carts, bricks, and bands

5. What acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick?
a. About 0.25 m/s2
b. About 0.50 m/s2
c. About 0.75 m/s2
d. About 1.00 m/s2

Answers

Answer 1

B.  The acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick is 0.5 m/s².

What is the applied force on an object?

The force applied on object is obtained by multiplying the mass and acceleration of the object.

According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.

Also, the applied force is directly proportional to the change in the momentum of the object.

Mathematically, the force acting on object is given as;

F = ma

a = F/m

where;

a is the acceleration of the objectm is the mass of the objectF is the applied force

From the trials, the acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick is 0.5 m/s².

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Carts, Bricks, And Bands5. What Acceleration Results When 2 Rubber Bands Stretched To 20 Cm Are Used

Related Questions

What is the frequency of a photon of EMR with a wavelength of 2.55x10*³m?1.18x1011 Hz8.50x10 12 Hz7.65x105 Hz1.18x105 Hz

Answers

In order to solve this equation, we will need to use the formula

[tex]f=\frac{c}{\lambda}[/tex]

where f = frequency, c is the speed of light and lamda is wavelength

c = 3x10^8 m/s

lamda = 2.55x10^-3 m

f = (3x10^8)/(2.55x10^-3) = 1.18x10^11 1/s

Timmy walks 5 m North, 3m West, and finally 1 m South. What is his displacement from his starting point?

Answers

Timmy walks 5 m North, 3m West, and finally 1 m South then his displacement from the starting point would be 5 meters in the northwest direction.

What is displacement?

Displacement describes this shift in location and it is calculated with the help of the initial and the final position of the object.

As given in the problem If Timmy walks 5 m North, 3m West, and finally 1 m South ,

The resultant displacement of the Timmy = √(4² + 3²)

                                                              = 5 meters

                                       

Thus, the resultant displacement of the Timmy would be 5 meters

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Popeye the Sailor man, who has a mass of 85 kg, ran ( at a constant rate ) up a flight of stairs that are 3.55 m high in 6 seconds. How many watts of power did he generate during his run ?

Answers

Given data

*The given mass of the Sailorman is m = 85 kg

*The given height is h = 3.55 m

*The given time is t = 6 s

*The value of the acceleration due to gravity is

[tex]g=9.8m/s^2[/tex]

The formula for the power generated by the Sailorman during his run is given as

[tex]\begin{gathered} P=\frac{W}{t} \\ =\frac{\text{mgh}}{t} \end{gathered}[/tex]

*Here W is the work done

[tex]\begin{gathered} P=\frac{85\times9.8\times3.55}{6} \\ =492.85\text{ W} \end{gathered}[/tex]

In a lightning discharge, 45 C of charge move through a potential difference of 1.0 x108 V in 0.030 s.A. What is the current of the lightning strike?B. How much energy is released by the lightning bolt?

Answers

Given:

Charge, Q = 45 C

Potential difference, V = 1.0 x 10⁸ V

Time, t = 0.030 s

Let's solve for the following:

• (A). What is the current of the lightning strike?

To find the current, apply the formula:

[tex]I=\frac{Q}{t}[/tex]

Where:

I si the current

Q is the charge = 45 C

t is the time = 0.030 s

Thus, we have:

[tex]\begin{gathered} I=\frac{45}{0.030} \\ \\ I=1500\text{ A} \end{gathered}[/tex]

Therefore, the current of the lightning strike is 1500 Amperes.

• (B). How much energy is released by the lightning bolt?

To find the amount of energy released, apply the formula:

[tex]E=V\times Q[/tex]

where:

E is the Energy released

V is the potential difference, V = 1.0 x 10⁸ V

Q is the charge = 45 C

Thus, we have:

[tex]\begin{gathered} E=1.0\times10^8\ast45 \\ \\ E=4.5\times10^9\text{ J} \end{gathered}[/tex]

Therefore, the energy released is 4.5 x 10⁹ Joules.

ANSWER:

(a). 1500 A

(b). 4.5 x 10 J

The source of the Sun’s heat and light energy is:A. combustion of helium gas.B. fusion of hydrogen nuclei.C. gravitational pressure.D. burning of fossil fuels.

Answers

To find

The source of the Sun’s heat and light energy is:

Explanation

The sun's core is very hot. So under pressure nuclear fusion takes place. Here hydrogen is changed to helium.

Conclusion

The correct option is

B. fusion of hydrogen nuclei.

A plane's average speed between 2 cities is 800 km/hr. If the trip takes 4.5 hours howfar does the plane fly?

Answers

Considering the average velocity can be written as:

[tex]v=\frac{\Delta s}{\Delta t}[/tex]

We can isolate the distance, and we get:

[tex]\Delta s=v*\Delta t\rightarrow\Delta s=800*4.5=3600km[/tex]

Then, the distance the plane flew is 3600km

Hey! I really need help with this question please :)

Answers

Answer: B

Explanation:

The formula for calculating the efficiency of a heat engine is expressed as

Efficiency = useful work done/Heat energy supplied x 100

From the information given,

Heat energy supplied = 500

useful work done = 50

Efficiency = 50/500 x 100

Efficiency = 10%

A plane is traveling with a velocity of 70 miles/hr with a direction angle of 24 degrees. The wind is blowing at 25 miles/hr with a direction angle of 190 degrees. What is the vertical component of the wind velocity? Round your answer to the nearest whole number.

Answers

Wind velocity:

25 m/h with a direction angle of 190°.

Vertical component:

25 sin 190 = -4.34 m/s = - 4 m/s

Which is negatively charged?A. protonB. nucleusC. electronD. neutron

Answers

Protons, electrons an neutrons are the particles that make up atoms.

Protons have a positive electric charge, electrons have a negative electric charge an neutrons are electrically neutral.

The nucleus of an atom is made of protons and neutrons, so, its electric charge is positive an proportional to th

A rock sample has a mass of 6 kg and a volume of 0.002 m3. Calculate the density of this rock sample.

Answers

Density = mass / volume

So 6/0.002=3,000 kg/m3.

The density of the rock sample is 3000 kg/m³.

Density is an important property of matter because it can be used to determine the weight of an object. For example, if you know the density of a rock sample and its volume, you can calculate its weight by multiplying the density by the volume. Density can also be used to determine the composition of a material. For example, if you know the densities of different materials, you can identify the material of a rock sample by measuring its density.

Density is a measure of how much mass is contained in a given volume.. The calculation of the density of the rock sample:

Density = Mass / Volume

Density = 6 kg / 0.002 m³

Density = 3000 kg/m³

As a result, the rock sample has a density of 3000 kg/m³.

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If an 800 kg roller coaster is at the top of its 50 m high track, it will have a potential energy 392,000 and a kinetic energy of 0J. This means the total mechanical energy is 392,000J. If the cart drops down to a new height of 10m, how much energy does the cart have now?

Answers

ANSWER:

313600 J

STEP-BY-STEP EXPLANATION:

We have that the gravitational potential energy is given by the following equation:

[tex]E_p=m\cdot g\cdot h[/tex]

We substitute and calculate the potential energy, knowing that g is the acceleration of gravity and is equal to 9.8 m/s^2:

[tex]\begin{gathered} E_p=800\cdot9.8\cdot10 \\ E_p=78400\text{ J} \end{gathered}[/tex]

We know that the total energy is 392,000 joules, so the energy it now carries would be the total minus the calculated potential energy:

[tex]\begin{gathered} E_k=392000-78400 \\ E_k=313600\text{ J} \end{gathered}[/tex]

The energy carried by the cart is 313600 J

Modern roller coasters have vertical loops like the one shown in the figure. The radius of curvature is smaller at the top than on the sides so the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats.
1. What is the speed of the roller coaster, in meters per second, at the top of the loop if the radius of curvature there is 14 m and the downward acceleration of the car is 1.1g? Note that g here is the acceleration due to gravity.
2. The beginning of this roller coaster is at the top of a high hill. If it started from rest at the top of this hill, how high, in meters, above the top of the loop is this initial starting point? You may assume there is no friction anywhere on the track.
3. If it actually starts 7.5 m higher than your answer to the previous part (yet still reaches the top of the loop with the same velocity), how much energy, in joules, did it lose to friction? Its mass is 1800 kg.

Answers

1 ) The speed of the roller coaster = 12.28 m / s

2 ) Height of the hill above the top of the loop = 7 m

3 ) Energy lost due to friction = 132 KJ

1 ) The speed of the roller coaster,

[tex]a_{c}[/tex] = v² / r

[tex]a_{c}[/tex] = Centripetal acceleration

v = Linear velocity

r = Radius

r = 14 m

[tex]a_{c}[/tex] = 1.1 g = 1.1 * 9.8

[tex]a_{c}[/tex] = 10.78 m / s²

v² = [tex]a_{c}[/tex] * r

v² = 10.78 * 14

v² = 150.9

v = 12.28 m / s

2 ) Initial starting point,

Considering hill as 1 and the loop as 2,

v1 = 0

h2 = 2 r = 2 * 14

h2 = 28 m

∑ [tex]F_{y}[/tex] = m [tex]a_{c}[/tex]

[tex]F_{N}[/tex] + [tex]F_{g}[/tex] = m [tex]a_{c}[/tex]

0 + m g = m v2² / r

v2² = g r

According to law of conservation of energy,

E1 = E2

m g h1 + 1 / 2 m v1² = m g h2 + 1 / 2 m v2²

m g h1 + 0 = 28 m g + 1 / 2 m g r

h1 = 28 +  1 / 2 ( 14 )

h1 = 35 m

Height of the hill above the top of the loop = h1 - h2

Height of the hill above the top of the loop = 35 - 28

Height of the hill above the top of the loop = 7 m

3 ) Energy lost due to friction,

h1 = 35 + 7.5

h1 = 42.5 m

m = 1800 kg

v2² = g r

v2² = 9.8 * 14

v2² = 137.2 m / s

Since energy is lost due to friction,

KE1 + U1 = KE2 + U2 + W

0 + m g h1 = 1 / 2 m v2² + m g h2 + W

( 1800 * 9.8 * 42.5 ) = ( 0.5 * 1800 * 137.5 ) + ( 1800 * 9.8 * 28 ) + W

749700 = 123750 + 493920 + W

W = 749700 - 617670

W = 132030 J

W = 132 KJ

Therefore,

1 ) The speed of the roller coaster = 12.28 m / s

2 ) Height of the hill above the top of the loop = 7 m

3 ) Energy lost due to friction = 132 KJ

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Round the answer for each problem to the correct number of significant figures.

a. (7.31 × 104) + (3.23 × 103)

b. (8.54 × 10–3) – (3.41 × 10–4)

c. 4.35 dm × 2.34 dm × 7.35 dm

d. 4.78 cm + 3.218 cm + 5.82 cm

e. 38,736 km ÷ 4784 km

Answers

After solving each problem to correct the number of significant figures :

(a) 76330

(b) 8199 × 10⁻³

(c) 74.81 dm.

(d) 13.8 cm

(e) 8.09 km

What is an Equation?

Equations are mathematical expressions that have two algebraic expressions on either side of an equals (=) sign. The expressions on the left and right are shown to be equal to one another, demonstrating this relationship. L.H.S. = R.H.S. (left hand side = right hand side) is a fundamental mathematical formula.

According to the question,

(a) (7.31 × 10⁴) + (3.23 × 10³)

(73100) + (3230) = 76330

(b) (8.54 × 10⁻³) – (3.41 × 10⁻⁴)

= 8199 × 10⁻³

(c) 4.35 dm × 2.34 dm × 7.35 dm

= 74.81 dm.

(d) 4.78 cm+3.218 cm+5.82 cm

= 13.8 cm

(e) 38,736 km ÷ 4784 km

= 8.09 km

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A car travels 400 km in the first 4.5 hours of a trip. It stops for an hour and then travels final 300 km in 2.5 hours. Find the average speed of the car.

Answers

Given data:

Distance traveled by car in t_1=4.5 hr is s_1=400 km.

Distance traveled by car in t_2=1 hr is s_2=0 km (as the car was stopped).

Distance traveled by car in t_3=2.5 hr is s_3=300 km.

The average speed is given as,

[tex]\begin{gathered} v_{avg}=\frac{\text{ total distance traveled}}{\text{total time taken}} \\ =\frac{s_1+s_2+s_3}{t_1+t_2+t_3} \end{gathered}[/tex]

Substitute all known values,

[tex]\begin{gathered} v_{avg}=\frac{(400\text{ km})+(0\text{ km})+(300\text{ km})}{(4.5\text{ hr})+(1\text{ hr})+(2.5\text{ hr})} \\ =87.5\text{ km/h} \end{gathered}[/tex]

Therefore, the average speed of the car is 87.5 km/h.

i am not sure the best way to solve this problem

Answers

ANSWER

14.11 s

EXPLANATION

We know that in total, the runner will run a distance of 100m. He runs at constant acceleration for a while and then his velocity gets constant until the end of the track - this means that in the last part, his acceleration is zero.

So we have two parts:

For the first part, we have the acceleration and time. If we set that the initial position is zero, as shown in the diagram above, and that the runner starts from rest - therefore, his initial velocity is zero - we can find the distance of the first part of the path, which we'll call x1:

[tex]x_1=x_0+v_0t+\frac{1}{2}at^2[/tex]

Since x0 and v0 are both zero, then those terms get cancelled:

[tex]x_1=\frac{1}{2}\cdot a\cdot t^2=\frac{1}{2}\cdot1.5\cdot6^2=27m[/tex]

So the first part of the track, where the runner is speeding up, has a distance of 27m. Therefore, the rest of the track where the runner runs at constant acceleration is:

[tex]100-27=73[/tex]

73m.

We want to find the time it took the runner to run the whole 100m. We know that he did the first part in 6 seconds. To find the time of the second part, we can use the distance we just found. Let's call it xf:

[tex]x_f-x_1=\frac{1}{2}at^2+v_0t[/tex]

We know that the acceleration in this part of the track is zero and the initial velocity for this part is the velocity the runner had when he reached 6 seconds - i.e. 27m:

[tex]73m=v_1\cdot t[/tex]

We don't know the time and we don't know the velocity, but we can find the second one using the formula for velocity for the first part of the track with t = 6s:

[tex]\begin{gathered} v_1=a\cdot t+v_0 \\ v_1=1.5\cdot6 \\ v_1=9m/s \end{gathered}[/tex]

Now we can find the time for the second part of the track:

[tex]\begin{gathered} 73m=9m/s\cdot t \\ t=\frac{73m}{9m/s} \\ t\approx8.11s \end{gathered}[/tex]

Therefore, the total time it took the runner to run 100m was:

[tex]\begin{gathered} t=6s+8.11s \\ t=14.11s \end{gathered}[/tex]

14.11 s

which of the following are independent of the mass of an object falling freely near earth's surface: (may have more than 1 answer) 1) acceleration of the object 2) gravitational force acting on the object 3) gravitational force acting on the object 4) magnitude of the gravitational field

Answers

As the object is falling freely, the acceleration of the object will be equal to the acceleration due to gravity.

It is given as,

[tex]g=\frac{GM}{R^2}[/tex]

Here, G is the univarshal gravitational constant and M is the mass of the Earth.

means acceleration of the object is constant and independent of the mass of the object.

so option 1 is correct.

now the gravitational force on that object is,

[tex]F=\frac{GMm}{R^2}[/tex]

here this is dependent on the mass of the object(m).

NOw the gravitational field means the force per unit mass and is given by,

[tex]E=\frac{GM}{R^2}[/tex]

Here we can se that this gravitational field is also independent of the mass of the object.

So, option 1 and 4 are correct.

my choice1A child pushes a toy truck up an inclined plane. The lifting force is...Select one:ut ofO a. the weight of the truck.O b. the length of the inclined plane.O c. the force the child uses to push the truck.uestionO d. equal to the total amount of work done.

Answers

When pushing a toy truck up an inclined plane, the force that makes the truck goes up is the force that the child uses to push the truck. (The force that pushes

Therefore the correct option is C.

A 4-kg ball traveling westward at 25 m/s hits a 15-kg ball at rest. The 4-kg ball bounces east at 8.0 m/s. What is the speed and direction of the 15-kg ball? What is the impulse of the second ball?

Answers

Given:

The mass of the first ball is,

[tex]m_1=4\text{ kg}[/tex]

The initial velocity of the first ball towards West is,

[tex]u_1=25\text{ m/s}[/tex]

The mass of thr second ball is,

[tex]m_2=15\text{ kg}[/tex]

the second object is initially at rest.

The final velocity of the first ball is,

[tex]v_1=-8.0\text{ m/s}[/tex]

we are taking West as positive.

Applying momentum conservation principle we can write,

[tex]m_1u_1+m_2\times0=m_1v_1+m_2v_2[/tex]

Substituting the values we get,

[tex]\begin{gathered} 4\times25+0=4\times(-8.0)+15\times v_2 \\ v_2=\frac{100+32}{15} \\ v_2=8.8\text{ m/s} \end{gathered}[/tex]

THe final velocity of the second ball is towards East and the magnitude is 8.8 m/s.

The impulse of the Second ball is,

[tex]\begin{gathered} I=m_2v_2-m_2\times0 \\ =15\times8.8 \\ =132\text{ kg.m/s} \end{gathered}[/tex]

Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet,
making contact over an area of 80.0cm2 with each foot. Both the ceramic and the
carpet are 2.00 cm thick and are 10.0ºC on their bottom sides. At what rate must
heat transfer occurs from each foot to keep the top of the ceramic and carpet at
33.0ºC?

Answers

The rate of heat transfer from each foot to keep the top of the ceramic and carpet at the desired temperature is 27.6 J/s.

What is thermal heat transfer?

Thermal heat transfer is defined as the movement of heat across the border of the system due to a difference in temperature between the system and its surroundings.

The rate at which heat transfer occurs from each foot to keep the top of the ceramic and carpet at the given temperature is calculated as follows;

Q/t = kA(T₂ - T₁)/d

where;

Q/t is the rate of heat transfer

k is heat transfer coefficientT₂ is the final temperatureT₁ is the initial temperatured is the distance between the wool and ceramics

The heat transfer coefficient of human skin = 0.3 W / m⁰C

Area = 80 cm² = 0.008 m²

thickness, d = 2 cm = 0.02 m

change in temperature, = 33⁰ C - 10 ⁰C = 23 ⁰C

Q/t = (0.3 x 0.008 x 23) / (0.002)

Q/t = 27.6 W = 27.6 J/s

Thus, the rate of heat transfer from each foot to keep the top of the ceramic and carpet at the desired temperature is 27.6 J/s.

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Name the instrument which is made on the basis of expansion of heat. ​

Answers

The instrument is the Thermometer

Answer:

It is a thermometer and it helps to see the temperature

In the setup of a cart pulled by a hanging mass, without friction, we have the following information: the mass of the cart is 0.53 Kg., and the hanging mass is 0.077 Kg.

Determine the acceleration of the cart (in m/s2).

Answers

The acceleration of the cart

a=1.2431m/s^2

What is acceleration?

Generally, The equation for Newton's second law of motion is

2nd law of motion,

Fnet=m a

on hanging mass,

m_1 g-T=m_1 a

m_1 g-m_2 a=m_1 a

Acceleration, [tex]$a=\frac{m_1 g}{m_1+m_2}$[/tex]

[tex]&a=\frac{0.077 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^2}{0.077 \mathrm{~kg}+0.53 \mathrm{~kg}} \\[/tex]

a=1.2431m/s^2

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which are neutrally charged, are found in thenucleus of the atom.

Answers

Given:

Nucleus of the atom

Required:

Neutra

12. How could extreme heat (resulting from Climate Change) affect human andanimal life?

Answers

ANSWER:

The answer is given in the step by step of the question

STEP-BY-STEP EXPLANATION:

Extreme heat can affect human and animal life in the following ways:

• Animals are sensitive to changes in temperature, as much or more than humans. Think how bad you feel when you have a fever. They also suffer this type of discomfort as a consequence of climate change.

,

• Climate change is pushing many species to the limit. They lack water to drink or suffer from temperatures in which they are not comfortable.

,

• Provocation of forest fires damaging the habitat of animals and humans

Two children are riding on a merry-go-round. Child A is at a greater distance from the axis of rotation than child B. Which child hasthe larger linear displacement?Select one:a. There is not enough information given to answer the question.b. Child Ac. They have the same non-zero linear displacementd. Child Be. They have the same zero linear displacement.

Answers

The linear displacement in a rotation is given by:

[tex]s=r\theta[/tex]

where r is the distance from the axis of rotation and theta is the angular displacement.

Since the linear displacement is proportional to the radius we conclude that child A has a larger linear displacement.

A projectile is fired at an upward angle of 55.0 ∘ from the top of a 155 m cliff with a speed of 165 m/s .
a)What will be its speed when it strikes the ground below? (Use conservation of energy and neglect air resistance.)

Answers

A projectile is fired at an upward angle and the speed of the object when it strikes the ground below will be 434.5 m/s.

What is Projectile motion?

A projectile is an object or particle that is thrown toward the surface of the Earth and moves along a curved route only under the influence of gravity. Galileo demonstrated that this curving path was a parabola, but in the unique situation where it is hurled straight up, it may also be a straight line.

According to the question,

[tex]h=v_0_yt+1/2gt^2[/tex]

-155 m = (165 × sin 55°)t - 0.5(9.8)t²

-155 = 135.16t - 4.9 t²

4.9 t² - 135.16t - 155 = 0

t = 27.5 seconds.

Now, the speed of the object when it strikes the ground will be,

[tex]v_f=v_i+gt[/tex]

= 165 + (9.8)(27.5)

[tex]v_f[/tex] = 434.5 m/s.

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A sample of unknown material weight 900N In air and and 400N when submerged in an alcohol solution with a density of 0.7 g/cm³.What is the density of the material ?

Answers

1.26 g/cm³

Explanation

Step 1

given

[tex]\begin{gathered} F_{g(air)}=900\text{ N} \\ F_{g(alchodol)}=400\text{ N} \\ \rho_{alcohol}=0.7\text{ }\frac{g}{cm^3} \end{gathered}[/tex]

unknown; the density of the material, so

[tex]\begin{gathered} F_B=F_{g(air)}-F_{g(alcohol)} \\ F_B=900\text{ N-400 N=500 N} \end{gathered}[/tex]

so, the proportion is

the ratio of the force equals the ratio of the density ,so

[tex]\begin{gathered} \frac{F_{g(air)}}{F_B}=\frac{\rho_{material}}{\rho_{alcholol}} \\ replace \\ \frac{900\text{ N}}{500\text{ N}}=\frac{\rho_{material}}{0.7\text{ }\frac{g}{cm^3}} \\ mutliply\text{ both sides by 0.7}\frac{g}{cm^3} \\ \frac{900\text{N}}{500\text{N}}*0.7\text{ }\frac{g}{cm^3}=\frac{\rho_{mater\imaginaryI al}}{0.7\text{\frac{g}{cm^{3}}}}*0.7\frac{g}{cm^3} \\ 1.26\frac{g}{cm^3}=\text{ density of the material } \\ \end{gathered}[/tex]

so, the density of the material is

1.26 g/cm³

I hope this helps you

A.Calculate the combined force of vector F ?B.Calculate the direction of the combined force vector F ?

Answers

Answer:

A. 282.93 N

B. 1.94 degrees

Explanation:

The combined force is found by first adding the three forces given.

We add the three forces by adding their x and y components separately and then combining the results to produce the total force,

The x component of a force is

[tex]\begin{gathered} \cos \theta=\frac{f_x}{F} \\ \Rightarrow f_x=F\cos \theta \end{gathered}[/tex]

Therefore, x components of the forces is

[tex]F_x=120\cos 65+100\cos 25+200\cos (-45)[/tex]

The y-component of the forces is

[tex]F_y=120\sin 120+100\sin 25+200\sin (-45)[/tex]

Now evaluating the above two components gives

[tex]F_x=282.77N[/tex][tex]F_y=9.597N[/tex]

Let us draw on big vector whose components are the above vectors.

The angle of the combined vector with respect to the x-axis is

[tex]\tan \theta=\frac{9.59}{282.77}[/tex][tex]\theta=\tan ^{-1}(\frac{9.59}{282.77})[/tex][tex]\boxed{\theta=1.94^o}[/tex]

which is our answer!

The magnitude of the combined vector is

[tex]F=\sqrt[]{F^2_x+F^2_y_{}}[/tex][tex]F=\sqrt[]{(9.59)^2_{}+(282.77)^2_{}}[/tex][tex]\boxed{F=282.93N}[/tex]

which is our answer!

Hence, to summerise:

A. 282.93 N

B. 1.94 degrees

Please help me with this!

Answers

The balloon goes 25 m east in first 10 s.

Then the wind blows the balloon 35 m west in 8 s.

a

The balloon travels 25 m in 10s.

b

the balloon travels a distance 35 m in next 8 s.

c

thhe total distance travelled by the balloon is,

[tex]\begin{gathered} d=25+35 \\ =60\text{ m} \end{gathered}[/tex]

d

Average speed in 10 s is,

[tex]\begin{gathered} s=\frac{25}{10} \\ =2.5\text{ m/s} \end{gathered}[/tex]

e

The average speed in next 8 s is,

[tex]\begin{gathered} s_8=\frac{35}{8} \\ =4.37\text{ m/s} \end{gathered}[/tex]

f

the average speed for the entire trip is,

[tex]\begin{gathered} s_{av}=\frac{25+35}{10+8} \\ =3.33\text{ m/s} \end{gathered}[/tex]

g

displacement during the first 10 s is.

[tex]d_1=(25m)\hat{i}[/tex]

h.

displacement during next 8 s is,

[tex]d_2=(-35m)\hat{i}[/tex]

i

The total displacement is,

[tex]\begin{gathered} d_1+d_2=(25-35)\hat{i}_{} \\ =-(10m)\hat{i} \end{gathered}[/tex]

j

the average velocity in 10 s is,

[tex]\begin{gathered} v_1=\frac{25m}{10}\hat{i} \\ =(2.5m)\hat{i} \end{gathered}[/tex]

k

The average velocity in 8 s is,

[tex]\begin{gathered} v_2=\frac{-35\text{ m}}{8\text{ s}}\hat{i} \\ =-(4.37m)\hat{i} \end{gathered}[/tex]

l

The average velocity entire the whole trip is,

[tex]\begin{gathered} v_{av}=\frac{d_1+d_2}{18} \\ =\frac{-10m\hat{i}}{18\text{ s}} \\ =-(0.55\hat{m/s)i} \end{gathered}[/tex]

Three resistors with resistances of 9 Ω, 18 Ω, and 30 Ω are in a series circuit with a 12 volt battery. What is the total resistance of the resistors in the circuit?

Answers

When resistors are in series, the total resistance is equal to the sum of the individual resistances of each resistor. Thus, the total resistance in this circuit is 9+18+30 = 57Ω

orce and Motion Unit TestUse the following scenario to answer the question.Taj and Micah chose to go bowling. Taj rolled the ball toward the pins first, knocking them all down.Which of the following is affecting these objects?point)O Gravity is affecting these objects.O An unbalanced force is affecting the objects.O Inertia is affecting these objects.O A balanced force is affecting the objects.

Answers

So lets go through all four answer choices.

The easiest to choose is whether gravity is affecting these objects. Assuming that there is some sort of gravity that would pull the pins down, gravity does affect these objects

Second is inertia.

We know that if an object has inertia, it will try to resist moving/coming to rest. In this case, we know that the pins have inertia because the pins fell over, so we know that the pins do have inertia

The last part is whether these objects have an unbalanced or balanced force. If a balanced force did exist, there would need to a force that would equally counteract the force of the bowling ball, which there isn't. Which means there is an unbalanced force affecting these objects.

Given that the user must choose one, the correct answer would be that an unbla

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