classroom and draw an approxim object Three forces are acting on an object (Figure 1.32) which is in equilibrium. Determine force A 1200 N Force A 51 39 42.00 1400 N Figure 1.32 Three forces, acting on​

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Answer 1
I’m not good with physics but I’m good with the theory what goes up must come down

Related Questions

4) What process produces carbon dioxide?
a) photosynthesis
b) replication
c) mutation
d) respiration

5) What happens during cytokinesis?
a) a spindle forms
b) chloroplasts release energy
c) the cytoplasm divides
d) chromosomes

plz answer asap

Answers

Answer:

d and c are the answers :)

A textbook is sitting at rest on a desk. Compared to the magnitude of the force of the textbook on the desk, how
would you describe the magnitude of the force of the desk on the book?
o less
O zero
O more
O equal

Answers

Equal because the book is not moving and the forces are balanced/equal

a car dealer sells 4 makes of cars in 5 colors with either standard or automatic transmissions. how many variations does the dealership offer?

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The number of variations the dealership have is 40 cars.

Data given;

The number of different product = 4The number of transmission = 2 (standard or automatic)The numbers of colors available = 5

Dealership variation

This is the total numbers of cars available in the dealer shop at the moment.

To get that, we simply multiply the numbers of colors available, the number of transmission and the makes of car available.

This is equal to

[tex]4*2*5=40[/tex]

From the above calculation, we can say that the variations the dealership offers is 40

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a narrow slit is illuminated with sodium yellow light of wavelength 589 nm. if the central maximum extends to ±30.0°, how wide is the slit?

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Answer:

because of the gravity of the earth

Scientific notation of 1,750

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1.75 x 10^3
hopes this helps if not im really sorry

Two carts, A and B, are connected by a spring and sitting at rest on a track. Cart A has a mass of 0.4 kg and Cart B has a mass of 0.8 kg. The spring is released and causes the two carts to push off from each other. Which of the following correctly compares the motion and forces acting on the two carts?

A. Cart A experiences a greater force but has the same speed as Cart

B. Cart A experiences a greater force and has a greater speed after the recoil.

C. Both carts experience the same force but Cart A has a greater speed after the recoil.

D. Both carts experience the same force and have the same speed after the recoil.

Answers

Both carts experience the same force but Cart A has a greater speed after the recoil.

The given parameters;

Mass of the cart A = 0.4 kgMass of the cart B = 0.8 kg

Apply the principle of conservation of linear momentum to determine the velocity of the carts after collision;

[tex]m_Av_0_A\ + m_Bv_0_B = m_Av_f_A \ + m_Bv_f_B\\\\the \ initial \ velocity \ of \ both \ carts = 0\\\\0.4(0) + 0.8(0) = 0.4v_f_A + 0.8v_f_B\\\\0 = 0.4v_f_A + 0.8v_f_B\\\\0.4v_f_A = -0.8v_f_B\\\\v_f_A= \frac{-0.8 v_f_B}{0.4} \\\\v_f_A = - 2 \ v_f_B \ \ m/s[/tex]

According to Newton's third law of motion, action and reaction are equal and opposite. The force exerted on cart A is equal to the force exerted on cart B but in opposite direction.

[tex]F_A = -F_B[/tex]

Thus, the correct statement that compares the motion and forces acting on the two carts is "Both carts experience the same force but Cart A has a greater speed after the recoil."

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A wave has frequency of 50 Hz and a wavelength of 10 m. What is the speed of the wave?
I need the Formula,Known,Substitute & Solve Answer with Units

Answers

Answer:

This is the answer that I got.

Explanation:

Hope it is right.

A 3 Kg ball at the end of a string is spun in a circle. If the length of the string is 2 m and the velocity of the ball is 6 m/s, what is the centripetal acceleration of the ball? What is the centripetal force on the ball?

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m is 3kg
r is 2m
v is 6m/s

Fc = mv^2 / r
Fc = (3x6^2)/2 =54

a = v^2/r
a = 6^2 /2 = 18

Exert a force on the object, what happens to their state of motion​

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The force might make the object move or bring the object to rest or change in direction of motion

A race car accelerates uniformly from 18.5 m/s in 2.47 seconds m. Determine the acceleration of the car in m/sec^2 and the distance

Answers

The change of velocity is from 18.5 to 46.1 m/s or 46.1 - 18.5 = 27.6 m/s. The time of change is 2.47 s. So the acceleration is 27.6 m/s / 2.47 s = 11.17 m/s^2. The distance traveled is average velocity times time of travel.

In a 200.0-m relay race (each leg of the race is 50.0 m long), one swimmer has a 0.450 second lead
and is swimming at a constant speed of 3.90 m/s towards the opposite end of the pool. What minimum
speed must the second swimmer have in order to catch up with the first swimmer by the end of the
pool?

Answers

The minimum  speed that the second swimmer must have in order to catch up with the first swimmer by the end of the  pool is; 4.04 m/s

We are given the following facts;

Both swimmers swim the same distance of 50 m for a leg of the relay race.Swimmer 1 swims at constant speed of 3.9 m/s. Swimmer 1 has a 0.450 second head start.

Formula for time taken is;

time = distance/speed

Time taken by swimmer 1 for one leg;

t₁ = 50/3.9

t₁ = 12.82 s

Since swimmer 1 has a head start of 0.45 s, then time that swimmer 2 must use in order to catch up is;

t₂ = 12.82 -0.45

t₂ = 12.37 s

Thus, speed of swimmer 2;

v = 50/12.37

v = 4.04 m/s

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c. Two persons A and B are pushing a block from opposite sides with
force of 5N and 6N respectively.
[3]
i) Find the magnitude of the resultant force on the block.
ii) Find the direction of resultant force with respect to A and B.

Answers

Answer:

1 N

Explanation:

1) You'll just have to sum the force vectors, and since they are opposite, you'll calculate the difference of their magnitudes, in absolute value.

Ftot = |F1 - F2| = 6N - 5N = 1N

2)The direction will be the same as the greater force's direction.

Which of the balls will exert the greatest force on object A?Why?

Answers

F = mass × Acceleration ( give Acceleration is 9.8 )

So force of 5 kg mass on A is

F = 5 × 9.8 = 49N

Force of mass 1 kg on A

F = 1 × 9.8 = 9.8 N

Force of mass 10 kg on A is

F = 10 × 9.8 = 98N

Clearly the 10 kg ball experts the most force cause it has more mass

A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from point D to point E. What is the location of point F, which partitions the directed line segment from D to E into a 5:6 ratio? Negative one-eleventh One-eleventh Two-fifteenths Fifteen-halves.

Answers

The location of the point F that partitions a line segment from D to E ([tex]\overline{DE}[/tex]), that goes from negative 4 to positive 5, into a 5:6 ratio is fifteen halves (option 4).  

We need to calculate the segment of the line DE to find the location of point F.

Since point D is located at negative -4 and point E is at positive 5, we have:

[tex] \overline{DE} = E - D = 5 - (-4) = 9 [/tex]

Hence, the segment of the line DE ([tex]\overline{DE}[/tex]) is 9.

Knowing that point F partitions the line segment from D to E ([tex]\overline{DE}[/tex]) into a 5:6 ratio, its location would be:

[tex] F = \frac{5}{6}\overline{DE} = \frac{5}{6}9 = 5*\frac{3}{2} = \frac{15}{2} [/tex]  

Therefore, the location of point F is fifteen halves (option 4).

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I hope it helps you!

Answer:

The above answer is actually incorrect - the real answer is B. 1/11

Explanation:

I'm adding a ss below for proof

Hope this helped!

I'd really appreciate Brainliest :)

If the western component of the wind is half of the south, what is the angle of the wind with the south?

Answer this question in details!

Answer is 26.57 or tan^-1 (1/2) or cot^-1 (2)

no links !​

Answers

Answer:

This above a triangle that models our situation.

Explanation:

We have a two componens., since we have a western componet and southern component. One travel in a southern direction. and the other travel in the west.

Let the component that travel in the south be the length of a.

According to the problem, the westard component is half of that so let that length be a/2.

Now we must find the angle of the wind in the South.

This means that what is angle that is opposite of the western componet because that angle is the most southward angle. So know we apply the tan property.

[tex] \tan(x) = \frac{opp}{adj} [/tex]

Our side opposite of the angle we trying to find is the western component and the side adjacent to it is the southern component. Also remeber since western and Southern negative displacements, we have

[tex] \tan(x) = \frac{ - \frac{a}{2} }{ - a} [/tex]

[tex] \tan(x) = - \frac{a}{2} \times - \frac{1}{a} = \frac{1}{2} [/tex]

Now we take the arctan or inverse tan of 1/2.

[tex] \tan {}^{ - 1} ( \frac{1}{2} ) = 26.57[/tex]

Based on the information about vector components and graph provided, the angle of the wind with the south is 26.57°.

What are the values of the western and southern component of the wind?

The western and southern component have a negative displacement as shown in the graph.

Let the southern component be -x

The western component is half of -x = -x/2

What is the angle of the wind with the South?

Since the angle is with the south, the trigonometric ratio to be used to find the angle is:

Tan θ = opp/adj

Tan θ = -×/2 /-x

Tan θ = 1/2

θ = tan^-1(0.5)

θ = 26.6°

Therefore, angle of the wind with the south is 26.57°.

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With an initial velocity of 20km per hour a car accelerated at 8m/s2 for 10s,what is the position of the car at the end of 10s

Answers

initial velocity=u=20km/h=5.5m/sAcceleration=a=8m/s^2Time=t=10s

[tex]\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2[/tex]

[tex]\\ \sf\longmapsto s=5.5(10)+\dfrac{1}{2}(8)(5.5)^2[/tex]

[tex]\\ \sf\longmapsto s=55+4(5.5)^2[/tex]

[tex]\\ \sf\longmapsto s=55+4(30.25)[/tex]

[tex]\\ \sf\longmapsto s=55+121[/tex]

[tex]\\ \sf\longmapsto s=176m[/tex]

calculate the resultant force acting on the object and state its direction

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[tex]\\ \sf\Rrightarrow F_{net}=F_1+F_2+F_3+F_4[/tex]

[tex]\\ \sf\Rrightarrow F_{net}=20N+20N+10N-10N[/tex]

[tex]\\ \sf\Rrightarrow F_{net}=40N[/tex]

Direction is east.

PLEASE HELP MEEEEE 80 points on the line and i need this done

Answers

Answer:

5) A 200n

F net=0n

6) B 50n

F net=0n

7) C 200n

F net= 900n, up

8) D 80n

F net=60n,left

9) E 300n

F net=60n,left

10)???

11) G 20n

F net=30n,Right

__________________________________

All the answers is hard

correct me if i'm wrong

HELP!!!!!!!!!!!!

9. Estimate the change in the thermal energy of water in a pond with a mass of
1000 kg and a specific heat of 4200 J/(kg. C) if the water cools by 1°C.

Answers

Answer:

-4,200,000 J

Explanation:

an airplane flies with a constant speed of 620 miles per hour. how far can it travel in 3 1/2 hours?

Answers

Answer:

hi there !

1 hour -------> 620 miles

3 1/2 ---------> ?

the number of miles the airplane travels in 3 1/2 hours

= 3 1/2 × 620 = "2170" miles

“Applications of thermal expansion in solids, liquids and gases.”
(Two for each states of matter)​

Answers

Thermal expansion is the tendency of matter to change its shape dimensions, density in response to a change in temperature.

The following are applications of thermal expansion in various matter.SolidsThe principle of thermal expansion is used in thermostat to regulate the temperature of heating device e.g Electric pressing Iron.Rivets are used to hold steel plates together very tightly.

 

  2. Gases

Hot-air balloons are an application of thermal expansionGases turbine

    3. Liquids

Engine CoolantsVolume of thermometer

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Mike drops a rock from the top of a suspension bridge. If the rock falls for 4.88 seconds,
how high is the bridge?

Answers

This question involves the concepts of the equations of motion in vertical direction.

The height of the bridge is "116.8 m".

We will use the second equation of motion in vertical direction to find out the height of the bridge.

[tex]h=v_it+\frac{1}{2}gt^2\\\\[/tex]

where,

h = hieght = ?

vi = initial speed = 0 m/s

t = time taken = 4.88 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]h=(0\ m/s)(4.88\ s)+\frac{1}{2}(9.81\ m/s^2)(4.88\ s)^2[/tex]

h = 116.8 m

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Assume ideal behavior. How many moles of are needed to fill a 2000L weather balloon at 210K and 1atm pressure? Round your answer to the nearest mole (use R = 0.082057 L atm mol-1K-1)

Answers

Considering the ideal gas law, 116.06 moles of are needed to fill a 2000 L weather balloon at 210 K and 1 atm pressure.

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

In this case, you know:

P= 1 atmV=2000 Ln= ?R= 0.082057 [tex]\frac{atmL}{molK}[/tex]T= 210 K

Replacing in the ideal gas law:

1 atm× 2000 L = n× 0.082057 [tex]\frac{atmL}{molK}[/tex]× 210 K

Solving:

[tex]n=\frac{1 atmx 2000 L}{0.082057 \frac{atmL}{molK}x 210 K}[/tex]

n=116.06 moles

Finally, 116.06 moles of are needed to fill a 2000 L weather balloon at 210 K and 1 atm pressure.

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ii. Which of these has a heating element with a low melting point?
a, electric bell
c. electric iron
b. fuse
d. all of these

Answers

The answer is Fuse, the Fuse has a low melting point
Letter B (Fuse) is the answer

Which of the following industries is the largest producer of primary air pollutants in the United States?

Answers

The largest producer of primary air pollution in the United States is what? electricity producing plants.

Answer:

electricity production

Explanation:

Recoil is noticeable if you throw a heavy ball while standing on roller skates. If instead you go through the motions of throwing the ball but hold onto it, your net recoil velocity will be

Answers

Answer:

Since there is no external force, there is no change (movement) in the center of mass. Internally, the center of mass might change position, but the external result is still zero net velocity.

The net recoil velocity must be zero.

the breaks on a 15,680 N car exert a stopping force of 640 N.

Answers

The brakes on a 15,680 N car exert a stopping force of 640 N. The car's velocity changes from 20.0 m/s to 0 m/s.

Which of the following statements cannot be supported by Kepler's laws of planetary motion?


A planet's distance from the sun will not be the same in six months.


A planet's speed as it moves around the sun will not be the same in six months.


The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.


The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets.

Answers

the second
option is the answer

Help me answer this question please!

Answers

Heat will be produced because of the friction from the tires

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.
Part A
How much work does the rope do on the wagon if you pull the wagon 200 m at a constant speed?

Answers

Angle (θ) = 60°Force (F) = 20 NDistance (s) = 200 mTherefore, work done= Fs Cos θ= (20 × 200 × Cos 60°) J= (20 × 200 × 1/2) J= (20 × 100) J= 2000 J

Answer:

2000 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

S O L U T I O N:

Here we've been provided with the information i.e

Force (F) = 20N

Distance (s) = 200m

Angle (θ) = 60°

Ans we have given to find out what is work done (W) = ?

The standard formula to calculate work done is given by,

[tex]:\implies\sf{w = f \times s \times cos \theta}[/tex]

[tex]:\implies\sf{w = 20 \times 200 \times cos60}[/tex]

[tex]:\implies\sf{w = 20 \times 200 \times \frac{1}{2} }[/tex]

[tex]:\implies\sf{w = 20 \times 100 \times 1}[/tex]

[tex]:\implies\sf{w = 2000J}[/tex]

Total work done done by rope is 2000J.
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