Two vehicles are approaching an intersection. One is a 2600 kgkg pickup traveling at 17.0 m/sm/s from east to west (the −x−x- direction), and the other is a 1300 kgkg sedan going from south to north (the +y−+y− direction at 24.0 m/sm/s ). Part A Find the xx -component of the net momentum of this system. pxpx = nothing kg⋅m/skg⋅m/s SubmitRequest Answer Part B Find the yy-component of the net momentum of this system. pypy = 0 kg⋅m/skg⋅m/s SubmitPrevious AnswersRequest Answer Incorrect; Try Again; 4 attempts remaining Part C What is the magnitude of the net momentum?

Explanation:

Given that,

Mass of the particle, m = 4 kg

Speed of the particle, v = 2.5 m/s

The radius of the circle, r = 2 m

We need to find the angular momentum about the center of the circle. The formula for the angular momentum is given by :

[tex]L=mvr[/tex]

Substitute all the values,

[tex]L=4\times 2.5\times 2\\\\L=20\ kg{\cdot}m^2s[/tex]

So, the angular momentum of the particle is 20 kg-m² s.

Answer:

Block of aluminum

Explanation:

The block of aluminum because of the conductivity, and is a metal, so it gets hotter faster

The temperature of wood and the aluminium will depend upon thier thermal coductivity.

The thermal conductivity of any material is defined as the amount of heat from a material can flow in a unit area per unit degree centigrade for a unit distance.

The thermal conductivity of wood is around 0.90 to 0.197 w/mk and that of aluminium is around 88 to 251 w/mk.

We can see that the aluminium block will absorb much amount of heat as compared to the wood block. Because the thermal conductivity of aluminium is much higher then the wood.

Hence temperature of wood and the aluminium will depend upon thier thermal coductivity.

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Answer:

100 milliliters

Explanation:

Answer:

2000 N.

Explanation:

From the question given above, the following data were obtained:

Load (L) = 8000 N

Mechanical advantage (MA) = 4

Effort (E) =?

The mechanical advantage of a machine is simply defined as:

Mechanical advantage = Load / Effort

MA = L / E

With the above formula, we can obtain the effort used to raise the load of 8000 N as follow:

Load (L) = 8000 N

Mechanical advantage (MA) = 4

Effort (E) =?

MA = L / E

4 = 8000 / E

Cross multiply

4 × E = 8000

Divide both side by 4

E = 8000 / 4

E = 2000 N

Thus, the effort used to raised the load is 2000 N.

Answer:

true..

Explanation:

Newton's 2nd law of motion is greater the mass of an object the more force it will take to accelerate the object

According to Newton’s second law of motion, weight depends on an object’s mass and the net force acting on the object. The given statement is true.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

The following equation help to understand the statement;

Force is defined as the product of mass and acceleration. Its unit is Newton.

F=ma

F=mg

W=mg

The weight of an item is determined by its mass and the net force exerted on it, according to Newton's second rule of motion. The provided assertion is correct.

Hence,the given statement is true.

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Answer:

[tex]6.28\times 10^{-5}\ \text{T}[/tex]

[tex]1.92\times 10^{-6}\ \text{Nm}[/tex]

Explanation:

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi 10^{-7}\ \text{H/m}[/tex]

[tex]I_l[/tex] = Current in circular loop = 13 A

[tex]r_l[/tex] = Radius of circular loop = 13 cm

[tex]N[/tex] = Number of turns = 58

[tex]r_c[/tex] = Radius of coil = 0.94 cm

[tex]I_c[/tex] = Current in coil = 1.9 A

[tex]\theta[/tex] = Angle between loop and coil = [tex]90^{\circ}[/tex]

Magnitude of magnetic field in circular loop

[tex]B_l=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B_l=\dfrac{4\pi 10^{-7}\times 13}{2\times 13\times 10^{-2}}\\\Rightarrow B_l=6.28\times 10^{-5}\ \text{T}[/tex]

The magnetic field produced by the loop at its center is [tex]6.28\times 10^{-5}\ \text{T}[/tex].

Torque is given by

[tex]\tau=\pi NI_cr_c^2B_l\sin\theta\\\Rightarrow \tau=\pi 58\times 1.9\times (0.94\times 10^{-2})^2\times 6.28\times 10^{-5}\sin90^{\circ}\\\Rightarrow \tau=1.92\times 10^{-6}\ \text{Nm}[/tex]

The torque on the coil due to the loop [tex]1.92\times 10^{-6}\ \text{Nm}[/tex].

Answer:

correct answer is C

Explanation:

In this exercise, you are asked to complete the sentences so that the sentence makes sense.

1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down

correct answer C

2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is

            y = v₀ t - ½ gt²

v₀ = 0

             y = -1/2 g t²

so the position is not proportional to the square of the time

correct answer is C

cold, dry, and stable air masses .

Answer:

 v = 4.8 10⁻⁴ m / s

Explanation:

To solve this exercise we can use the concepts of energy. In this case the potential energy is transformed into kinetic energy

          U = K

          q V = ½ m v²

          v = [tex]\sqrt { \frac{2qV}{m} }[/tex]

in the exercise they indicate the value of the charge q₁ = 1 pC = 1 10⁻¹² C

let's calculate

          v = [tex]\sqrt{ \frac{2 \ 1 \ 10^{-12} 120 \ 10^3}{500 ^{2} }[/tex]

          v = 4.8 10⁻⁴ m / s

Answer:

C

Explanation:

if you measure your body's flexibility then you can keep track of how flexible you have gotten over time