What is the net force (magnitude and direction) of the system below?

What Is The Net Force (magnitude And Direction) Of The System Below?

Answers

Answer 1

The net force (magnitude and direction) of the system given in the question is 40 N horizontal to the right

How to determine the net force

Case 1 (Net force between up and downward force)

Force up (Fu) = 50 NForce down (Fd) = 30 NNet force 1 (F1) = ?

F1 = Fu - Fd

F1 = 50 - 30

F1 = 20 N up

Case 2 (Net force between right and left)

Force right (Fr) = 60 NForce left (Fl) = 20 NNet force 2 (F2) = ?

F2 = Fr - Fl

F2 = 60 - 20

F2 = 40 N right

SUMMARY

Net force between up and down = 20 N upNet force between right and left = 40 N right

From the above, the net force between right and left (i.e 40 N) is greater than the net force between up and down (i.e 20 N)

Thus, the net force of the system will be 40 N horizontal to the right

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Related Questions

A uniform stick has a mass of 4.42 kg and a length of 1.23 m. It is initially lying flat at rest on a frictionless horizontal surface and is struck perpendicularly by a puck imparting a horizontal impulsive force of impulse 12.8 N s at a distance of 46.4 cm from the center. Determine the subsequent motion of the stick.

Answers

10.7 rad/s is the final angular velocity of the stick.

Given:

Mass of the stick = 4.42 kg

Length of the stick = 1.23m

Force of impulse (I) = 12.8 N s

The linear velocity of the stick, [tex]v=\frac{I}{m}[/tex]

                                                  [tex]v=\frac{12.8 N.s (\frac{1 kg m/s^2}{1 N}) }{4.42 kg}[/tex]

                                                  [tex]v[/tex]  [tex]= 2.89 m/s[/tex]

Therefore, the final linear velocity of the stick is 2.89 m/s

∴[tex]w=\frac{12 Ir}{ml^{2} }[/tex]

[tex]w=\frac{12 ( 12.8 N.s ) ( 46.4 cm)}{(4.42 kg) (1.23 m)^2}[/tex]

[tex]w= \frac{12 (12.8 N.s) (46.4 cm) (\frac{10^-^2 m}{1 cm} )}{(4.42 kg) (1.23m)2}[/tex]

[tex]w=10.7 rad/s[/tex]

Therefore, 10.7 rad/s is the final angular velocity of the stick.

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A light bulb receives 100 J of energy electrically. It transfers 30 J of energy to the surroundings by heat. What is the efficiency of the light bulb?

Answers

Answer:

The efficiency of the light bulb is 30%.

Explanation:

Efficiency = output/input × 100%

Data

Energy input = 100J

Energy output = 30J

Efficiency = ?

Efficiency = Energy output/Energy input × 100%

Efficiency = 30J/100J × 100%

Efficiency = 0.3 × 100%

Efficiency = 30%

Therefore the efficiency of the light bulb is 30%.

A type of electromagnetic radiation that has a higher frequency than visible light is Question 15 options: ultraviolet infrared radiowaves microwaves two of the above

Answers

Ultraviolet rays are Electromagnetic waves that have higher frequencies than visible light, so they carry more energy.

What are Ultraviolet rays?Johann Wilhelm Ritter, a German physicist, first noticed that paper soaked in silver chloride browned more quickly when exposed to Ultraviolet rays than when it was exposed to violet light itself in 1801.For the most part, humans cannot see ultraviolet rays. The cornea filters out shorter wavelengths, while the lens of the human eye primarily inhibits radiation in the 300–400 nm range. Additionally, UV color receptor adaptations are not present in humans. The retina's photoreceptors are sensitive to near-UV, though, and those without lenses (a condition known as aphakia) see near-UV as whitish-blue or whitish-violet.The body produces vitamin D, which is necessary for life when exposed to UV light, more especially UVB. UV rays are necessary for humans to maintain proper vitamin levels.

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which two forms of energy do burning substances produce

Answers

Answer:thermal (heat) energy and radiant (light) energy.

Explanation:

Answer:

Answer: We use the chemical energy in fuels by burning them and transforming them into other types of energy: thermal energy, as when we burn fuel for heat; and kinetic energy, as when we burn gasoline to power our car's motion

Explanation:

hope it will help you

What is weight??????????????????

Answers

Answer:

Weight is a body's relative mass, or the quantity of matter contained by it.

Explanation:

Hope this helps

Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero

Answers

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of  Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1​

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

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If the distance between two charged particles is increased 2.69 times, the ratio of new to old electric force is _______ times the old value. (Round your answer to two decimal places).

Answers

Answer:

F = K Q1 Q2 / R^2       force between 2 charged partices

F2 / F1 = (R1 / R2)^2 = (1 / 2.69)^2 = .139

F2 = .139 F1

An object on Earth weighs 150 N. What is its mass?
Group of answer choices

15 g

150 kg

15 kg

150 g

Answers

Answer:

[tex] \tt \: mass \: = \frac{weight}{acceleration \: due \: to \: gravity} [/tex]

[tex] \longrightarrow \tt \: \frac{150}{10} [/tex]

[tex] \longrightarrow \boxed{ \tt{15 \: kg}}[/tex]

Our final answer is 15 kg .

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A toy car starts from the rest and accelerates at 1.50m/s2 [E] for 2.25s. What is the final velocity, of the car

Answers

3.375m/s is the final velocity of the car.

How do you find final velocity?

The final velocity depends on how large the acceleration is and the distance over which it acts.

Initial velocity of an object, you can multiply the acceleration due to a force by the time the force is applied and add it to the initial velocity to get the final velocity.

According to the question,

A toy car starts from the rest and accelerates

So the acceleration = 1.50m/s²

Time =  2.25s

[tex]x=x_{0} + vt[/tex]

[tex]x = 0 + ( 1.50m/s^2*2.25s)[/tex]

[tex]x = 3.375m/s[/tex]

The final velocity, of the car is 3.375 m/s.

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Question 17
The driver of a car traveling along a straight road with a
speed of 72KM ph observes a signboard which give the
speed limit to be 54KM ph. The signboard is 70m ahead
when the driver applies the brakes, calculate the
acceleration of the car which will cause the car to pass the
signboard at the stated speed limit?

Answers

So, the acceleration of the car is -1.25 m/s². In other word, the car is also decelerating by 1.25 m/s².

Introduction

Hi ! I will help you to discuss about "deceleration in a straight line movement". Please note in advance that deceleration is acceleration which has a negative value. When an object decelerates, the object will continue to move until it reaches a certain speed (which is less than before) or until it stops. The higher the deceleration value, an object that is moving will stop faster and cover a shorter distance.

Formula Used

In this opportunity, I will give you the following equation to express the relationship between final velocity and initial velocity, acceleration, and distance.

[tex] \boxed{\sf{\bold{(v_t)^2= (v_0)^2 + 2 \times a \times s}}}[/tex]

With the following condition:

[tex] \sf{v_t} [/tex] = final velocity of an object (m/s)[tex] \sf{v_0} [/tex] = initial velocity of an object (m/s)a = acceleration that happen (m/s²)s = the shift or distance of the object (m)

Problem Solving

We know that:

[tex] \sf{v_0} [/tex] = initial velocity of an object = 72 km/h = 20 m/s[tex] \sf{v_t} [/tex] = final velocity of an object = 54 km/h = 15 m/ss = the shift or distance of the object = 70 m

Note :

1 m/s = 3.6 km/h. So 10 m/s = 36 km/h

What was asked ?

a = acceleration that happen = ... m/s²

Step by step :

[tex] \sf{(v_t)^2 = (v_0)^2 + 2 \times a \times s} [/tex]

[tex] \sf{15^2 = 20^2 + 2 \times a \times 70} [/tex]

[tex] \sf{225 = 400 + 140 \times a} [/tex]

[tex] \sf{140 a = -175} [/tex]

[tex] \sf{a = \frac{-175}{140}} [/tex]

[tex] \boxed{\sf{\bold{a = -1.25 \: m/s^2}}} [/tex]

Conclusion

Here, we see that the acceleration is -1.25 m/s². In other words, the car is also decelerating by 1.25 m/s².


Calculate the time needed for a 0.600 kg hammer to reach the surface of the Earth
if dropped from 10.0 m on Earth and on the Moon. The gravitational strength on
the Moon is 1.6 N/kg.

Answers

The time needed for the hammer to reach the surface of the Earth is 3.54 s.

Time of motion of the hammer

The time of motion is calculated as follows;

t = √(2h/g)

where;

h is height of fallg is acceleration due to gravity

t = √(2 x 10 / 1.6)

t = 3.54 s

Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.

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Assume the first dot for animal A represents t = 0. At which time does animal B overtake animal A?

Answers

The time in which the animal B overtake animal A according to the motion map is at 2 seconds.

What is a Motion map?

This is referred to as one-dimensional plots built off of a position line and depicts velocity and acceleration.

The time in which Animal B is ahead of animal A is at 2 seconds which therefore makes it the most appropriate choice.

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A box weighing 660 N is sliding across a cement floor and the coefficient of sliding
friction between the box and the floor is 0.15. If the force pushing the box is 500 N, what
is the acceleration of the box?
O 6 m/s²
O 2 m/s²
O 10 m/s²
pts
O 9 m/s²

Answers

The acceleration of the box is 6 m/s²; option A.

What is the acceleration of the box?

The acceleration of the box is determined as follows:

Frictional force = 660 * 0.15 = 99 N

Net force = 500 - 99 = 401 N

Force = mass * acceleration

Mass of the box = 660/9.8 = 67.3 kg

Acceleration = net force/mass

Acceleration = 401/67.3

Acceleration = 6 m/s²

In conclusion, the acceleration of the box is determined from the net force and the mass of the box.

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g Three projectiles are launched from the same level on flat ground. Each reaches the same maximum height before returning to the ground, but projectile 3 (in blue) has the largest range, followed by projectile 2 (in green). Projectile 1 has the smallest range (in red). Rank the three from largest to smallest according to:

Answers

the correct answer would be projected largest range because projectile has the smallest one

The constant G in Newton's equation _______. View Available Hint(s)for Part A The constant G in Newton's equation _______. was measured by Newton shows gravity to be a relatively huge force produces equilibrium makes the units of measurement consistent

Answers

The constant G in Newton's equation makes the units of measurement consistent was measured by Newton shows gravity to be a relatively huge force produces equilibrium makes the units of measurement consistent.

Newton's law:

The three fundamental laws of classical mechanics known as Newton's laws of motion describe how an object's motion and the forces acting on it interact.

Newton's First lawNewton's Second lawNewton's Third law

Newton's first law, an object will not change its motion unless a force acts on it.

Newton's second law, the force on an object is equal to its mass times its acceleration.

Newton's third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

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1. A large ball was let go on a hill and started rolling down with a constant acceleration of 4.2 m/s². What was the velocity of the ball after 12 s?​

Answers

Answer:

The velocity after 12s is 50.4m/s.

Explanation:

In acceleration formula make velocity the subject.

acceleration(a) = velocity(v)÷time(t)

velocity (v) = acceleration(a)×time(t)

V = 4.2m/×12s

V = 50.4m/s

Therefore the velocity after 12s is 50.4m/s.

Verification of laws of reflection of sound (Record Work)

Answers

The Angle of Incidence is equal to the Angle of Reflection.

What is the Verification of laws of reflection of sound?

Take a wooden drawing board and fix a white sheet of paper on it. In the middle of paper draw a straight line and Mark a point B on it. Draw a perpendicular . Place a mirror on line such that the flat side of the mirror is along the line. Hold the mirror in the mirror holder.

Fix two steel pins P and Q on the straight line AB at least 10 cm apart. Look for the images of the pins P and Q and fix two pins P a such that P', Q', and images of P and Q are all in the same straight line. Remove the pins and draw small circles around the pinpricks.

Remove the mirror and Join P'Q' which produce the straight line to meet at B. Measure ∠ABN = i and ∠CBN = r. It is found that ∠i=∠r. This proves that the Angle of Incidence is equal to the Angle of Reflection.

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A substance that has 1500 liters and which has a density of 250 kg / m3 determine the mass of said substance

Answers

【Answer】Therefore, the mass in said substance is 375 kilograms.

                   

                               [tex]\red{ {\hspace{50 pt}\above 1.2pt}\boldsymbol{\mathsf{Procedure}}{\hspace{50pt}\above 1.2pt}}[/tex]

This is an exercise on fluids and their fundamental characteristics.

We start to solve, obtaining the data:

DATA:v = 1500 lt = 1.5 m³d = 250kg/m³m = ?

                                   Conversion from liters to m³

                                      [tex]\boldsymbol{1500\not{l}*\dfrac{1 \ m^{3} }{1000\not{l} }=1.5 \ m^{3} }[/tex]

To calculate mass: multiply density by volume.

                                       [tex]\boldsymbol{m=d*v \ \ \to \ \ \ Formula}[/tex]

We clear our data in the formula:

                                       [tex]\boldsymbol{m=250\dfrac{kg}{\not{m^{3}}}*1.5\not{m^{3}} }[/tex]

                                        [tex]\boldsymbol{m=375 \ kg}[/tex]

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. A microwave oven operates at 2.4GHz with an intensity inside the oven of 2500 W/m^2. What are the amplitudes of the oscillating electric and magnetic fields

Answers

Answer: A microwave oven operates at 2.4GHz with an intensity inside the oven of 2500 W/m^2. Then, the amplitudes of the oscillating electric and magnetic fields are, [tex]B_0=4.57*10^{-6} T and E_0=1.37*10^{3} V/m[/tex]

Explanation: To find the correct answer, we have to know more about the EM Waves.

What are EM waves?Electromagnetic waves can be defined as the waves, that can interact with the matter and are created as a result of vibrations of electric and magnetic fields.Maxwell theoretically predicted the existence of electromagnetic waves.A time varying magnetic field is a source of electric field.Similarly, an oscillating electric field gives rise to an oscillating magnetic field.The amplitudes of electric and magnetic fields can relate as,

                          [tex]E_0=cB_0[/tex]

The expression for intensity of electromagnetic waves in terms of amplitude of magnetic field can be written as,

                        [tex]I=\frac{1}{2u_0} cB_0^{2}[/tex]

How to solve the problem?From the above expression of intensity, we can find the expression for amplitude of magnetic field as,

                          [tex]B_0=\sqrt{\frac{2u_0I}{c} }=\sqrt{\frac{2*4*3.14*10^{-7}*2500 }{3*10^{8} } } =4.57*10^{-6} T[/tex]

Thus, from the amplitude of magnetic field, we get, the amplitude of electric field as,

                              [tex]E_0=cB_0=3*10^{8} *4.57*10^{-6} =1371 V/m[/tex]

Thus, from the above calculations, we get the amplitudes of the oscillating electric and magnetic fields are, [tex]B_0=4.57*10^{-6} T and E_0=1.37*10^{3} V/m[/tex]

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25 points please help PLEASE
Journal prompt to be answered in 2 fully developed paragraphs
Prompt: What are some products (or programs) that you could purchase to help your performance in your current physical activity? How would the product (or program help)? Do you really think it is effective? Use specific examples from your experience.

Answers

Answer:

você tem um camping de em Rio do Rio de Janeiro ou em lojas em SP e SP e RJ e região metropolitana de Porto Velho de Porto Velho Rio

Long distance running is the physical activity I engage in now, and there are a number of items and programs that can greatly improve my performance.

A GPS powered watch is one such item which I think is quite useful. I can monitor my speed, distance, heart rate and other important information while running with this device. This data lets me change my pace, track my effort, and make sure I stay within my desired heart rate range. The watch also provides post-run analysis, which I can use to identify my weaknesses and modify my training appropriately. In my opinion, GPS running watches have changed the game by allowing me to run harder and push my physical limits.

A strength training program designed for runners has also been very helpful in enhancing my performance. The program's primary emphasis is on functional workouts that target the major running muscle groups, such as the core, glutes and legs.

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what provides the centiperal force to a satellite revolving around the earth?​

Answers

Answer:

the gravitational force

A stone of mass 150g is rotated in a horizontal circle at 10m/s which is attached to the end of a 1m long. what will be the acceleration of the stone and it's centripetal force?​

Answers

force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N

Answer:

Acceleration: [tex]100\; {\rm m\cdot s^{-2}}[/tex] assuming that the radius of the rotation is [tex]1\; {\rm m}[/tex].

Centripetal force: [tex]15\; {\rm N}[/tex].

Explanation:

In a circular motion, if the tangential velocity is [tex]v[/tex] and the radius of the motion is [tex]r[/tex], the centripetal acceleration of the motion would be [tex]a = (v^{2} / r)[/tex].

In this question, it is implied that for this circular motion, [tex]v = 10\; {\rm m\cdot s^{-1}}[/tex] while [tex]r = 1\; {\rm m}[/tex]. Thus, the (centripetal) acceleration would be:

[tex]\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(10\; {\rm m\cdot s^{-2}})^{2}}{1\; {\rm m}} \\ &= 100\; {\rm m \cdot s^{-2}}\end{aligned}[/tex].

Note that the unit of mass in this question is gram, whereas the standard unit for mass should be [tex]{\rm kg}[/tex] (so as to leverage the fact that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].) Apply unit conversion: [tex]m = 150\; {\rm g} = 0.150\; {\rm kg}[/tex].

Using that fact that [tex](\text{net force}) = (\text{mass}) \, (\text{acceleration})[/tex]:

[tex]\begin{aligned} (\text{net force}) &= (\text{mass}) \, (\text{acceleration}) \\ &= 0.150\; {\rm kg} \times 100\; {\rm m\cdot s^{-2}} \\ &= 15\; {\rm kg \cdot m \cdot s^{-2}} \\ &= 15\; {\rm N}\end{aligned}[/tex].

With a mass of 111.1 kg, is moving with a velocity of 3.000 m/s to the right. Mass is initially at rest and has a mass of 222.2 kg. After the elastic collision, bounces off with a velocity of 1.000 m/s to the left. Calculate the velocity of after the collision.

Answers

Let us assume the positive \hat{x} direction is towards the right.

During any collision, the total momentum is conserved.

That means, the total momentum before the collision is the same as the total momentum after the collision.

p before = P after

Given,

Mass 1, m_1 = 111.1kg

Mass 2, m_2 = 222.2kg

Initial Velocity of Mass 1, 1,1 3.00.î m/s

Initial Velocity of Mass 2, \overrightarrow v_{2,i} =0

Velocity of Mass 1 after collision, \overrightarrow v_{1,f} =-1.000\hat{x}\hspace{0.1cm}m/s

The total Momentum of the system before the collision is,

\overrightarrow p_{before} =m_1\overrightarrow v_{1,i} + m_2\overrightarrow v_{2,i}

=> \overrightarrow p_{before} =(111.1)(3.00\hat{x}) + 0

=> \overrightarrow p_{before} =333.3\hspace{0.1cm}\hat{x}\hspace{0.1cm}kgm/s

The total Momentum of the system after the collision is,

\overrightarrow p_{after} =m_1\overrightarrow v_{1,f} + m_2\overrightarrow v_{2,f}

=> \overrightarrow p_{after} =(111.1)(-1.000\hat{x}) + (222.2)\overrightarrow v_{2,f}

=> \overrightarrow p_{after} =(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f}

We know the total momentum before the collision is the same as the total momentum after the collision.

p before = P after

=>333.3\hat{x} =(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f}

=>(-111.1\hat{x}) + (222.2)\overrightarrow v_{2,f} = 333.3\hat{x}

=> (222.2)\overrightarrow v_{2,f} = 333.3\hat{x} + 111.1\hat{x}

=> (222.2)\overrightarrow v_{2,f} = 444.4\hat{x}

=> \overrightarrow v_{2,f} = \frac{444.4}{222.2}\hspace{0.1cm}\hat{x}\hspace{0.1cm}m/s

=> \overrightarrow v_{2,f} = 2.00\hspace{0.1cm}\hat{x}\hspace{0.1cm}m/s

Therefore the velocity of m_2 after collision is v_{2,f} = 2.00\hspace{0.1cm}m/s going to the right

Please check the attached file for a brief answer.

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Determine the launch speed of a horizontally launched projectile that lands 26.3m from the base of a 19.3m high cliff.

Answers

The launch velocity of the projectile is 13.28 m/s.

What is projectile motion?

The motion of an object thrown in the air under the force of gravity is known as projectile motion.

Since the object is launched horizontally, its initial velocity along the vertical direction is zero. From the second kinematic equation,

s=u*t+(1/2)at^2.

where s is the displacement, t is the time, u is the initial velocity and a is the acceleration. Since the height is decreasing, so it will be taken negative.

For the vertical motion, s=-19.3 m, a=-9.8 m/s^2 and u=0. Put the values in the above equation and solve it.

-19.3 = (0)*t+(1/2)*(-9.8)*t^2

19.3 = (1/2)*(9.8)*t^2

t=1.98 s

Since the velocity along the horizontal direction is constant, the displacement along the horizontal direction is given by the formula,

X=vt

where X is the horizontal displacement, v is the initial horizontal velocity and t is the time.

For the horizontal motion, X=26.3 m and t=1.98 s. Put the values in this equation and solve it.

26.3=v*(1.98)

v=13.28 m/s

The launch velocity is equal to the initial horizontal velocity, so it is equal to 13.28 m/s.

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In an antique automobile, a 6-V battery supplies a total of 48 W to two identical headlights in parallel. The resistance of each bulb is:

Answers

Answer:

The resistance of the each bulb is 1.75 ohm.

Explanation:

The Power P supplied by the battery is given by the formula,

P=VI

where V is the voltage and I is the current.

Then the total current through given parallel circuit is I=P/V

Given P=48 W and V=6 V the value of I is

I=48 W/6 V

I= 8 A

Since both bulbs have same resistance and are connected in parallel so the current through each circuit will be half of the total current. Hence the current through each circuit I₀ is 4 A.

The Resistance R is calculated using formula R=V/I₀.

So the resistance is

R=6 V/4 A

R=1.75 ohm

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Answer:

The resistance of the bulb is 1.5 ohm.

Explanation:

The power P delivered to the whole circuit is given by the formula,

P=VI

Given  P= 48 W and V= 6V, the current in the circuit is

I=P/V

I=48/6= 8 A.

Since the bulbs are identical and are in parallel, therefore the current through each parallel circuit is half of the total current. Therefore current I1 through each bulb is 4 A.

The resistance is given by the formula,

R=V/I1

Hence R=6/4 = 1.5 ohm

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a car travelling at 5m/s is brought to rest in a distance of 20m.Find it's decelaration and time taken.​

Answers

Answer:

time taken= 8s

deceleration = -0.625ms^-2

Explanation:

s = 1/2(u+v)t (linear motion formula)

20 = 1/2 ( 5+0 ) t

t=time taken, find t

t= 8s

a= (v-u) / t

a= (0-5) / 8s

= -0.625 ms^-2

Suppose the carts collided on a surface that had a slight incline to it. Would you expect the momentum to be conserved

Answers

Momentum is conserved when carts are collided on a slanting plane.

To find the answer, we need to know about the conversation of momentum.

What's the conversation of momentum?Conservation of linear momentum says the total momentum before the collision and after the collision remains the same. Mathematically, m1u1+m2u2 = m1v1+m2v2How is the momentum conserved when collision occurs on a slanting plane?On a slanting plane, the velocity has two components, horizontal component horizontal component Vertical componentSo, its momentum has also similar two components. The momentum is conserved along horizontal direction and vertical direction separately.

Thus, we can conclude that the momentum is conserved when carts are collided on a slanting plane.

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If oxygen, which has a electronegativity of 3.5, bonds with hydrogen, which has an electronegativity of 2.1, the bond between the two atoms will be classified as a polar____ bond

Answers

Given the electronegativity of O is 3.5 and H is 2.1,  the bond between the two atoms will be classified as a polar covalent bond.

How are bonds classified?

According to the difference in electronegativity (ΔEN).

If ΔEN ≤ 0.4, the bond is nonpolar covalent.If 0.4 < ΔEN ≤ 1.9, the bond is polar covalent.If ΔEN > 1.9, the bond is ionic.

Oxygen has an electronegativity of 3.5 and hydrogen of 2.1. The difference in electronegativity is:

ΔEN = EN(O) - EN(H) = 3.5 - 2.1 = 1.4

Given the electronegativity of O is 3.5 and H is 2.1,  the bond between the two atoms will be classified as a polar covalent bond.

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All objects fall at the same rate, what is the acceleration of the force of gravity?

Answers

Answer:

9.8 m/s

Explanation:

If these objects are in free fall, meaning that their downward movement is only due to the force of gravity, then acceleration due to gravity is about 9.8 metres per second per second.

Select odd one out and also give the reason:
a) Gravitational force , magnetic force , electrostatic force

Answers

Gravitational force is not related to other force because it is related to gravity.

How magnetic force and electrostatic force are related?

Electrostatic force and magnetic force are link to each other because both have charges. Electrostatic force is type of force that is present between two electrically charged particles. They can either be a repulsive or attractive force.

So we can conclude that Gravitational force is not related to other force because it is related to gravity.

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