Answer:nh
0.750 kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board.
Water which contains large amounts of calcium dissolved within it results from
a) factory sewage outfalls.
b) water dissolving materials in the ground.
c) volcanic activity.
d) increase in number of wells in area.
Answer: B
Explanation:
The other options have nothing to do with the water apart from transporting it.
Please show your workmkmmkekdmnesnndndndndn
Answer:
Explanation:
Remark
The question really is how long does the projectile stay in the air if it was just dropped from a height of 46 meters.
Givens
vi = 0
a = 9.81 m/s^2
h = 46 m
t = ?
Solution
d = vi*t + 1/2 a t^2
46 = 0 + 1/2 * 9.81 * t^2 multiply both sides by 2
92 = 9.81 * t^2 divide by 9.81
t^2 = 9.38 Take the square root of both sides
sqrt(t^2) = sqrt(9.38)
t = 3.06 seconds.
In 3.06 seconds the object will hit the ground, thus stopping it. How far horizontally does it go in that time.
d = r * t
d=?
r = 102 m/s
t = 3.06 seconds
d = 102 * 3.06
d = 312.4 meters.
Heat rash is caused by chemical exposure Truth or false
Answer:
This is false.
Explanation:
Heat rash develops when some of your sweat ducts clog. Instead of evaporating, perspiration gets trapped beneath the skin, causing inflammation and rash. Heat rash is usually self-limited, meaning it resolves on its own without treatment. Over-the-counter treatments such as calamine, hydrocortisone cream, itch preparations (such as Benadryl spray), or sunburn lotions can be used as skincare to treat the itching and burning symptoms. Heat rash usually goes away on its own within three or four days so long as you don't irritate the site further. Heat rash happens when the sweat glands get blocked. The trapped sweat irritates the skin and leads to small bumps.
Answer:
FALSE
Explanation:
Heat rash develops when blocked pores or sweat ducts trap perspiration under your skin. Symptoms range from superficial blisters to deep, red lumps. Some forms of heat rash feel tingly or intensely itchy.
Suppose that scientists observe violent gas eruptions on a planet with an acceleration due to gravity of 3.8 m/s2. The jets throw sand and dust about 75 m above the surface.
What is the speed i of the material just as it leaves the surface?
Scientists estimate that the jets originate as high‑pressure gas speeds through vents just below ground at about 160 km/h. How much energy per kilogram of material does the jet lose to non conservative forces as the high‑speed matter forces its way to the surface and into the air?
Answer:
linarkuoouonpnpjjñihy
Explanation:
Which is the largest number? O A. 9.9 x 106 O B. 4.5 x 107 O c. 5.5 x 105 O D. 5.8 x 108
Answer:
107
Explanation:
It's the biggest numberwhat is 4 differences between saturated unsaturated and supersaturated solutions
Answer:
Unsaturated Solution: Less amount of salt in water, clear solution, no precipitation. Saturated Solution: The maximum amount of salt is dissolved in water, Colour of the solution slightly changes, but no precipitation. Supersaturated Solution: More salt is dissolved in water, Cloudy solution, precipitation is visible.
How can I create a whirlpool of water?
Answer:
To make a small scale whirlpool of water, take a cup and fill it with water and stir. It will create a vortex that pulls particles down.
Or you could see this when draining a tub of water.
12. Suppose the polar bears were running on land instead of swimming. if the polar bear runs
at a speed of about 5.5m/s how far will it travel in 55 seconds?
the polar bear will travel 302.5m
a block starts from rest and begins sliding down an incline. The block reaches a speed of 12 meters per second as it slides a distance of 50 meters. calculate the blocks rate of acceleration
Vf² = V₁² + 2ad
(12 m/s) ² = (0 m/s) ² + 2 (a) (50m)
a = 1.44 m/s ²
Answer:
[tex]\boxed {\boxed {\sf 1.44 \ m/s^2}}[/tex]
Explanation:
We are asked to find the block's rate of acceleration.
We are given the final velocity, initial velocity, and distance. Therefore, we will use the following kinematic equation:
[tex]{v_f}^2={v_i}^2+2ad[/tex]
The block starts at rest or 0 meters per second. The block reaches a final velocity of 12 meters per second. The distance traveled is 50 meters.
[tex]v_i[/tex]= 0 m/s [tex]v_f[/tex]= 12 m/s [tex]d[/tex]= 50 mSubstitute the values into the formula.
[tex](12 \ m/s)^2 = (0 m/s)^2 + 2a (50 \ m)[/tex]
Solve the exponents.
(12 m/s)² = 12 m/s * 12 m/s = 144 m²/s²(0 m/s)² = 0 m/s * 0 m/s = 0 m²/s²[tex]144 \ m^2/s^2= 0 m^2/s^2 + 2a (50 \ m)[/tex]
The 0 can be subtracted from both sides, or simply canceled.
[tex]144 \ m^2/s^2 = 2a (50 \ m)[/tex]
Multiply 2 and 50 meters.
[tex]144 \ m^2/s^2 = (2 *50 \ m)* a[/tex]
[tex]144 \ m^2/s^2 = (100 \ m)* a[/tex]
We are solving for a, so we must isolate the variable. It is being multiplied by 100 meters. The inverse of multiplication is division, so we divide both sides by 100 meters.
[tex]\frac {144 \ m^2/s^2}{100 \ m}= \frac{(100 \ m)*a}{100 \ m}[/tex]
[tex]\frac {144 \ m^2/s^2}{100 \ m}=a[/tex]
[tex]1.44 \ m/s^2 =a[/tex]
The block's rate of acceleration is 1.44 meters per second squared.
The term that is used to quantity the amount of heat in the evaporator is
Answer:
Enthalpy
The quantity of heat contained in a material as measured from a particular base or point.
HELP
A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s . Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision.
Answer:
Hey I go to GOC send me your number
Explanation:
When converted to a household measurement, 9 kilograms is approximately equal to a
Answer:
D) 19.8 lbs
Explanation:
1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.
1kg is also equal to 2.205 lbs. 9*2.205=19.8416
9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system
a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.
This leaves us with 19.8 lbs
Light of 650 nm wavelength illuminates two slits that are 0.20 mm
apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?
PLEAsE URGENT. thank you
We have that for the Question "Light of 650 nm wavelength illuminates two slits that are 0.20 mm apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?" it can be said that the distance to the screen
d=1.168m
From the question we are told
Light of 650 nm wavelength illuminates two slits that are 0.20 mm
apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?
Generally the equation for the distance is mathematically given as
[tex]d=\frac{0.33*10^{-2}*0.23*10^{-3}}{650*10^{-9}}\\\\d=1.160m[/tex]
d=1.168m
Therefore
The distance to the screen
d=1.168m
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A boy 11.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the instant the ball is thrown. If the boy throws the ball upward at 50.0 ∘ above the horizontal, at 8.00 m/s .
How fast must the dog run to catch the ball just as it reaches the ground?
How far from the tree will the dog catch the ball?
12414253
Explanation:
Pls help thx need asap!!
Answer:
10N right
Explanation:
First, we need to find the base of the triangle to find the force acting on the right of the object.
We have been given the hypotenuse and we need to find the adjacent, so we are using cosSOH, CAH, TOAcos(θ) = adjacent ÷ hypotenuseNext, sub in the values (we don't know the adjacent so i've called it x):
cos(60) = x ÷ 60To find x, we need to multiply both sides by 60
60 × cos(60) = xx = 30 NNow we know all the forces, we can work out the net force:
Both of the 40N cancel out because they are opposite forcesThe force on the right of the object is 10N stronger than the force on the left (30N - 20N), so the object would move to the rightStarting from rest, a student pulls a (25.0 A) kg box a distance of (4.50 B) m across the floor using a (134 C) N force applied at 35.0 degrees above horizontal. The coefficient of kinetic friction is 0.220 between the floor and the box. Find the work done by the pulling force. Give your answer in joules (J) and with 2 significant figures.
A 25.0 kg box is moved 4.50 m by a force of 134 N applied at 35.0° above horizontal. The work done by the pulling force is 4.9 × 10² J.
Starting from rest, a student pulls a 25.0 kg box a distance of 4.50 m across the floor using a 134 N force applied at 35.0 degrees above horizontal.
We can calculate the work (W) done by the pulling force using the following expression.
[tex]W = F \times d \times cos\theta = 134 N \times 4.50 m \times cos35.0\° = 4.9 \times 10^{2} J[/tex]
where,
F: puling forced: displacementθ: angle between F and dThe friction between the floor and the box. does not influence the work done by the pulling force.
A 25.0 kg box is moved 4.50 m by a force of 134 N applied at 35.0° above horizontal. The work done by the pulling force is 4.9 × 10² J.
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make a layout plan of your dream fishpond following parts?
Answer:
Do you have. a picture or full page for you're question its saying following parts
PLEASE ANSWER ASAP IM GONNA MARK AS BRAINLIEST
What is the value of the force measured by the spring balance?
In these images taken a few days apart, the light part of the Moon appeared to get smaller
over time. Why did this happen? (1)
a) The Moon moved farther away from the sun so less sunlight reached the Moon's
surface.
b) Earth moved so its shadow was blocking more of the Moon so the student astronomer
was able to see less of it.
c) The Moon rotated so that less of the light-colored rock on the Moon's surface faced
Earth
d) The Moon moved so that the student astronomer was able to see less of the half that
facer the sun
Answer:
Explanation:
d) The Moon moved so that the student astronomer was able to see less of the half that faced the sun.
A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonconducting 2.50 cm wire. A uniform electric field of magnitude 1.85×108N/C is directed parallel to the wire, as shown in the figure. (Figure 1)
Part A
Find the tension in the wire.
Express your answer with the appropriate units.
Part B
What would the tension be if both charges were negative?
Express your answer with the appropriate units.
(a) The tension on the wire when the two charges have opposite signs is 383.5 N.
(b) The tension on the wire if both charges were negative is 3.640.25 N.
The given parameters;
first charge, q₁ = 8.75 μC second charge, q₂ = -6.5 μC electric field, E = 1.85 x 10⁸ N/Cdistance between the two charges, r = 2.5 cm(a)
The attractive force between the charges is calculated as follows;
[tex]F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N[/tex]
The force on the negative charge due to the electric field is calculated as follows;
[tex]F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N[/tex]
The tension on the wire is the resultant of the two forces and it is calculated as follows;
[tex]T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N[/tex]
(b) when the two charges are negative
The repulsive force between the two charges is calculated as follows;
[tex]F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N[/tex]
The force on the first negative charge due to the electric field is calculated as follows;
[tex]F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N[/tex]
The force on the second negative charge due to the electric field is calculated as follows;
[tex]F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N[/tex]
The tension on the wire is the resultant of the three forces and it is calculated as follows;
[tex]T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N[/tex]
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HELP! I need some assistance
Answer:
I'm a Filipino hahahah I'm not
11. A plow pushes 100 kg of snow with 300 N of force. How much is the pile of snow
accelerated?
Answer:
Explanation:
This is an application of Newton's second Law.
Formula
F = m * a
F = 300 N
m = 100 kg
a = ?
F = m * a
300N = 100 kg * a Divide by 100
300N/100kg = a
a = 3 m/sec^2
What is an involuntary and a voluntary muscle action?
Answer:
Involuntary muscle action are muscles that move WITHOUT conscious control, and voluntary muscle action are muscles that move WITH conscious control.
Explanation:
A example of a involuntary muscle action would be your heart. Your heart is a muscle beats without you thinking about it and is very vital to your body. A example of a voluntary muscle would be the muscles that are attached to your bones and allow you to move your body, and it's voluntary because your body won't move on it's own.
Hope this helps.
A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introduction of 430 J of heat caused the system to expand, doing 238 J of work against a constant pressure of 101 kPa. What is the value of for this process
Answer: You do not specify what is being asked for. ∆E? ∆H?
∆E = (430 - 238) J = 192 J
∆H = 430 J
Explanation:
If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.
Therefore ∆H = 430 J
If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w
The question states that 238 J of work are done AND the system expanded
(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)
Therefore, ∆E = (430 - 238) J = 192 J
What happens to the size of an Object when it is heated ?
Answer:
the object melts so the size get smaller
giúp em toàn bộ nhé ............
Explanation:
nnnnkgcbchg of bbjbjk oh jjzjjzkedkkdkfjjsjdsjfkjfdkkdowkronqfojofj see j FC
Bill is sitting on a tree limb that is 4 meters above the ground with a water balloon. He wants to drop the water balloon so that it falls on his friend Tim who is approaching on a bike at a constant speed of 8 meters per second. How far away should Tim be from a point directly under bill when bill drops the balloon?
Answer:
wow poor bill
Explanation:
2. A 6.0 kg mass is pulled along a horizontal surface where the coefficient of friction is 0.20.
a) What is the friction force acting on the mass?
b) What force is needed to accelerate it at 0.68 m/s^(2)?
Answer:
Below
Explanation:
a) To find the force of friction acting on the mass we can use this formula :
friction = coefficientoffriction x normal force
First we need to find the normal force of this object :
Fn = mg
= (6.0)(9.8)
= 58.8 N
Now we can find the force of friction acting on the object :
friction = (0.20)(58.8N)
= 11.76 N
= 12 N (we round this to 2 sig figs)
b) To find the force needed to accelerate this object at 0.68 m/s^2 use this :
F = ma
0.68 = x - 11.76 / 6
x = 15.84
x = 16 N
Hope this helps! Best of luck <3
Which phrase describes an irregular galaxy O has a round shape O does not have a specific shape O has arms that extend from the center O is larger than other types of galaxies
Answer:
I think it does not have a specific shape
Explanation:
Irregular galaxies have no definite shape, which means that the second option is correct. They are definitely not round.
The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name.
They are actually smaller than the other types of galaxies. This makes them prone to collisions.
The phrase that describes the irregular galaxy is that it does not have a specific shape. The correct option is B.
What is irregular galaxy?A galaxy with an undefined shape is an irregular galaxy with a large number of young stars and a large amount of gas and dust that is typically found near larger galaxies.
An irregular galaxy is a catch-all term for any galaxy that does not neatly fit into one of the Hubble classification schemes.
They have no discernible shape or structure and could have formed as a result of collisions, close encounters with other galaxies, or violent internal activity.
They lack beautiful spiral arms, but they do have dark patches of gas and dust. Some irregular galaxies appear to be the result of two galaxies colliding.
Thus, the correct option is B.
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Incisors (front teeth) act as wedges in the human body. Which change would increase the mechanical advantage of incisors?
1 a decrease in the distance of the teeth from the fulcrum,
2 an increase in the distance of the teeth from the fulcrum,
3 an increase in the thickness at the widest part of the teeth and no change in the length of the teeth,
4 an increase in the length of the teeth and no change in the thickness at the widest part of the teeth
To increase the mechanical advantage of incisors (front teeth) so that they act as wedges in the human body is through 2. an increase in the distance of the teeth from the fulcrum.
The decrease in the distance will decrease the mechanical advantage of incisors. The increase in the thickness at the widest part of the teeth will not necessarily support the incisors to become more effective wedges.
Thus, incisors will acquire some mechanical advantage by increasing the force that it exerts. This advantage can be achieved by increasing the distance of incisors from the fulcrum.
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Answer:
4 aka D
Explanation: