I need to know how to do these and the formula too.

I Need To Know How To Do These And The Formula Too.

Answers

Answer 1

Answer:

a = 4m kgm/s & b. v = 4 m/s

Explanation:

a:

A is approaching in the opposite direction of B

So let's take the velocity of A as negative.

Here we'd need to solve this using conservation of momentum.

m1u1 + m2u2 = m1v1 + m2v2

So initial momentum = (m X (-4)) + (m X 8) = -4m + 8m = 4m

b:

After collision sphere A stops, so v = 0 for sphere A

Using the conservation of momentum formula, we get

(m X (-4)) + (m X 8) = (0 X m) + (m X v) [let the final velocity of B be v]

⇒ 4m = vm

⇒ v = 4 m/s


Related Questions

When you look at an image of yourself in a plane mirror, the image is:A.real.B.enlarged.C.reduced.D.virtual.

Answers

Since the light does not pass through the object the image will be virtual . [Option D]

A race car traveling northward on a straight, level track at a constant speed travels 0.739 km in 18.0 s. The return trip over the same track is made in 23.0 s.What is the average velocity (in m/s) for the total trip?answer in : m/s

Answers

The average velocity of an object is given by:

[tex]\begin{gathered} v=\frac{\Delta x}{t} \\ \text{ where }\Delta x\text{ is the displacement and }t\text{ is the time } \end{gathered}[/tex]

In this case we know that the car travelled a track and the return on it; this means that the displacement of the car is zero and therefore, the average velocity is zero

A 1-cm high object is placed 10 cm in front of a concave mirror with a focal length 15 cm. Determine the image distance and the magnification. Enumerate the characteristics of the image produced. Must follow GRESA format.Given:Required:Equation(s) Needed:Solution:(a)(b)(c)Final answer:L-ocationO-rientation:S-izeT-ype:

Answers

Given:

object size: 1 cm

object distance (do): 10 cm

mirror focal length (f): 15 cm

Required:

Image distance (di),

magnification

image characteristics.

Equations needed:

[tex]\begin{gathered} \frac{1}{d_i}+\frac{1}{d_o}=\frac{1}{f} \\ \text{magnification}=-\frac{d_i}{d_o} \end{gathered}[/tex]

Solution:

[tex]\begin{gathered} \frac{1}{d_i}+\frac{1}{10}=\frac{1}{15} \\ \frac{1}{d_i}=\frac{1}{15}-\frac{1}{10} \\ \frac{1}{d_i}=\frac{2}{30}-\frac{3}{30} \\ \frac{1}{d_i}=\frac{-1}{30}_{} \\ d_i=-30\text{ cm} \\ \\ \text{magnification}=-\frac{-30}{10}=3 \end{gathered}[/tex]

Location: 30 cm "inside" the mirror (di = -30)

Orientation: Upright (magnification > 0)

Size: 3 cm (object size multiplied by magnification)

Type: virtual image (di < 0)

If you have to face south east to see a star, which is the azimuth?A.90 degreesB.135 degreesC.180 degreesD.195 degrees

Answers

The azimuth is the angle measured in clockwise from the north.

The given system can be represented as,

From the diagram,

The angle 90 degree from the north point towards the east.

The angle 180 degree from the north point towards the south.

The angle 195 degree from the north point towards south west.

The angle 135 degree from the north is in between the east and the south.

Thus, the azimuth to see the star in the south east is 135 degrees.

Hence, option B is the correct answer.

Can you please help me with #6 (the 6 was wrong in the question it meant #5 as you can see I tried to change it)

Answers

[tex]\begin{gathered} \text{Question 5)} \\ Q=52,000J \\ m=2kg \\ C=130\text{ J/Kg K} \\ \Delta T=\text{?} \\ Q=Cm\Delta T \\ \text{Solving }\Delta T \\ \Delta T=\frac{Q}{Cm} \\ \Delta T=\frac{52,000J}{(130\text{ J/Kg K)(}2kg\text{)}} \\ \Delta T=200K,\text{ but Kelving and }degree\text{ Celsius have the same change} \\ \text{Hence} \\ \Delta T=200\text{ \degree{}C} \\ \\ \text{Question 6} \\ T1=350\text{ K} \\ \Delta T=200\text{ K} \\ T2=\text{?} \\ \Delta T=T2-T1 \\ \text{Sol}v\text{ing T2} \\ T2=\Delta T+T1 \\ T2=200\text{ K}+350\text{ K} \\ T2=550\text{ K} \\ \text{The final temperature is }550\text{ K} \end{gathered}[/tex]

You drop a rock down a well that is 11.5m deep. How long does it take the rock to hit the bottom of the well?
a. 2.3 s
b. 2.8 s
c. 1.5 s
d. 3.1 s

Answers

A 2.3 s have a good day

A beam of microwaves has a frequency of 3.0*10^ 11Hz. Radio waves have a frequency of 3.0*10^9 Hz. Calculate the wavelength of each. Which type of radiation has the longer wavelength?

Answers

We will have the following:

Assuming both are traveling in vacuum we will have the following:

Microwaves:

[tex]\lambda_m=\frac{3.00\cdot10^8}{3.0\cdot10^{11}}\Rightarrow\lambda_m=1\cdot10^{-3}m[/tex]

Radio:

[tex]\lambda_r=\frac{3.0\cdot10^8}{3.0\cdot10^9}\Rightarrow\lambda_r=1\cdot10^{-1}m[/tex]

So, radio waves have a longer wavelength.

According to Bernoulli's principle, what type of relationship exists between a fluid's velocity and its pressure?O an inverse relationshipO a direct relationshipO a balanced relationshipO an unbalanced relationship

Answers

To find:

The relation between the pressure and the velocity of the fluid.

Explanation:

The Bernoulli's principle states that the increase in the velocity of the fluid occurs simultaneously with the decrease in the static pressure of the fluid.

That is, the velocity of the fluid is inversely proportioinal to the pressure of the fluid.

Final answer:

Thus the relationship between a fluid's velocity and its pressure is an inverse relationship.

Therefore, the correct answer is option A.

A cart of mass m is attached to a perfect spring and oscillates on a frictionless surface, as shown. The amplitude of the oscillation is given by x. The kinetic energy of the cart has a maximum value at(a) position A(b) position B(c) position C(d) positions A and C

Answers

Kinetic energy is the energy an object possess during motion. Therefore, the kinetic energy of the cart reaches it maximum kinetic energy as the cart passes through the equilibrium position. Therefore, the position is at C.

Do you know the answer for these? [ showing the work] I'm trying to figure out the answers to the bullet points under your mission section.

Answers

Answer:

Explanation:

• Orbital Radius,:

We are already told that the altitude of the satellite is 300,000 meters. Having this information in hand, it is easy to find the radius of orbit using the following relation:

[tex]R_{\text{satellite}}=A+R_{\text{earth}}[/tex]

where A is the altitude.

Now we know that A = 300,000 m and earth radius = 6.37 * 10^6 m; therefore, the orbital radius of the satellite is

[tex]R_{\text{satellite}}=3\cdot10^6+6.37\cdot10^6[/tex]

[tex]\boxed{R_{\text{satellite}}=9.37\times10^6m}[/tex]

which is our answer!

• Velocity:

We are told that the velocity of the satellite is given by

[tex]v=\sqrt[]{G\frac{m_E}{R_{\text{satellite}}}}[/tex]

where G is the gravitational constant and m_E is the mass of the earth.

Substituting the numerical values for these constants gives

[tex]v=\sqrt[]{(6.67\times10^{-11})\frac{5.98\times10^{24}}{9.37\times10^6}}[/tex]

Using a calculator we evaluate the above to be:

[tex]\boxed{v=6.52\cdot10^3m/s}[/tex]

which is around 6.5 km per second!

• Orbital Period:

The orbital period T of the satellite is given by

[tex]T=2\pi\sqrt[]{\frac{R^3_{satellite}}{Gm_E}}[/tex]

putting in the numerical values for the constants gives

[tex]T=2\pi\sqrt[]{\frac{(9.37\times10^6)^3}{(6.67\times10^{-11})(5.98\times10^{24})}}[/tex][tex]\boxed{T=9024s\approx2.5hr}[/tex]

Hence, the period of satellites orbit is only 2.5 hours! This means that we can see the same satellite multiple times in the night sky if it is observable!

• Orbital Path:

The problem with satellites is that since they are travelling so fast, they don't get to observe one location on earth for a long time. One solution to this is to place the satellites into something called the geosynchronous orbit. In such an orbit, the period of the satellite matches the earth's period of rotation. This way, when observed from the earth, the satellite looks stationary, but in fact, it is travelling with the earth in the same orbital period. Such a satellite can be launched to observe locations along the arctic and the antarctic circles to obtain substantial data.

A bumper car with a mass of 150 kg is moving forward at 10m/s. It collides with a stationary bumper car with a mass of 209 kg. After the collision, the 200 kg bumper car moves forward at 8.0 m/s. Does the 150 bumper car continue to move forward or backward after the collision?

Answers

Given,

Mass of the bumper car is 150 kg.

It moves with a speed 10 m//s.

The mass of the stationary bumper car is 200 kg.

After collision,

The 200 kg bumper moves it a velocity of 8.0 m/s.

By conservation of mass,

[tex]\begin{gathered} 150\times10=200\times8+150v \\ \Rightarrow v=-0.66\text{ m/s} \end{gathered}[/tex]

Thus the 150 kg bumper car moves backward after collision.

Hey I need help with my homework not understanding the problem can someone please help

Answers

Answer:

a. 4 g/mL

Explanation:

First, we need to find the volume of the object. The volume is equal to the volume of the water displaced. So, it can be calculated as the difference in the volumes

75 mL - 50 mL = 25 mL

Then, the volume of the object is 25 mL and the mass is 100 g. Now, we can calculate the density as follows

density = mass/volume

density = 100 g/25 mL

density = 4 g/mL

Therefore, the answer is

a. 4 g/mL

What is the total mechanical energy of a 3 kg object traveling at 7 m/s if its potential energy is 42 J?

Answers

The mechanical energy of the a 3kg object at 7m/s and having a potential energy of 42J is 115.5 J.

What is the mechanical energy of the object?

Mechanical energy is simply energy accumulated due to performing some particular work. It is the sum of the potential energy and kinetic energy of an object. It is expressed as;

Mechanical energy = Potential energy + Kinetic energy

Given the data in the question;

Mass of the object m = 3kgVelocity of the object v = 7m/sPotential energy of the object = 42J

First, we determine the kinetic energy of the object.

Kinetic energy = 1/2 × mass × ( velocity )²

Plug in the given values

Kinetic energy = 1/2 × 3kg × ( 7m/s)²

Kinetic energy = 1/2 × 3kg × 49m²/s²

Kinetic energy = 73.5kgm²/s²

Kinetic energy = 73.5J

Now, the mechanical energy of the object will be;

Mechanical energy = Potential energy + Kinetic energy

Plug in the values

Mechanical energy = 42J + 73.5J

Mechanical energy = 115.5 J

Therefore, the mechanical energy is 115.5 Joules.

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Answer:

The answer is 115.5 guyssss

Explanation:

I need help with all these all in da bottomA bike moves 50 m in 5 seconds. calculate the speed

Answers

Answer:

Step 1

S = ?

ΔD = 50 m

ΔT = 5 s

Step 2

S = ΔD/ΔT

Step 3

S = ΔD/ΔT

Step 4

S = (50 m)/(5 s)

Step 5

S = 10 m/s

Explanation:

Step 1. variables.

The variables are the speed S, the distance traveled ΔD, and the time ΔT. Taking into account the given information, the speed is unknown, the distance traveled is 50 m and the time is 5 s. so:

S = ?

ΔD = 50 m

ΔT = 5 s

Step 2. Formula

We will use the following formula

S = ΔD/ΔT

Step 3. Solve for the unknown variable

Since the unknown variable is S, the equation is the same, so

S = ΔD/ΔT

Step 4. Plug in number

Now, we need to replace ΔD = 50 m and ΔT = 5 s on the equation

S = (50 m) /(5s)

Step 5. Calculate the answer

Then, the speed is equal to

S = 10 m/s

Therefore, the answer is:

10 m/s

42) When you weigh yourself on good old terra firma (solid ground), your weight is 142 lb. In an elevator your apparent weight is 121 lb. What are the direction and magnitude of the elevator’s acceleration?

Answers

We are given the following information

Actual weight = 142 lb.

Apparent weight = 121 lb.

Direction of the elevator's acceleration:

Notice that the apparent weight (121 lb.) is less than the actual weight (142 lb.)

This means that the direction of the elevator's acceleration must be downward since a downward acceleration

results in an apparent weight less than the actual weight.

Magnitude of the elevator’s acceleration:

There are two forces acting on you in the elevator.

The applied force of scale acts upward and the force due to gravity acts downward.

Let us apply Newton's 2nd law of motion to find the magnitude of the elevator’s acceleration.

[tex]\begin{gathered} \sum F_y=m\cdot a \\ W_a-W=m\cdot a \\ a=\frac{W_a-W}{m} \\ a=\frac{W_a-W}{\frac{W}{g}} \\ a=\frac{W_a-W}{W}\cdot g \\ a=\frac{121-142}{142}\cdot9.81 \\ a=-1.5\; \; \frac{m}{s^2} \end{gathered}[/tex]

Notice that the negative sign represents a downward acceleration as we discussed earlier.

When we are interested in the magnitude, we only consider the numerical value, not the direction.

Therefore, the magnitude of the elevator’s acceleration is 1.5 m/s²

Technically, how would you know if any work was done on an object? Work referring to the physics Term

Answers

ANSWER and EXPLANATION

We want to know how to tell if work is done on an object.

Work is a result of a force acting on an object that thereby causes a displacement of the object. It is defined mathematically as:

[tex]W=F\cdot d[/tex]

where F = force; d = displacement

Therefore, we say that work is done on an object when a force is applied to the object and it causes the object to move a certain distance.

True or False ?7. When two spheres with equal masses collide centrally and elastically, the spheres exchange their kinetic energies. 8. In each impact, the momentum of the colliding body system is marginally the same before and immediately after the impact, but varies during the impact. 

Answers

(7)

When two spheres with equal masses collide centrally and elastically, the spheres exchange their kinetic energies. This statement is a false statement.

Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 56.46 N when they are separated by 50.91 cm. What is the magnitude of the charges in microCoulombs?

Answers

Answer:

q = 40.3 microCoulombs

Explanation:

The force of attraction between two charges is given by the formula:

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

The force of attraction. F = 56.46 N

The separation, r = 50.91 cm

r = 50.91/100

r = 0.5091 m

The electrostatic constant is:

[tex]k=9\times10^9Nm^2C^{-2}[/tex]

Solve for the magnitude of charge q

[tex]\begin{gathered} F=\frac{kq^2}{r^2} \\ \\ 56.46=\frac{9\times10^9\times q^2}{0.5091^2} \\ \\ 56.46\times0.5091^2=9\times10^9\times q^2 \\ \\ q^2=\frac{56.46\times0.5091^2}{9\times10^9} \\ \\ q^2=1.63\times10^{-9} \\ \\ q=\sqrt{1.63\times10^{-9}} \\ \\ q=4.03\times10^{-5}C \\ \\ q=40.3\mu C \end{gathered}[/tex]

i) How many grams of CO₂ is formed when 73.9g of propane is burnt? C3H8+ O2 CO2 + H₂O​

Answers

73.9g CO₂ is formed when 73.9g  or 1 mole of propane is burnt

C₃H₈+ 5O₂ ⇄ 3CO₂ + 4H₂O​

1 mole C₃H₈ gives 3 moles of CO₂

44 g C₃H₈ gives 44 g CO₂

73.9g  C₃H₈ gives 73.9g CO₂

Quantifying the amount of a substance is made easier with the help of the mole concept. It is understood that even a single gram of a pure element has a sizable number of atoms when dealing with particles at the atomic (or molecular) level. In this regard, the mole notion is frequently used. The "mole," which is a count of a large number of particles, is the most commonly used unit of measurement.

The letter NA is typically used to indicate the value 6.02214076*1023, also known as the Avogadro constant. Atoms, molecules, monoatomic and polyatomic ions, as well as other particles, are among the elementary entities that can be represented in moles (such as electrons).

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Lighting seeks the most direct path to the ground. True O False

Answers

Lighting is basically the discharge of electricity that does not care where it strikes. It basically seeks the shortest or the most direct path to the ground.

Therefore, the given statement is a true statement.

Answer:

TRUE

Explanation:

what is the theory E=MC^2?

Answers

E = M C ^2 is Albert Einstein's theory of special relativity.

It states that the mass multiplied by the speed of light squared equal to the kinetic energy of that body.

m= mass

C = speed of light

Negatively charged particles in the out most energy levels of the electron cloud ?

Answers

Electrons are negatively charged particles in the out most energy levels of the electron cloud

An atom is made up of 3 particles , which are electron , proton and neutron . Where electron is a negatively charged particle and revolves around the nucleus in definite orbits whereas protons and neutrons are positively charged particles and neutral respectively ., which resides inside the nucleus

Electrons are negatively charged subatomic particles found in the outermost regions of atoms .

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A spring is compressed between two carts, one with a mass of 5kg and the other with a 15 kg. Which will move away with higher velocity? Assume the momentum of the first cart is equal to the momentum of the second cart. V1and V2are the velocities of cart 1 and cart 2 respectively

Answers

Given,

The mass of the cart 1, m₁=5 kg

The mass of the cart 2, m₂=15 kg

The velocity of cart 1 is V₁

The velocity of cart 2 is V₂

The momentum of both carts is the same.

The momentum of an object is the product of the mass of the object and its velocity.

As the momentum of both carts is the same we have,

[tex]m_1V_1=m_2V_2[/tex]

On substituting the known values,

[tex]\begin{gathered} 5V_1=15V_2 \\ \Rightarrow V_1=3V_2 \end{gathered}[/tex]

Thus the velocity of cart 1 is three times the velocity of cart 2.

Therefore, cart 1 will move away with a higher velocity.

A model rocket is launched with an initial upward velocity of 215 ft/s. The rockets height h in feet) after t seconds is given by the following.h=215t-16t^2Find all values for t for which the rockets height is 97 feet.Round your answer(s) to the nearest hundredth.

Answers

Given,

The initial velocity of the rocket, u=215 ft/s

The equation which gives the height of the rocket at any given instant of time,

[tex]h=215t-16t^2[/tex]

The height of the rocket, h=97 ft.

The given equation is a quadratic equation. On rearranging the above equation and substituting the value of h,

[tex]\begin{gathered} 16t^2-215t+h=0 \\ 16t^2-t+97=0 \end{gathered}[/tex]

A quadratic equation is written as,

[tex]ax^2+bx+c=0[/tex]

Comparing the two equations,

x=t, a=16, b=-215, c=97.

The solution of a quadratic equation is given by,

[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]

On substituting the known values in the above equation,

[tex]\begin{gathered} t=\frac{-(-215)\pm\sqrt[]{(-215)^2-4\times16\times97}}{2\times16} \\ =\frac{215\pm\sqrt[]{40017}}{32} \\ =12.97\text{ or}\pm0.47\text{ } \end{gathered}[/tex]

Thus, the values of t are 12.97 s or 0.47 s.

That is, the rocket will be at a height of 97 feet at t=12.97 s or t=0.47 s

Sound travels with an average speed of 340m/s. Lightning from a distant storm cloud is seen almost immediately. If the sound reaches our ear 2s later. how far is the storm​

Answers

Hello..!

To calculate its distance we apply the following data.

Data:

V = 340m/sT = 2sD = ?

Then, we apply the formula that is.

Formula:

D = V • T

Finally we develop the problem.

Developing:

D = (340m/s) • (2s)D = 680m

The storm is at a distance of 680 meters

Answer:

Formula

D = V • T

Resolve:

D = (340m/s) • (2s)D = 680m

Hope this helped

What is the wavelength of a photon of EMR with a frequency of 3.24x10¹3 Hz?9.26x1020 m9.72x1021 m1.08x105 Hz9.26x10-6 m

Answers

The wavelength is given by:

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

where v is the velocity and f is the frequency. In this case the velocity is that of light since this is an electromagnetic radiation and the frequency is given, then we have:

[tex]\begin{gathered} \lambda=\frac{3\times10^8\text{ m/s}}{3.24\times10^{13}\text{ Hz}} \\ \lambda=9.26\times10^{-6}\text{ m} \end{gathered}[/tex]

Therefore, the wavelength is:

[tex]\lambda=9.26\times10^{-6}\text{m}[/tex]

PLEASE HELP MEEEE!!!! ill give 50 points!!
an astronaut travels to the volcanic moon of Jupiter, lo. there she finds that her scale reading has decreased from 490N(Earth) to 90N(lo). what is the gravitational force constant (g) of lo?

Answers

The gravitational force constant (g) of lo is the scale reading has decreased from 490N(Earth) to 90N(lo) is 1.8 m / s²

F = m g

F = Force

m = Mass

g = Acceleration due to gravity

In Earth,

F = 490 N

g = 9.8 m / s²

m = F / g

m = 490 / 9.8

m = 50 kg

In Io,

F = 90 N

g = F / m

g = 90 / 50

g = 1.8 m / s²

Therefore, the gravitational force constant (g) of lo is 1.8 m / s²

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A car speeds up uniformly while rounding a turn. The car's angular speed increases from 0.54 radians per second to 0.96 radians per second as it turns through 1.40 radians. What is the car's angular acceleration? Include units in your answer. Answer must be in 3 significant digits.

Answers

Given data:

* The initial angular speed of the car is,

[tex]\omega_i=0.54\text{ rad/s}[/tex]

* The final angular speed of the car is,

[tex]\omega_f=0.96\text{ rad/s}[/tex]

* The angular displacement of the car is,

[tex]\theta=1.4\text{ radians}[/tex]

Solution:

By the kinematics equation, the angular acceleration of the car in terms of the angular displacement is,

[tex]\omega^2_{\text{f}}-\omega^2_i=2\alpha\theta[/tex]

where,

[tex]\alpha\text{ is the angular acceleration,}[/tex]

Substituting the known values,

[tex]\begin{gathered} 0.96^2-0.54^2=2\times\alpha\times1.4 \\ \alpha=\frac{0.96^2-0.54^2}{2\times1.4} \\ \alpha=\frac{0.9216-0.2916}{2.8} \\ \alpha=\frac{0.63}{2.8} \end{gathered}[/tex]

By simplifying,

[tex]\alpha=0.225rads^{-2}[/tex]

Thus, the angular acceleration of the car is 0.225 radians per second squared.

What is the wavelength range of electromagnetic radiation thehuman eye can detect?a 4 x 10-7 m to 8 x 10-7 mb 9 x 10-8 m to 11 x 10-9 mC 1x 10-4 m to 3 x 10-5 md 6 x 10-5 m to 7 x 10-5 m

Answers

The wavelength range of electromagnetic radiation the human eye can detect is:

a 4 x 10-7 m to 8 x 10-7 m

An object has a kinetic energy of 12,000 J and a mass of 48.7 kg. Find the velocity of the object.I need help with this problem pls

Answers

Given:

The kinetic energy of the object is K = 12000 J.

The mass of the object is m = 48.7 kg

To find the velocity of the object.

Explanation:

The velocity of the object will be

[tex]\begin{gathered} K=\frac{1}{2}mv^2 \\ v=\sqrt{\frac{2K}{m}} \\ =\sqrt{\frac{2\times12000}{48.7}} \\ =22.199\text{ m/s} \end{gathered}[/tex]

Thus, the velocity of the object is 22.199 m/s.

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