Two identical items, object 1 and object 2, are dropped from the top of a 50.0m50.0m building. Object 1 is dropped with an initial velocity of 0m/s0m/s, while object 2 is thrown straight downward with an initial velocity of 13.0m/s13.0m/s. What is the difference in time, in seconds rounded to the nearest tenth, between when the two objects hit the ground

Answers

Answer 1

Answer:

Δt = 1.1 s

Explanation:

Given information:

H= 50.0 m

g= 9.8 m/s²

Object 1v₀ = 0

   [tex]H = \frac{1}{2} * g* t^{2}[/tex]

Solving for t, we get:

t₁= 3.2 s

Object 2v₀ = 13 m/s

We can find the final velocity for the object when it hits the ground, using the following expression:

[tex]v_{f}^{2} - v_{o}^{2} = 2*g*H[/tex]

Solving for vf, we get:

vf = 33.9 m/s

Applying the definition of acceleration, being this acceleration the one due to gravity (g), we can write the following equation:

[tex]v_{f} = v_{o} + g*t[/tex]

(Assuming the downward direction to be positive).

Solving for t, we get:

t₂ = 2.1 s

So the difference in time when both objects hit the ground, it's simply

Δt = t₂ - t₁ = 3.2 s - 2.1 s = 1.1 s


Related Questions

A 20m length wire 1.5mm in diameter has a resistance of 2.5 ohm what is the resistance of a 35m length of wire 3mm in diameter made of the same material?​

Answers

From the calculations, the resistance of the material is 1.1 ohm.

What is the resistance of the wire?

Given that;

R α l/A

R = ρl/A

R = resistance

l = length

A = Area

ρ = resistivity

Now;

A = πr^2

A = 3.142 * (1.5 * 10^-3/2)^2

A = 1.77 * 10^-6 m^2

ρ = RA/l

ρ =  2.5 *  1.77 * 10^-6 /20

ρ = 2.2 * 10^-7 ohm/m

Now;

R =ρl/A

A =  3.142 * (3  * 10^-3/2)^2

A = 7.1 * 10^-6 m^2

Thus

R = 2.2 * 10^-7 * 35/ 7.1 * 10^-6

R = 1.1 ohm

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Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus.
What is the strength of the nucleus' electric field at the orbital radius of the electrons?
What is the kinetic energy of an electron in a circular orbit around the gold nucleus?

Answers

Answer:

1. [tex] E = 1.14 \cdot 10^{13} N/C [/tex]

2. [tex]E_{k} = 9.1 \cdot 10^{-17} J[/tex]      

Explanation:

1. The strength of the nucleus' electric field (E):

[tex]E = \frac{kq}{r^{2}}[/tex]

Where:

k: is the Coulomb constant = 9x10⁹ Nm²/C²

q: is the proton charge = 1.6x10⁻¹⁹ C

r: is the radius = 10⁻¹⁰ m

[tex]E = \frac{kq}{r^{2}} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*79*1.6 \cdot 10^{-19} C}{(10^{-10} m)^{2}} = 1.14 \cdot 10^{13} N/C[/tex]

2. The kinetic energy (Ek) of an electron is the following:

[tex] E_{k} = \frac{1}{2}mv^{2} [/tex]    

Where:

m is the electron's mass = 9.1x10⁻³¹ kg

v: is the speed of the electron

We can find the speed of the electron by equaling the centripetal force (Fc) and the electrostatic force (Fe):

[tex] F_{c} = F_{e} [/tex]  

[tex] \frac{mv^{2}}{r} = \frac{kq^{2}}{r^{2}} = qE [/tex]

[tex] v^{2} = \frac{qEr}{m} = \frac{1.6 \cdot 10^{-19} C*1.14 \cdot 10^{13} N/C*10^{-10} m}{9.1 \cdot 10^{-31} kg} = 2.00 \cdot 10^{14} m^{2}/s^{2} [/tex]                  

Now, we can find the kinetic energy:

[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*2.00 \cdot 10^{14} m^{2}/s^{2} = 9.1 \cdot 10^{-17} J [/tex]    

I hope it helps you!

A race car accelerates from rest to a velocity of +90 m/s over a distance of 423m. Determine the acceleration of the race car.

Answers

Answer:

9.57m/s²

Explanation:

Given parameters:

Initial velocity  = 0m/s

Final velocity  = 90m/s

Distance  = 423m

Unknown:

Acceleration of the race car  = ?

Solution:

To solve this problem, we should apply one of the appropriate motion equations;

      V²  = U²   + 2as

Where V is the final velocity

           U is the initial velocity

           a is the acceleration

           s is the distance

  90²  = 0²  + 2 x a x 423

  8100 = 846a

      a  = 9.57m/s²

how do you Convert 50 g to kg in an equation for physics

Answers

Answer:

Divide by 1000

Explanation:

Suppose a signal from a vibrating motor is to be sampled discretely with a digital data acquisition system to determine the RPM of the motor. It is known that the maximum possible RPM is 1800. What is the minimum sampling frequency (in Hz) with which the signal can be sampled in order to measure the RPM of the motor

Answers

Answer:

f > 60 Hz

Explanation:

According to Nyquist's Sampling Theorem, to be fully reconstructed without any aliasing,  the signal must be sampled at least more than twice during the period of the maximum frequency of the signal.In this case, the signal to be sampled has only one frequency.However, as we have the information in RPM, we need to convert this to cycles/sec (Hz) first, as follows:

       [tex]f = \frac{1800 rev}{min} * \frac{1 min}{60 sec} = 30 Hz[/tex]

Per Nyquist, fs > 2*30 Hz

        ⇒ fs > 60 Hz

You measure the radius of a sphere as (6.45 ± 0.30) cm, and you measure its mass as (1.79 ± 0.08) kg. What is the density and uncertainty in the density of the sphere, in kilograms per cubic meter?

Answers

Answer:

[tex](1630.13\pm 300.10)\ kg/m^3[/tex]

Explanation:

Given that,

The radius of a sphere is (6.45 ± 0.30) cm

Mass of the sphere is (1.79 ± 0.08) kg

Density = mass/volume

For sphere,

[tex]d=\dfrac{m}{V}\\\\d=\dfrac{m}{\dfrac{4}{3}\pi r^3}\\\\d=\dfrac{1.79\ kg}{\dfrac{4}{3}\pi (6.4\times 10^{-2}\ m)^3}\\\\d=1630.13\ kg/m^3[/tex]

We can find the uncertainty in volume as follows :

[tex]\dfrac{\delta V}{V}=3\dfrac{\delta r}{r}\\\\=3\times \dfrac{0.3\times 10^{-2}}{6.45\times 10^{-2}}\\\\=0.1395[/tex]

Uncertainty in mass,

[tex]\dfrac{\delta m}{m}=\dfrac{0.08}{1.79}\\\\=0.0446[/tex]

Now, the uncertainty in density of sphere is given by :

[tex]\dfrac{\delta d}{d}=\dfrac{\delta m}{m}+\dfrac{\delta V}V}\\\\=0.0446+0.1395\\\\\dfrac{\delta d}{d}=0.1841\\\\\delta d=0.1841\times d\\\\\delta d=0.1841\times 1630.13\\\\\delta d = 300.10\ kg/m^3[/tex]

Hence, the density pf the sphere is [tex](1630.13\pm 300.10)\ kg/m^3[/tex]

What must be true if a wave's wavelength is short?
A
Its frequency is low.
B
It does not carry energy.
с
Its frequency is high.
D
The waves are visible.

Answers

C because it’s frequently is high

Answer:

im sure its A

Explanation:

A 849-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 35.0 m/s. What is the average force exerted on the car during this time

Answers

Answer:

The average force exerted on the car during this time is 5,943 N

Explanation:

Given;

mass of the car, m = 849 kg

initial velocity of the car, u = 0

time of motion of the car, t = 5.00 s

final velocity of the car, v = 35 m/s

The average force exerted on the car during this time is given by;

[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]

Therefore, the average force exerted on the car during this time is 5,943 N

Answer:

5943N

Let's say (+x) = eastward

Average horizontal acceleration

ax = vx -v0x/5.00s

= 35.0m/s-0/5.00s

= +7.09m/s

From here we apply the second law of newton

During this period average horizontal force acting on car

Summation x = max = (849kg)(+7.09m/s²)

= 5943N

+5.943x10³N

= 5.94kN east ward.

a man weighing 490 n on earth weighs only 81.7 n on the moon. His mass on the moon is__kg. (Use g=9.8 m/s2

Answers

Answer:

m = 50 [kg]

Explanation:

In order to solve this problem we must be clear about the difference between weight and mass. Weight is the product of mass by the acceleration of the planet or the star. While the mass is always preserved it never changes regardless of where it is located.

So for the earth we have:

g = gravity acceleration = 9.8 [m/s^2]

m = mass [kg]

W = weigth = 490 [N]

therefore the mass will be:

m = W/g

m = 490/9.8

m = 50 [kg]

Now it is important to remember that the mass will be the same on the moon or on the earth, but the weight will be different, because the gravity acceleration of the moon is different from the gravity acceleration on earth

So the gravity on the moon is equal to:

81.7 = 50 * gm

gm = 1.634 [m/s^2]

How long is a day in Neptune

Answers

Answer: the long day in neptune would be .18383562 years!

Explanation:also for every day is 16 hours

Which of the following is true about the following lever?

Answers

Answer:

It will rotate counter-clockwise.

Explanation:

The reason is that there is more Nm on the left side which will lift the lever towards the left side.

The lever would rotate counterclockwise as the torque on the left is higher, so, option B is correct.

What is torque?

The force which causes the object to rotate about any axis is called perpendicular distance.

Torque is a twisting or turning force that frequently results in rotation around an axis, which may be a fixed point or the center of mass. The ability of something rotating, such as a gear or a shaft, to overcome turning resistance is another way to think of torque.

Given:

The force on the left side, f = 40 N,

The force on the right side, F = 80 N,

The distance of the pivot from the left side, d = 9 m,

The  distance of the pivot from the right side, d = 3 m,

Calculate the torque on both side as shown below,

Torque on the left side = 40 × 9

Torque on the left side = 360 Nm

Torque on the right side = 80 × 3

Torque on the right side = 240 Nm,

Here, the torque on the left side is more,

Thus, the lever will rotate counterclockwise.

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Define reflection of sound?​

Answers

The reflection of sound is the movement of sound waves bouncing off of a surface and back into another direction, hope this helped :)

What is the force of a 12 kg object that is accelerating 6 m/s

Answers

We are given:

Mass of object (m) = 12 kg

acceleration (a) = 6 m/s²

Solving for the Force:

From newton's second law of motion:

F = ma

replacing the variables

F = 12*6

F = 72N

the equation of motion is given for a particle when s is in meters and t is in seconds. Find the acceleration after 4.5 seconds s

Answers

Answer:

Explanation:

The question is incomplete.

The equation of motion is given for a particle, where s is in meters and t is in seconds. Find the acceleration after 4.5 seconds.

s= sin2(pi)t

Acceleration = d²S/dt²

dS/dt = 2πcos2πt

d²S/dt² = -4π²sin2πt

A(t) = -4π²sin2πt

Next is to find acceleration after 4.5 seconds

A(4.5) = -4π²sin2π(4.5)

A(4.5) = -4π²sin9π

A(4.5) = -4π²sin1620

A(4.5) = -4π²(0)

A(4.5) = 0m/s²

a box with a Constance velocity has a 5 N of force applied to it from all sides and direction. what will happen to the motion of the box as result?
A-the object will come to rest
b-the velocity of the object will remain the same
c- the velocity of the object will decrease
d- the velocity of the object will increase

Answers

Answer:

  b-the velocity of the object will remain the same

Explanation:

Forces from opposite sides cancel each other, so there is no net force on the box that would affect its motion. The velocity of the box will remain unchanged.

__

(The box may be crushed, but it will continue in the same direction at the same speed.)

Answer:

b the velocity of the object will remain the same

Explanation:

use your brain:)

Light of wavelength 580 nm falls on a slit that is 3.70×10−3mm wide. Part A Estimate how far the first brightest diffraction fringe is from the strong central maximum if the screen is 10.0 m away.

Answers

Answer:

Explanation:

wavelength λ = 580 x 10⁻⁹ m

slit width d = 3.7 x 10⁻⁶ m

distance of screen D = 10 m

distance of first bright fringe =  1.5 x λ D / d

= 1.5 x 580 x 10⁻⁹  x 10 / 3.7 x 10⁻⁶

= 2351.34  x 10⁻³ m

= 2351.34 mm .

The feeling of weightlessness occurs because _____________________.

there is no supporting force under your mass.
there is no gravity present.
there is only a small amount of gravity present.

Answers

Answer:

there is only a small amount of gravity present.

Explanation:

this is because the only force acting upon your body during free fall is the force of gravity which is a non contact force.

There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?

Answers

Answer:

We are given:

V1 = 5.5L          T1 = -38 C   or   235 k

V2 = 1.3L           T2 = T

From the gas equation:

PV = nRT

Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:

V / T = k  (where k is a constant)

so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables

Changing the variables for the first case:

V1 / T1 = k   (where k is the same constant) ----------------(1)

Similarly,

V2 / T2 = k  (again, k has the same value)----------------(2)

From (1) and (2):

k is the common value

V1 / T1 = V2 / T2

Replacing the variables

5.5 / 235 = 1.3 / T

T = 1.3 * 235 / 5.5

T = 55.54 k

Therefore, at 55.54 K the gas will have a volume of 1.3L

Peter is running laps around a circular track with a diameter of 100 meters. If it takes Peter 12 minutes to run 4 laps, how quickly is he running (in meters per second)?

Answers

Answer:

v = 1.74 m/s

Explanation:

Given that,

Diameter of a circular track, d = 100 m

Distance covered for the 4 laps,

[tex]D=4\pi d\\\\D=4\pi \times 100\\\\D=1256.63\ m[/tex]

Time, t = 12 minutes = 720 s

We need to find the velocity of the peter. It can be calculated as follows :

[tex]v=\dfrac{D}{t}\\\\v=\dfrac{1256.63\ m}{720\ s}\\\\v=1.74\ m/s[/tex]

So, the speed is running with a velocity of 1.74 m/s.

Peter is running at 1.7453 m/sec.

Given to us,

Diameter of the circular track, D = 100 meters,

Number of laps Peter run, L = 4 laps,

Time taken by Peter, t = 12 minutes,

1 lap = circumference of the circle,

4 laps = 4 x circumference of the circle,

As we know, the circumference of a circle is given by πD.

So, 4 laps = 4 x circumference of the circle,

[tex]\begin{aligned}4 laps &= 4\times \pi \times D\\&= 4 \times \pi \times 100\\& = 1,256.6370\ meters\\\end{aligned}[/tex]

Also, we know that 1 minute has 60 sec.

so, 4 minutes = (4 x 60) seconds

Further, speed is given [tex]\bold{(\dfrac{Distance}{Time} )}[/tex]

Thus,

[tex]\begin{aligned}speed &= \dfrac{Distance\ coverd\ by\ Peter}{Time\ taken\ by\ Peter}\\&=\dfrac{1,256.6370}{12\times 60}\\&=1.7453\ m/sec \end{aligned}[/tex]

Hence, Peter is running at 1.7453 m/sec.

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How long does it take a P-wave to travel 7,000 km? ______ minutes b) How long does it take an S-wave to travel 7,000 km? ______ minutes c) How long does it take a Love wave to travel 7,000 km? ______ minutes d) How long does it take a Rayleigh wave to travel 7,000 km? ______minutes

Answers

Answer:

A. 8.64 secs.

B. 14.58 secs.

C. 26.002 secs.

D. 33.46secs.

Explanation:

A. P wave would travel 7000km

p-wave travels on a speed of 13.5km/s

= 7000km/13.5km/s

= 8.64 secs.

B. S-wave time to travel 7000km

s-wave travels on a speed of 8km/s

= 7000km/8km/s

= 14.58 secs.

C Love wave travels at a speed of 10,000m/s ( 2.7778 m/s ).

= 7000km to miles

= 4349.598m/2.788m/s

= 26.002 secs.

D. Rayleigh wave to travel 7,000 km

10,000m/s ( 2.1667 m/s ).

= 7000km to miles

= 4349.598m/2.1667m/s

= 33.46secs.

The acceleration of the spacecraft in which the Apollo astronauts took off from the moon was 3.4 m/s2 m / s 2 . On the moon, g g = 1.6 m/s2 m / s 2 . what's the apparent weight

Answers

Complete Question

The acceleration of the spacecraft in which the Apollo astronauts took off from the moon was 3.4 m/s2. On the moon, g = 1.6 m/s2.  What was the apparent weight of a 75 kg astronaut during takeoff?

Answer:

The value is  [tex]N = 375 \ N[/tex]

Explanation:

From the question we are told that

  The acceleration is  [tex]a = 3.4 \ m/s^2[/tex]

   The acceleration due to gravity in the moon is [tex]g = 1.6 m/s^2[/tex]

   The mass of the astronaut is  [tex]m = 75 \ kg[/tex]

   

Generally the apparent weight is mathematically represented as

      [tex]W = ma + mg[/tex]

=>    [tex]W = 3.4 * 75 + 1.6 * 75[/tex]

=>    [tex]W = 375 \ N[/tex]

A person drops a marble from the top of a skyscraper. After falling four floors the marble has gained a certain speed. How many more floors will the marble have to fall to triple this speed?

a. 8
b. 12
c. 32
d. 48

Answers

Answer:

B. 12

Explanation:

4 x 3 = 12

Calculate the ratio of the mechanical energy at B and mechanical energy at a (eb,ea) and (ec,ea). What do these ratios tell you about the conservation of energy?

A) is the mechanical conserved between a and b? explain

B) is the mechanical energy conserved between b and c ?explain

Answers

Answer:

Yes at A the mechanical energy is conserved.

Yes at B the part of mechanical energy is conserved potential energy and  kinetic energy and some is lost as frictional force.

Explanation:

Ratio = Eb/ Ea=  1058.3 J/2940 J= 0.3599

Ratio = Ec/ Eb= 0J/ 1058.3 J= 0

At point A the skater is at rest  or it is the starting point and the whole energy is due to the position of the  skater i.e= mgh = 50 *9.8*6=  2940 J

Since there's no movement there is no Kinetic energy = 0 J

Yes at A the mechanical energy is conserved.

At point B the skater has traveled for some of the distance . It has potential energy and  kinetic energy.

Yes at B the part of mechanical energy is conserved as potential energy and  kinetic energy.

The total Mechanical energy = 1058.3 J

At point B Total Mechanical energy = PE+ KE

1058.3J = 980 J + 78.3 J

1058.3 J = mgh + 1/2mv²

      = 50*2*9.8 + 1/2 *50*(8.85)²

       = 980 J + 78.3 J

As the total energy of the system must remain the same some of the mechanical energy is lost as frictional force at point B .

2940 J-1058.3 J= 1881.7

At Point C the skater has arrived at the end point and the height , speed, PE, KE and ME  all are zero.

(a) The ratio of the mechanical energy at B and mechanical energy at A is 0.36.

(b) The ratio of the mechanical energy at C and mechanical energy at A is 0.

(c) mechanical energy is conserved between a and b.

(d) mechanical energy is not conserved between b and c.

The given parameters;

mechanical energy at A, [tex]E_a = 2,940 \ J[/tex]mechanical energy at B, [tex]E_b =1,058.3 \ J[/tex]mechanical energy at C, [tex]E_c = 0[/tex]

The ratio of the mechanical energy at B and mechanical energy at A;

[tex]ratio = \frac{E_b}{E_a} = \frac{1058.3}{2940} = 0.36[/tex]

The ratio of the mechanical energy at C and mechanical energy at A;

[tex]ratio = \frac{E_c}{E_a} = \frac{0}{2940} = 0[/tex]

The change mechanical energy between A and B from the given position;

[tex]\Delta E = mg(h_b - h_a) - \frac{1}{2}m(v_b^2 - v_a^2)\\\\ \Delta E = 50\times 9.8(2-6) \ - \ \frac{1}{2} \times 50(8.85^2 - 0)\\\\\Delta E =- 1960 + 1960\\\\\Delta E = 0 \ J[/tex]

Thus, we can conclude that mechanical energy is conserved between a and b.

The change mechanical energy between A and B from the given position;

[tex]\Delta E = mg(h_c - h_b) - \frac{1}{2}m(v_c^2 - v_b^2)\\\\ \Delta E = 50\times 9.8(0-2) \ - \ \frac{1}{2} \times 50(0^2 - 8.85^2)\\\\\Delta E = -980 + 1960 \\\\\Delta E = 980 \ J[/tex]

Thus, we can conclude that mechanical energy is not conserved between b and c.

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I need help with this please

Answers

The answer is Rubber

an air filled parallel plate capacitor has a capacitor of 1.3 pico farad. the separation of the plates is doubled and wax is inserted between them, the new capacitance is 2.6 pico farad. find the dielectric constant of the wax​

Answers

you get 3.9 because i used to have this question

What is Triangle ill send a picture​

Answers

A triangle is just a shape...
It’s a shape lol
.......

A tennis ball is dropped from a roof 16 meters from the ground. How long does it take
for the ball to reach the ground?

Answers

3.3 seconds would be the right answer

The time taken for the tennis ball dropped from the roof-top to reach the ground level is 1.8 seconds.

What is Motion?

Motion is simply the change in position of an object or particle over time.

From the Second Equation of Motion;

s = ut + (1/2)gt²

Where s is the distance from ground level, u is initial velocity, t is time elapsed and g is acceleration due to gravity ( g = 9.8m/s² ).

Given the data in the question;

Since the ball was initially at rest before it was dropped.

Initial velocity u = 0Height or distance from gound level s = 16mTime taken to reach the gound t = ?

We substitute our values into the expression above.

s = ut + (1/2)gt²

16m = ( 0 × t ) + ( (1/2) × 9.8m/s² × t² )

16m = 0.5 × 9.8m/s² × t²

16m = 4.9m/s² × t²

t² = 16m / 4.9m/s²

t² = 3.2653s²

t = √(3.2653s²)

t = 1.8s

Therefore the time taken for the tennis ball dropped from the roof-top to reach the ground level is 1.8 seconds.

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The car alarm on a stationary car emits sound waves with a frequency of 450 Hz. If you are moving away from the stationary car at 20 m/s, what alarm frequency do you perceive

Answers

Answer:

Explanation:

The frequency of wave is directly proportional to velocity

f = kV

k = f/V

f1/V1 = f2/V2

Given

f1 = 450Hz

V1 = 343m/s

f2 = ?

V2 = 20m/s

Substitute into the formula

450/343 = f2/20

Cross multiply

343f2 = 450×20

343f2 = 9000

f2 = 9000/343

f2 = 26.24Hz

1. Which statement is true of culture?
O Culture do not change or evolve
O Culture refers to the degree in which resources are used to address for problems.
O Culture has little affect on a persons life
Food can reflect culture

Answers

I think it is either A or B. I’m mostly leaning towards B.

In each of the given pairs, choose which element will have the bigger atom. Give reasons for your choice. (a) Mg (atomic number 12) or Cl (atomic number 17) (b) Na (atomic number 11) or K (atomic number 19)

Answers

Answer:

Mg (atomic number 12)

K (atomic number 19)

Explanation:

The size of an atom is estimated in terms of its atomic radius.

The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance (d) between two nuclei in the solid - state of metals.

Across a period, atomic radii decrease progressively from left to right. This is due to the progressive increase in the nuclear charge without increase in the number of electronic shells. Down a group, atomic radii increase progressively due to the successive shells of electrons being added which have been compensated for by the increase in nuclear charge.

Cl is further right of Mg in the third period

K is below Na in the first group

Other Questions
What is the rule of law? How does the rule of law apply in your daily life? can someone give me an introduction paragraph for teen angst pleasseeeee I need it for today!! will mark brainiest :) the designating factors of a climate include How is the custom of shaking hands different in France? A runner sprinted for 117 yards (yd). How many feet (ft) is this? First fill in the blank on the left side of the equation using one of the ratios. Then write your answer on the right side of the equation. ft mi yd 1760 yd 1 mi 3 ft 1 ft 12 in 1 yd Ratios: 1 mi 1760 yd 1 yd 3 ft 12 in 1 ft . in 117 yd 1 ft 0 x IK IT LOOKS CONFUSING SO LOOK AT THE PICTURE HELP ME PLEASE!!!!:) Why was the resolution of the federal governments fiscal problems so critical to the success of the new nation? 1. Price: $40.00, 10% tax, 15% tip PART A: Which statement identifies the central idea of the text?A)If it werent for Francis Pizarros, the Inca Empire would have continued to expand and prosper.B)Despite the riches they possessed, the Incas lived simple lives.C)The Incas were a highly advanced civilization that were able to expand their net of influence far and wide.D)Warfare was the most common and successful way in which the Incas were able to expand their empire.PART B: Which detail from the text best supports the answer to Part A?A)"The Inca controlled perhaps 10 million people, speaking a hundred different tongues. It was the largest empire on earth at the time."B)"Members of a community supported one another. The Inca tailored this practice of reciprocity give-and-take to their own needs."C)"All were required to pay taxes as tribute. The Mita was a labor tax, which required the head of every household to work for the state fo 19. (01.03 MC)Mountain ranges most affected early human tribal groups by.......(A) providing them with meeting places(B) keeping them separate most of the time(C) stopping the flow of water from higher elevations(D) serving as places to build cities Which must be present in order for factoring by grouping to apply? - The ratios of the first two terms and the second two terms must be positive. - The ratio of the first two terms must be equivalent to the ratio of the last two terms. - There must be at least three terms. - The first and the last terms must be perfect squares. I NEED HELP PLEASE, ITS DUE TODAY The movement of matter across a cell membrane against a concentration difference is called active transport , and it requires? Which statement describes a characteristic of speaking as compared towriting?A. Speaking is more formal than writing.B. Speaking is more lasting than writing.C. Speaking is more organized than writing.OdD. Speaking is more immediate than writing. QuestionWhat is the unit of wavelength?AhertzB.metremetre per secondDsecondBD Write a brief summary of what Confucianism is.Write a brief summary of what Daoism is.Write a brief summary of what legalism is. What did the Pendleton Act do answer quick 50 points A car is driving down the road at 30 MPH. A bug hits the windshield with a force of 50N. With what force does the car hit the bug?Group of answer choices80N20N50N300N What is a constraint in math? For what situations would they be useful? Idk how to do this soooo helppp