ayuda!!!!!!!!!!! es para ahora :((

Answer: 100 V

Explanation:

Given

Resistance [tex]R=1\ k\Omega[/tex]

Potential difference [tex]V=100\ V[/tex]

If the resistor is placed across the potential difference, then it receives a voltage of 100 V as it parallel to the source.

Voltage drop across resistor is 100 V.

Answer:

2

Explanation:

in reflection the incident ray and reflected rays are equal an angle

Answer:

by using v = u + at equation we can find "a"

14 = 60 - 2.7a

2.7a = 60 - 14

2.7a = 46

decceleration = 17.03

Answer:

the positive charge is the period number

Explanation:

I might be wrong

Answer:

The positive charge is the group number.

Explanation:

Answer:

The water table is the upper surface of the zone of saturation. The zone of saturation is where the pores and fractures of the ground are saturated with water. It can also be simply explained as, the upper level, below which the ground is saturated.

Answer:

I think it's D!!

cuz protons are in the atoms

Melting: the substance changes back from the solid to the liquid

Evaporation: the process by which water changed from a liquid to a gas.

Boiling: the process by which a liquid turns into a vapor when it is heated to it's boiling point.

Condensation: the substance changed from a gas to a liquid

Freezing: the substance changed from a liquid to solid.

Answer:

Melting is a process that causes a substance to change from a solid to a liquid.

Evaporation is the process of turning from liquid into vapour.

Boiling is the rapid vaporization of a liquid, which occurs when a liquid is heated to its boiling point, the

Condensation is the process of water vapor turning back into liquid water

Freezing is a phase transition where a liquid turns into a solid when its temperature is lowered below its freezing point

Answer:

C) 2.00 m/s^2

Explanation:

F = m*a

150N = 75kg(a)

a = 150N/75kg

a = 2.0m/s²

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. ... The gravitational potential energy is equal to its weight times the height to which it is lifted. PE = kg x 9.8 m/s2 x m = joules. The 9.8 us the gravitational acceleration constant.

so the answer is "the gravitational potential energy is decreasing"

Answer:

B. low melting points and low densities

Explanation:

Alkali metals are any of the monovalent elements found in Group IA of the periodic table. They readily lose their one valence electron to form ionic compounds with nonmetals. Examples of alkali metal are Potassium (K), Lithium (L), and Sodium (Na).

Compared to most other metals, the chemical properties that alkali metals have are low melting points (28.5°C) and low densities that is typically less than 1 grams per cubic centimeters.

1. set up the equation

2. multiply the variables

3. analyze your results

Answer:

From the base unit of length, we can define volume, and from the base units of length, mass, and time, we can define energy. ... Volume. Since volume is length cubed, its SI derived unit is m3

Explanation:

Hope it helps

Answer:

(a) [tex]\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}[/tex]

(b) P =  0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

[tex]P = \sigma AT^4[/tex]

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

[tex]\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}[/tex]

(b)

Now, for 90% radiator blackbody at 2000 K:

[tex]P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4[/tex]

P =  0.816 Watt

Answer:

It is streamlined

Explanation:

The speed of a car as well as its air resistance and fuel consumption has a lot to do with its shape.

A car that is streamlined glides through air with less resistance, moves with a higher speed and consumes less fuel.

The essence of streamlining is to reduce the aerodynamic drag on the car thereby increasing its speed, decreasing its air resistance and fuel consumption.

Answer:

it is streamlined

Explanation:

Answer:

Pascal Law's says that:

If the area of one end of a U-tube is A, and the area of the other end is A'. then if we apply a force F in the first end (the one of area A), the force experienced at the other end must be:

F' = F*(A'/A).

b) Now we can apply this to our particular case:

if the area of one end is 0.01m^2, and the area of the other end is 1m^2

Then we have:

A = 0.01m^2

A' = 1m^2

So, if now we apply a force F in the first end, the force experienced at the other end will be:

F' = F*(1m^2/0.01m^2) = F*100

This means that the force in the other end must be 100 times the force in the first end.

Answer:

guard cell

hope it helps

Answer:

50m/s2

Explanation:

u=0m/s

a=10m/s

v=?

t=5s

v=u+at

=0+(10×5)

=0+50

=50m/s

Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of

a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²

There are two forces acting on the car in this situation:

• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n

• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)

By Newton's second law, the net forces in the perpendicular and parallel directions are

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) = ma

==>   sin(θ) = a/g   ==>   θ = arcsin(a/g) ≈ 17.8°

(Notice that in the paralell case, the positive direction points toward the center of the curve.)

When rounding the curve at 31.8 m/s, the car's radial acceleration changes to

a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²

and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) + f = ma

The first equation gives

n = mg cos(θ)

and substituting into the second equation, we get

mg sin(θ) + µmg cos(θ) = ma

==>   µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12

Answer:

Explanation:

You are in the chapter on Physics about uniform circular motion and gravity. This is a centripetal force problem in particular, and the equation for that is

[tex]F_c=\frac{mv^2}{r}[/tex] where

[tex]F_c[/tex] is the centripetal force needed to keep the car moving in its circular path,

m is the mass of the car,

v is the velocity with which the car is moving, and

r is the radius of the circle that the car is moving around.

For us, the centripetal force is supplied by the friction keeping the car on the road, altering the equation to become

[tex]f=\frac{mv^2}{r}[/tex] and friction is defined by

f = μ[tex]F_n[/tex] (the coefficient of friction multiplied by the weight of the car).

Going on and getting buried even deeper,

[tex]F_n=mg[/tex] which says that the weight of the car is equal to its mass times the pull of gravity. Putting all that together, finally, we have the equation we need to solve this problem:

μ·m·g = [tex]\frac{mv^2}{r}[/tex] and we solve this for μ:

μ = [tex]\frac{mv^2}{mgr}[/tex] and it just so happens that the mass of the car cancels out. (I'll tell you why the mass of the car doesn't matter at the end of this problem). Filling in and solving for the coefficient of friction:

μ = [tex]\frac{31.8^2}{(9.8)(75.0)}[/tex] to 2 significant figures is

μ = 1.4

The mass of the car doesn't affect whether or not the car can stay on the curve. Even though a car with a greater mass will have a greater frictional force, that doesn't mean that it's easier for that car to stay on the road; a larger mass only means that a larger centripetal force is needed to keep it moving in a circle. This makes the gain in friction become offset by the fact that a larger centripetal force is necessary. Thus,

On a flat curve, the mass of the object experiencing circular motion does not affect the velocity at which it can stay on the curve.

Answer:

Ions are formed by the addition of electrons to, or the removal of electrons from, neutral atoms or molecules or other ions; by combination of ions with other particles; or by rupture of a covalent bond between two atoms in such a way that both of the electrons of the bond are left in association with one of the ..

Answer:

Ions are formed when atoms lose or gain electrons, simply when electrons from the metal transfers electrons from its outermost shell( forming a positively charged metallic ion) to the outermost shell of the non-metallic atom( forming a negatively charged ion)

Answer:

D. 2.7 N

Explanation:

Applying

F = kq'q/r²................ Equation 1

Where F = force, k = coulomb's constant, q' = first charge, q = second charge, r = distance between the charge

From the question,

Given: q' = +3.0×10⁻⁷ C, q = +4.0×10⁻⁷C, r = 2.0×10⁻² m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 1

F = ( +3.0×10⁻⁷)(+4.0×10⁻⁷)(8.98×10⁹)/(2.0×10⁻²)²

F = 26.94×10⁻¹ N

F = 2.694 N

F ≈ 2.7 N

Answer:

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Answer:

Q = 8608.8 J

Explanation:

Given that,

The mass of ice, m = 34 g

The temperature changes from -2°C to 118°C​.

The specific heat of ice, c = 2.11 J/g°C

The heat energy needed,

[tex]Q=mc\delata T\\\\Q=34\times 2.11\times (118-(-2))\\Q=8608.8\ J[/tex]

So, 8608.8 J of energy is needed.

Answer:

[tex]{ \bf{momentum = mass \times velocity}} \\ \\ { \tt{35 = m \times 10}} \\ { \tt{mass = 3.5 \: kg}}[/tex]

Answer:

88.34 N directed towards the center of the circle

Explanation:

Applying,

F = mv²/r................... Equation 1

F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.

But,

v = 2πr/t................... Equation 2

Where t = time, π = pie

Substitute equation 2 into equation 1

F = m(2πr/t)²/r

F = 4π²r²m/t²r

F = 4π²rm/t²............. Equation 3

From the question,

Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s

Constant: π = 3.14

Substitute these values into equation 3

F = 4(3.14²)(0.7)(0.8)/0.5²

F = 88.34 N directed towards the center of the circle

Answer:

is from 9 x 107 m/sec to 27 x 107 m/s

Answer:

-148,6 MJ

Explanation:

Dado que la liberación de calor al medio ambiente es;

H = mcθ

Dónde;

m = masa de agua

c = capacidad calorífica específica del agua

θ = aumento de temperatura

H = 1220 kg × 4200 × [-29-0]

H = -148,6 MJ