A single covalent bond is stronger than a single hydrogen bond so why does a group of polar molecules tend to have a higher boiling point than a group of non polar molecules

The fastest time trial for the first quarter checkpoint is 2.02 s.

The slowest time trial for the first quarter checkpoint is 2.7 s.

The range of the times measured for the checkpoint is 0.68 s.

The given parameters;

The fastest time trial for the first quarter checkpoint is the least measured time value.

fastest time trial = least time measured

fastest time trial =  2.02 s

The slowest time trial for the first quarter checkpoint is the highest measured time value.

slowest time trial = 2.7 s

The range of the times measured for the checkpoint is difference between the fastest time and slowest time.

Range = fastest time - slowest time

Range = 2.7 - 2.02

Range = 0.68 s

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Answer:

2.02, 2.15, 0.13

Explanation:

A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2.15, 2.05, 2.02, 2.07.

Use the table to answer the questions.

What is the fastest time trial for the first quarter checkpoint?

2.02

seconds

What is the slowest time trial for the first quarter checkpoint?

2.15 seconds

What is the range of times measured for this checkpoint?

0.13 seconds

Answer: A

Explanation: An object in free fall only is affected by gravity, so the acceleration is 9.8 m/s²

Explanation:

[tex] \rule{999pt}{66646pt}[/tex]

Answer:

Approximately [tex]4.18 \times 10^{16}\; \rm N[/tex].

Explanation:

Consider two objects of mass [tex]m_{1}[/tex] and [tex]m_{2}[/tex]. Let [tex]r[/tex] denote the distance between the center of mass of each object. Let [tex]G[/tex] denote the gravitational constant. ([tex]G \approx 6.67 \times 10^{-11}\; {\rm m^{3}\cdot kg^{-1}\cdot s^{-2}}[/tex].)

By Newton's Law of Universal Gravitation, the size of gravitational attraction between these two objects would be:

[tex]\begin{aligned}F &= \frac{G\, m_{1}\, m_{2}}{r^{2}}\end{aligned}[/tex].

In this question, [tex]m_{1} = 5.98\times 10^{24}\; {\rm kg}[/tex] and [tex]m_{2} = 6.24 \times 10^{23}\; {\rm kg}[/tex] are the mass of the two planets. The distance between the two planets is [tex]r = 7.83 \times 10^{10}\; \rm m[/tex] (approximately the same as the distance between the center of mass of planet Earth and the center of mass of Mars.)

Apply Newton's Law of Universal Gravitation to find the size of gravitational attraction between the two planets:

[tex]\begin{aligned}F &= \frac{G\, m_{1}\, m_{2}}{r^{2}} \\ &= \frac{1}{(7.83 \times 10^{10}\; {\rm m})^{2}} \\ &\quad \times (6.67 \times 10^{11}\; {\rm m^{3} \cdot kg^{-1} \cdot s^{-2}}) \\ &\quad \times (5.98 \times 10^{24}\; {\rm kg}) \\ &\quad \times (6.42 \times 10^{23}\; {\rm kg}) \\ &\approx 4.18 \times 10^{16}\; {\rm kg \cdot m \cdot s^{-2}} \end{aligned}[/tex].

Since  [tex]1\; {\rm kg \cdot m \cdot s^{-2}} = 1\; {\rm N}[/tex], the size of gravitational attraction between the two planets would be approximately [tex]4.18 \times 10^{16}\; {\rm N}[/tex].

Answer:

It can rotate in any direction. The sign of the angular acceleration depends on how you set the reference system, it can be both negative or positive.

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# be careful#

The resultant velocity of  the boat is equal to 2 m/s due south.

Given the following data:

To determine the resultant velocity of  the boat:

Resultant velocity can be defined as the sum (addition) of each of the vector velocity acting on a physical or an object. Thus, it is simply a combination of two or more single vector velocity.

Note: When the velocities are acting in the same direction, you will add them up while you will subtract when the velocities are acting in opposite directions.

In this scenario, the velocities are acting in opposite directions. Thus, we would subtract as follows:

[tex]V = 3 - 1[/tex]

V = 2 m/s due south.

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Answer:

C because it make sens

C the light wave traveled through ice and then through a Dimond.

Answer:

60n would be the needed force

The force needed to push a box weighing 30 N up a ramp that has an ideal mechanical advantage of 3 is equal to 10 N.

The mechanical advantage can be described as the ratio of the input force to the output force. The mechanical advantage of any machine can be determined by the ratio of the forces involved to do the work.

The ratio of the resistance force to the effort is called the actual mechanical advantage which will be comparatively less. The efficiency of a machine is always determined by equating the ratio of its output to its input.

The efficiency of the machine is equal to the ratio of the actual mechanical advantage (M.A.) and theoretical mechanical advantage. Mechanical advantage can be defined as the force produced by a machine to the force applied to it.

Given the load = 30 N and the ideal mechanical advantage = 3

Mechanical advantage = Load/ Effort

Input force or effort = Load/ M.A.

Force = 30/3

Input Force = 10 N

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the compare software

hope it's help

Answer:

you ask or answer?

Explanation:

thanks for point

The magnitude of the magnetic force acting on the charge which moves to the right is 0 N.

Given to us,

the charge [tex]q[/tex] =  [tex]5\times 10^{-7}[/tex] C,

the velocity [tex]v[/tex] = [tex]2.6\times 10^5[/tex]  m/sec,

the magnetic field [tex]B[/tex] = [tex]10^{-2}[/tex] T,

angle between the direction of v and B [tex]\theta[/tex] = 0,

Magnetic force is as important as the electrostatic or Coulomb force. The magnitude of the magnetic force F on a charge q moving at a velocity of v in a magnetic field of strength B is given by

[tex]\begin{aligned}F&=qvB\ sin\Theta\\&= 5\times10^{-7}\times2.6\times10^5\times10^{-2} \times sin(0)\\&= 0\ N\\\end{aligned}[/tex]

Hence, the magnitude of the magnetic force acting on the charge which moves to the right is 0 N

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Answer:

The mass of the object is 11.2 Kg.

Hope you could get an idea from here.

Doubt clarification - use comment section.

The mass of an object that has a weight of 110 N is 11.2kg.

The weight of an object is defined as the force that acts on the object due to gravity. It is a vector quantity in some cases such as when the force of gravity acts on an object but it is also a scalar quantity where the force of gravity has a magnitude.

Weight is the gravitational force of attraction on an object due to the presence of another heavy object, such as the Earth or the Moon. It is expressed by multiplying the mass by the gravitational acceleration.

In above case,

Weight (W) = 110 N

Acceleration due to gravity (g) = 9.8 m/s^2

Let the mass of the object be m.

By using the formula,

W = mg,

[tex]110 N = 9.8 m/s^2 *m\\m = 110 N/ 9.8 m/s^2[/tex]

so, m = 11.2 Kg

Thus, the mass of an object that has a weight of 110 N is 11.2kg.

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Answer:

not able to understand the question

Answer:

Approximately [tex]6.59\times 10^{3}\; \rm m\cdot s^{-1}[/tex].

Explanation:

Look up the radius of the earth: approximately [tex]6.371 \times 10^{3}\; \rm km[/tex].

The radius of the orbit of this satellite is the of the radius of the earth (at ground level) plus the height of the satellite relative to ground level:

[tex]\begin{aligned}r &\approx 6.371 \times 10^{3}\; {\rm km} + 550\; {\rm km} \\ &= 6.921 \times 10^{3}\; \rm km\end{aligned}[/tex].

Convert the units of both distance and time to standard units:

Orbital radius:

[tex]\begin{aligned}r &\approx 6.921 \times 10^{3}\; {\rm km} \\ &= 6.921 \times 10^{3}\; {\rm km} \times \frac{10^{3}\; \rm m}{1\; \rm km} \\ &= 6.921 \times 10^{6}\; \rm m\end{aligned}[/tex].

Orbital period:

[tex]\begin{aligned}t &= 110\; \text{minute} \\ &= 110\; \text{minute} \times \frac{60\; \text{second}}{1\; \text{minute}} \\ &= 6.6 \times 10^{3}\; \text{second}\end{aligned}[/tex].

Calculate the circumference of this orbit:

[tex]\begin{aligned}& 2\, \pi\, r \\ \approx\; & 2 \, \pi \times 6.921 \times 10^{6}\; {\rm m} \\ \approx\; & 4.35 \times 10^{7}\; \rm m\end{aligned}[/tex].

Calculate the orbital speed (tangential) of this satellite:

[tex]\begin{aligned}v &= \frac{2\, \pi\, r}{t} \\ &\approx \frac{4.35 \times 10^{7}\; \rm m}{6.6 \times 10^{3}\; \rm s} \\ &\approx 6.6 \times 10^{3}\; \rm m \cdot s^{-1}\end{aligned}[/tex].

Answer:

Scientific investigation is a quest to find the answer to a question using the scientific method.

Explanation:

the scientific method is a systematic process that involves using measurable observations to formulate, test or modify a hypothesis.

Answer:

functionalist theory,feminist theory. discipline of sociology

Answer:

E.visible lights

Explanation:

hope its attachments

Answer:

3m

Explanation:

I can't make a drawing right know, but it's all around vectors. First of all the ball travels down 5 meters, so we have a vector pointing down, then it is reflected and travels 2 meters up. You have to sum the vectors so you obtain that the distance traveled is: 5-2 meters = 3m

Answer:

1.1x10¹⁶ kg

Explanation:

Let m =  mass of asteroid y.

Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.

Given:

F = 6.2x10⁸ N

d = 2100 km = 2.1x10⁶ m

Note that

G = 6.67408x10⁻¹¹ m³/(kg-s²)

The gravitational force between the asteroids is

F = (G*m*(m/3))/d² = (Gm²)/(3d²)

or

m² = (3Fd²)/G

    = [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²))

   = 1.229x10³² kg²

m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)

The final temperature of water is equal to 50.9999°C

Given the following data:

To determine the final temperature of water:

Mathematically, quantity of heat is given by the formula;

[tex]Q=mc\theta[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]5000=2.38892 \times 10^{-2}\times 4186 \times \theta\\\\5000=100.0001912 \theta\\\\ \theta=\frac{5000}{100.0001912} \\\\ \theta=49.9999^{\circ}C[/tex]

For the final temperature:

[tex]\theta = T_2 - T_1\\\\T_2 = \theta+T_1\\\\T_2 = 49.9999 + 50[/tex]

Final temperature = 50.9999°C

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By Newton's second law, the net force acting on the block in the vertical direction is

∑ F [ver] = n - mg = 0

where n = magnitude of normal force and mg = weight of the block. It follows that n = mg.

When the block is at rest, the applied force X will not be enough to move the box until it can overcome the maximum mag. of static friction. If µ[s] is the coefficient of static friction, then the maximum mag. of the frictional force is

f = µ[s] n = µ[s] mg

The net horizontal force would be

∑ F [hor] = X - µ[s] mg = 0

so a minimum force of X = µ[s] mg is required to get the block moving. Any mag. smaller than this and the block stays at rest/in equilibrium.

Once the mag. of X exceeds µ[s] mg, the block will begin to move. At that point, if the coefficient of kinetic friction is µ[k], then the net force on the block is

∑ F [hor] = X - µ[k] mg = 0

so a minimum force of X = µ[k] mg would be needed to keep the block moving at constant speed, or otherwise X = µ[k] mg + ma if the block is accelerating with mag. a.

The principles here are captured in the attached plot.

Answer:

Density of the mixture = 855.56kgm-3

Explanation:

Density = Mass / Volume

Volume of Liquid X = 0.5m³

Density of Liquid X = 900kgm-3

Mass of Liquid X = Density × Volume

= 900kgm-3 × 0.5m³ = 450kg

Volume of Liquid Y = 0.4m³

Density of Liquid Y = 800kgm-3

Mass of Liquid Y = Density × Volume

= 800kgm-3 × 0.4m³= 320kg

As X and Y are mixed, we add their masses and volumes together:

Mass = 770kg

Volume = 0.9m³

Now we can find the density of the mixture:

Density = 770kg / 0.9m³ = 855.56kgm-3 (rounded to the 2nd decimal)

Answer:

Well, I think you're talking about kinematics, especially uniform rectilinear motion. We know that there is a specific equation for that:

S = Vt + S0

With S being the distance, V the velocity, t the time and S0 the initial distance (initial displacement).

From this you can calculate t, if that's what you want.

Answer:

The force of the car is 15000N.

Explanation:

The unit of force is Newtons (N), so based on the question, the force is 15000 Newtons.

Answer:

[tex]\huge\boxed{\sf S = 18.75 \ meters}[/tex]

Explanation:

Given Data:

Initial Velocity = Vi = 0 m/s (rest)

Final Velocity for 6 seconds = Vf = 25 m/s

Time (1) = T1 = 6 seconds

Time (2) = T2 = 3 seconds

Required:

Distance for 3 seconds = S = ?

Solution:

For 6 seconds, the acceleration will be:

[tex]\displaystyle a = \frac{Vf-Vi}{t} \\\\a = \frac{25 - 0}{6} \\\\a = 25 / 6\\\\\boxed{a = 4.167 \ m/s^2}[/tex]

Since, acceleration is constant, it will be the same at 3 seconds as well.

Using second equation of motion to find Distance (S) with time being 3 seconds:

[tex]\displaystyle S= Vit+\frac{1}{2} at^2\\\\S = (0)(3)+ \frac{1}{2} (4.167)(3)^2\\\\S = \frac{1}{2} (4.167)(9)\\\\S = \frac{37.5}{2} \\\\\boxed{S = 18.75 \ meters}\\\\\rule[225]{225}{2}[/tex]

Hope this helped!

Answer:

they are machined with shape characteristics which maximize the fatigue life of a metal.

Explanation:

they are highly polished to provide the surface characteristics which enable the best fatigue life.