g A 320-g air track cart is traveling at 1.25 m/s and a 270-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the center of mass of the two carts

Answer:

speed = 72.75km/hr

Explanation:

Let us first determine the Kinetic energy (KE) of the 18000kg truck

[tex]KE = \frac{1}{2} mv^2\\where:\\m = mass = 18000\\v = velocity = 21km/hr\\\therefore KE = \frac{1}{2} \times 18000 \times (21)^2\\= 9000 \times 441\\KE = 3,969,000\ Joules\\= 3,969\ KJ[/tex]

Next we will substitute this value of kinetic energy into the KE equation for the 1500kg compact car

[tex]KE = \frac{1}{2} mv^2\\3969000 = \frac{1}{2} \times 1500 \times v^2\\3969000 = 750 \times v^2\\v^2 = \frac{3969000}{750} = 5292\\v = \sqrt{5292} \\v = 72.75km/hr[/tex]

The speed at which the 1500 kg compact car have the same kinetic energy as the truck is 73 km/h

The kinetic energy of an object is known to be the energy at work and in motion. It is usually expressed as;

[tex]\mathbf{K.E = \dfrac{1}{2}mv^2}[/tex]

To determine the speed at which the car will have the same kinetic energy as the truck, we can say that:

[tex]\mathbf{\implies \dfrac{1}{2}m_c v_c^2= \dfrac{1}{2}m_t v_t^2}[/tex]

∴

[tex]\mathbf{\implies \dfrac{1}{2}\times 1500\times v_c^2= \dfrac{1}{2}\times 18000 \times 21^2}[/tex]

[tex]\mathbf{v_c^2= \dfrac{3969000}{ 750}}[/tex]

[tex]\mathbf{v_c^2=5292}[/tex]

[tex]\mathbf{v_c=\sqrt{5292}}[/tex]

[tex]\mathbf{v_c\simeq 73 km/h}[/tex]

Therefore, we can conclude that the speed at which the 1500 kg compact car have the same kinetic energy as the truck is 73 km/h

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Answer:

The average force ≅ 519.44 N.

Explanation:

Impulse = change in momentum of a body

i.e Ft = m(v - u)

where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.

m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s

So that,

F x 0.00360 = 0.055(34 - 0)

F x 0.00360 = 0.055 x 34

                    = 1.87

F = [tex]\frac{1.87}{0.0036}[/tex]

 = 519.4444

The average force exerted on the ball by the club is approximately 519.44 N.

Answer:

Work done, W = 24.9001 J

Explanation:

Given that,

Force, F = (2xî + 4yĵ) Where x and y are in meters

This force acts on an object as the object moves in the x-direction from the origin to x = 4.99 m.

We need to find the work done by the force on the object. It is given by :

[tex]W=\int\limits^a_b {F dr} \\\\W=\int\limits^{4.99}_0 {(2xi+4yj)dx} \\\\W=\int\limits^{4.99}_0 {2xi\ dx} \\\\W=x^2|_0^{4.99}\\\\\text{Applying limits}\\\\W=(4.99)^2-0^2\\\\W=24.9001\ J[/tex]

So, the work done is 24.9001 J.

Answer:

The time taken before the phone hits the ground is 1.4 s.

Explanation:

Given;

height of the rock wall, h = 10 m

speed of the climber, v = 0.25 m/s

The time taken before the phone hits the ground is given;

h = ut + ¹/₂gt²

10 = 0.25t + ¹/₂(9.8)t²

20 = 0.5t + 9.8t²

9.8t² + 0.5t - 20 = 0

solving this quadratic equation;

t = 1.4 s

Therefore, the time taken before the phone hits the ground is 1.4 s.

Answer:

7.2 × 10^5 N

Explanation:

The first step is to convert 1400 km/hr to m/s

= 1,400×1000/3600

= 1,400,000/3600

= 388.88 m/s

The acceleration can be calculated as follows

a= v-u/t

= 388.88/2.1

= 185.18

Therefore the required net force can be calculated as follow

= 388.88 × 185.18

= 7.2 × 10^5 N

Answer:

The average speed is 22.2 km/h

Explanation:

Average Speed

Given an object travels a total distance d and took a total time t, then the average speed is:

[tex]\displaystyle \bar v=\frac{d}{t}[/tex]

The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:

[tex]\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h[/tex]

Then he drives d2=7 km at v2=43 km/h taking a time of:

[tex]\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h[/tex]

The total time is

t=0.467 h + 0.163 h = 0.63 h

The total distance is

d = 7 km + 7 km = 14 km

The average speed is:

[tex]\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h[/tex]

The average speed is 22.2 km/h

Answer: wow that’s hard

Explanation:

Answer:

3.56s

Explanation:

Ya that's the right answer.

Explanation:

so sorry

don't know but please mark me as brainliest please

Answer:

The value  is  [tex]\lambda = 5.30 *10^{-10 } \ m[/tex]

Explanation:

From the question we are told that

    The velocity of the electron is  [tex]v = 1.37 *10^{6} \ m/s[/tex]

    The mass of the electron is  [tex]m = 9.11 *10^{-28} \ g = 9.11 *10^{-31} \ kg[/tex]

Generally the  deBroglie wavelength  is mathematically represented as

       [tex]\lambda = \frac{h}{mv}[/tex]

Here h is the Planck'c constant with value  [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

So

      [tex]\lambda = \frac{6.62607015 * 10^{-34} }{9.11 *10^{-31} * 1.37 *10^{6}}[/tex]

=>   [tex]\lambda = 5.30 *10^{-10 } \ m[/tex]

Answer:

[tex]5\sqrt{2} \ N[/tex]

Explanation:

It is given that,

A 5-newton force directed east and a 5-newton force directed  north.

We need to find the resultant of the two forces.

As the two forces are acting in perpendicular to each other. The resultant of two such forces is given by :

[tex]F=\sqrt{5^2+5^2} \\\\=5\sqrt{2} \ N[/tex]

So, the resulatnt force is [tex]5\sqrt{2} \ N[/tex]

Answer: I don't  do   adopt me   but I do have a Ro.blox username its  Nvm i cant but my user name

Explanation:

Answer:

adopt me sucks but my best pet was an albino monkey I think or a king monkey

Explanation:

Answer:

manual

Explanation:

A manual distraction is when a driver takes their hand(s) off the wheel for any reason, for any amount of time. Without both hands on the wheel, your reaction time suffers, and so does your ability to steer.

Answer:

7 because salt lake and Southis weat

Answer:

4.5s

Explanation:

u=30m/s

v=50m/s

s=180m

a=constant(given)

By third equation of motion,    v  

2

=u  

2

+2as

(50)  

2

=(30)  

2

+2a(180)

a=  

3

40

​  

m/s  

2

By first equation of motion,    v=u+at

50=30+(  

9

40

​  

)t

t=  

2

9

​  

=4.5sec

Answer: what’s the answer

Explanation:

Answer:

6.601 N/m.

Explanation:

Generally, the force on a wire carrying current in a magnetic field can be determined by:

F = BIL Sinθ

where: F is the force, B is the value of the magnetic field, I is the value of current in the wire and θ is the angle between the direction of current and magnetic field.

For force per meter of length, we have;

[tex]\frac{F}{L}[/tex] = BI Sinθ

Given: I = 8.05 A, B = 0.82 T and θ = [tex]90^{o}[/tex](since they are perpendicular).

Then;

[tex]\frac{F}{L}[/tex] = 0.82 x 8.05 x Sin [tex]90^{o}[/tex]

  = 6.601

  = 6.601 N/m

the force per meter length is 6.601 N/m.

The required magnitude of force per meter of wire is of 6.601 N/m.

Given data:

The magnitude of current in a wire is, I = 8.05 A.

The strength of magnetic field is, B = 0.82 T.

We known that, in a magnetic field there is always a force acting on a wire kept in the region. This force is known as magnetic force, and the expression for the magnetic force is given as,

[tex]F = B \times I \times L \times sin \theta[/tex]

Now,

For the force per unit length, and current being perpendicular to magnetic field,

[tex]\dfrac{F}{L} = B \times I \times sin90\\\\\dfrac{F}{L} = 0.82 \times 8.05 \times sin90\\\\\dfrac{F}{L} = 6.601 \;\rm N/m[/tex]

Thus, we can conclude that the required magnitude of force per meter of wire is of 6.601 N/m.

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Answer:

ice (solid), water (liquid) and vapor (gas)

Explanation:

Answer:

B. symbolic

Explanation:

The mental image formed in Terell's mind is a symbolic mental image.

A symbolic mental image helps to bring perception to imagery formed on the mind.

Answer:

Terell lives in Washington, D.C., where he can see the Potomac River from his bedroom window. When he sees this river, Terell thinks of swimming, boating, fun, and friends. What type of mental image is this?

symbolic

Explanation:

Answer:

The value is  [tex]I = 0.0932 ML ^2[/tex]  

Explanation:

From the question we are told that

  The rotational inertia about one end is [tex]I_R = \frac{1}{3} ML^2[/tex]

   The location of the axis of rotation considered is [tex]d = 0.4 L[/tex]

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is  [tex]0.4 M[/tex]

Generally the length of the rod from the its beginning to the axis of rotation consider is

      [tex]k = 1 - 0.4 L = 0.6L[/tex]

Generally the mass of the portion  of the rod from the its beginning to the axis of rotation consider is

    [tex]m = 1- 0.4 M = 0.6 M[/tex]

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

     [tex]I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2[/tex]

    [tex]I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2[/tex]

Generally the rotational inertia about the axis of rotation consider for the second  portion of the rod is

     [tex]I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2[/tex]

=> [tex]I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2[/tex]

=>  [tex]I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2[/tex]

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

  [tex]I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2[/tex]

=>   [tex]I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ][/tex]

=>   [tex]I = 0.0932 ML ^2[/tex]  

Answer:

Lunar and solar eclipses occur with about equal frequency. Lunar eclipses are more widely visible because Earth casts a much larger shadow on the Moon during a lunar eclipse than the Moon casts on Earth during a solar eclipse. As a result, you are more likely to see a lunar eclipse than a solar eclipse.

For a total eclipse to take place, the sun, moon and Earth must be in a direct line. The second type of solar eclipse is a partial solar eclipse. This happens when the sun, moon and Earth are not exactly lined up. The sun appears to have a dark shadow on only a small part of its surface.

For a lunar eclipse to occur, the Sun, Earth, and Moon must be roughly aligned in a line

Answer:

D . The arrows on plates A and B would show the plated moving away from each other

Explanation:

The correct arrow in this instance will show that the plates A and B would be moving away from each other.

At the mid-ocean ridge, two plates are moving apart and pulling away from one another.

Answer: Work done is  - 501.56 J

Explanation:

Given that;

External pressure P = 15.0 atm

Volume V1 = 0.120 liters

Volume V2 = 0.450 liters

Work done = ?

we know that; Work = -Pdv

where P is pressure and dv is change in volume

so we substitute our values into the equation

Work = -15.0 × ( 0.450 - 0.120)

= -15 × 0.33

= - 4.95 atm/L

we know that;

1 atm.L = 101.325 J

so

- 4.95 atm/L = 101.325 J × -4.95 atm/L ÷

= - 501.56 J

Therefore Work done is  - 501.56 J

Answer:

a. Object A

Explanation:

The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.

The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.

So that,

For object A on the moon, mass = 25 lbs

For object B on the earth, W = 25 N,

W = m x g

25 = m x 10                (g = 10 m/[tex]s^{2}[/tex])

m = [tex]\frac{25}{10}[/tex]

   = 2.5 lbs

Mass of object B is 2.5 lbs.

Therefore, the mass of the object A is more than that of B.

The force required to hold it completely submerged under water is 0.252 N

As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.

Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation

Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m

Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³

Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg

Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N

Volume of water displaced = 4/3 π r³ = 2.805 e-5

Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N

F = 0.275 - 0.023 = 0.252 N

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The force required to hold it completely submerged under water is 0.25 N

The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³

The density of water [tex]\rho_w[/tex] = 1000 kg/m³

Diameter = 3.77 cm = 0.0377 m

radius of ball = 0.0377/2 = 0.01885 m

The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]

Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:

The force is required to hold it completely submerged under water (F) is:

[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]

F = 0.25 N

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Answer:

baby yoda

Explanation:

Answer:

Explanation:

To get the focal length, we will use the lens formula;

1/f = 1/u + 1/v

f is the focal length

u is the object distance

v is the image distance

Given

since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.

u = 35cm

v = -2.0

1/f = 1/35-1/2

1/f = 2-35/70

1/f = -33/70

f = -70/33

f = -2.12 cm

f = -0.0212m

Power of a lend is the reciprocal of its focal length

Power of the lens = 1/f

P = 1/-0.0212

P = -47.17dioptres

The power of the lens is -47.17D

Answer:

E=mc2

Explanation:

It explains that in their interiors, atoms fuse together, creating energy.

To solve this we must be knowing each and every concept related to energy and its different types. Therefore, the famous equation that describes the energy of stars is E=mc².

In physics, energy is the capacity to accomplish work. It can be potential, kinetic, thermal, electric, chemical, radioactive, or in other forms.

There is also heat and work—energy inside the process of being transferred by one body to the other. Energy is always classified according to its type once it has been transmitted. As a result, heat transported could become thermal energy, whereas labor done may emerge as mechanical energy. The famous equation that describes the energy of stars is E=mc²

Therefore, the famous equation that describes the energy of stars is E=mc².

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