that will make us feel successful. 0 D. The best things come in small packages.
A similar theme in both "The Stag at the Pool" and "Hans in Luck" is A. The things we think are valuable sometimes turn out not to be valuable after all.
In "The stag at the pool", a thirsty stag went to the pool to drink. As he bent over, he admired his antlers but despised his legs. He saw a lion and ran but his antlers got stuck in a tree and he was caught by the lion.
The theme in "Hans in Luck" shows that one doesn't need possessions to be happy. The character in the story had several possessions but was still unhappy. Both themes showed that the things we think are valuable sometimes turn out not to be valuable after all.
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A flywheel, rotating about its axis at a rate of 4 rev/s is acted upon by a torque of 25 Nm for 10 sec. If the wheel has moment of inertia of 1.2 kgm^2, what would be the speed of the wheel at the end in rev/s?
The applied torque increases the angular speed by the application of an
angular acceleration.
The speed after 10 seconds is approximately 37.16 rev/s.Reasons:
The speed of the flywheel at the axis = 4 rev/s
The torque applied, T = 25 N·m
The time the torque is applied, t = 10 sec
Moment of inertia of the flywheel, I = 1.2 kg·m²
Required:
The speed at the end of the 10 seconds
Solution:
T = I·α
Where;
α = Angular acceleration
[tex]\displaystyle \alpha = \frac{T}{I}[/tex]
Therefore;
[tex]\displaystyle \alpha = \frac{24 \ N\cdot m}{1.2 \ kg \cdot m^2} = \mathbf{20\frac{5}{6} \ s^{-2}}[/tex]
The rotational speed, ω = ω₀ + α·t
Which gives;
[tex]\displaystyle \mathrm{The \ angular \ speed, } \ \omega = \frac{2 \cdot \pi \times 4 \ rad }{s} = \frac{8 \cdot \pi \ rad }{s}[/tex]
ω₀ = 8·π rad/s
Which gives;
[tex]\displaystyle \omega = \mathbf{8 \cdot \pi +2 \frac{5}{6} \times 10} = 233.47[/tex]
The speed of the wheel in revolution per second is therefore;
[tex]\displaystyle Speed \ in \ rev/s = \frac{8 \cdot \pi +2 \frac{5}{6} \times 10}{2\cdot \pi} \approx 37.16[/tex]
The speed after 10 seconds is approximately 37.16 rev/s.
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An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 5.60 mm before stopping. How far does the lighter fragment slide
Answer:
M1 + M2 = 0 total momentum before explosion = momentum after explosion where M1 is the lighter fragment
M1 V1 + 7 M1 V2 = 0
V2 = -V1 / 7
The lighter fragment will slide 7 times as far - 39.2 mm because it must have 7 times as much velocity - assuming the distance slid is proportional to the original velocity
What is the difference between real and apparent weightlessness?
Answer:
In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.
Explanation:
Hope this helps:)
why does air pressure decrease with increasing altitude?
How much distance does a car travel with a speed of 2m/s in 15 min?
Explanation:
1 minute is 60 seconds so you multiply 60 * 15 and then multiply that answer * 2
The figure shows a rear view of a loaded two-wheeled wheelbarrow on a horizontal surface. It has balloon tires and a weight W = 684 N, which is uniformly distributed. The left tire has a contact area with the ground of AL = 6.20 × 10-4 m2, whereas the right tire is underinflated and has a contact area of AR = 9.20 × 10-4 m2. Find (a) the force from the left tire, (b) the pressure from the left tire, (c) the force from the right tire, (d) the pressure from the right tire that each tire applies to the ground.
Answer:
Explanation:
Summing moments about the CG to zero will show that the two normal forces are equal and have a value of FL = FR = 684/2 = 342 N
Pressure on the left
PL = 342 / 6.20e-4 = 551,612.9032... = 5.5e5 Pa
Pressure on the right
PR = 342 / 9.20e-4 = 371,739.1304... = 3.7e5 Pa
A) The force from the left tire is; FL = 342 N
B) The pressure from the left tire is; PL = 551613 N/m²
C) The force from the right tire is; FR = 342 N
D) The pressure from the right tire is; PR = 371739 N/m²
We see that;
FL and FR are upward forces
W is the downward force.
We know that in equilibrium;
Sum of upward forces = sum of downward forces
Thus;
FL + FR = W
We are given W = 684 N
Since W is at the center, it means that FL = FR. Thus;
FL = FR = 684/2
FL = FR = 342 N
We are given;
Contact area of left tire; AL = 6.2 × 10⁻⁴ m²
Contact area of right tire; AR = 9.2 × 10⁻⁴ m²
Formula for pressure is;
Pressure = Force/Area
Pressure on the left tire;
PL = FL/AL
PL = 342/(6.2 × 10⁻⁴)
PL = 551613 N/m²
Pressure on the t right tire;
PR = FR/AR
PR = 342/(9.2 × 10⁻⁴)
PR = 371739 N/m²
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