Val is comparing the gravitational force and the electromagnetic force
Which describes the electromagnetic force ONLY? Check all that apply

First write down all your known variables:

vi = 25m/s

a = 7.0m/s^2

t = 6.0s

vf = ?

Then choose the kinematic equation that relates all the variables and solve for the unknown variable:

vf = vi + at

vf = (25) + (7.0)(6.0)

vf = 67m/s

The final velocity of the motorcycle is 67m/s.

Mass of the sound reflector is given as

[tex]m_1 = 170 kg[/tex]

mass of the chain itself is

[tex]m_2 = 8 kg[/tex]

now here since this system is suspended at rest and no acceleration is given for the system so in order to find the tension in the chain we will just use force balance equation

As per force balance we can say that weight of the chain and reflector downwards will be counter balanced by tension in the string

Now here we will have

[tex]T = m_1g + m_2g[/tex]

since chain is given with some mass so tension at the top end of the chain will be

[tex]T = 170* 9.8 + 8 * 9.8[/tex]

[tex]T = 1744.4 N[/tex]

So here tension in the chain will be 1744.4 N at the top end and this tension will decrease as we move down along the length of the chain.

The pet store would be the reference point because it is where he started and it will not move. Hope this helped.

The pet store would be the reference point because it is where he started motion and it will not move.

The phenomenon of an item changing its position with respect to time is known as motion in physics. In mathematics, displacement, distance, velocity, and acceleration are used to explain motion.

A reference point is a location or object that is used as a point of comparison to ascertain whether something is moving. When an object shifts in relation to a fixed point, it is said to be in motion. Good reference points are things that are fixed in relation to Earth, like a building, a tree, or a sign.

The pet store would be the reference point because it is where he started motion and it will not move.

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Let the salmon jump with speed v at an angle of 33.3 degree

So here we can find out the components of his velocity in x and y direction

[tex]v_x = v cos33.2[/tex]

[tex]v_y = v sin33.2[/tex]

now the horizontal displacement of the salmon is 3.15 m so he cover this horizontal distance with constant speed as there is no acceleration in x direction

here we can say

[tex]v_x * t = 3.15[/tex]

[tex]v cos33.2 * t = 3.15[/tex]

[tex]v*t = 3.76[/tex]

now in Y direction it is an accelerated motion as it is accelerated due to gravity

[tex]y = v_y * t + \frac{1}{2}at^2[/tex]

[tex]-0.45 = v sin33.2* t - \frac{1}{2}*9.8* t^2[/tex]

now we will plug in v*t = 3.76

[tex]-0.45 = 3.76 * sin33.2 - 4.9 * t^2[/tex]

[tex]4.9 t^2 = 2.51[/tex]

[tex]t^2 = 0.512[/tex]

[tex]t = 0.72 s[/tex]

now speed v is given by equation above

[tex]v*t = 3.76[/tex]

[tex]v*0.72 = 3.76[/tex]

[tex]v = 5.25 m/s[/tex]

so here he must have to jump with minimum speed of 5.25 m/s

10% of the income.

Answer:

20% of their income

Explanation:

I got a five on Plato

A) gravity is the attraction of any mass to any other other mass .

D) gravity is a force that exist between any two masses in the universe

both are ok ...

Answer: A sealed calorimeter is a closed system because thermal energy is not transferred to the environment.

A closed system is an isolated system where there is no transfer of energy with its surroundings. Also the system is not subjected to any external force whose source is external to the system.

Answer:C. Because thermal energy has not transferred to the environment.

Explanation:

Answer

6.6 km

The description of the problem is shown in the attached figure, where the line "d" represents the final displacement vector.

First, the trekker walked 3.3km in a 40 ° direction, as shown in the figure. We can write this vector in its Cartesian coordinates:

[tex]-3.3sin (40)x + 3.3cos (40)y[/tex]

Then the hiker walked 3.4 km in a 60 degree northwest direction.

We can write this as a vector in its Cartesian coordinates:

[tex]-3.4sin (60)x + 3.4cos (60)y[/tex].

When adding this two vectors we will obtain the final displacement "d"

[tex]d = [- 3.3sin(40) -3.4sin (60)]x + [3.3cos(40) + 3.3cos (60)]y\\[/tex]

[tex]d = -5.07x +4.23y\\[/tex]

To obtain the magnitude of this vector we calculate its module:

[tex]\sqrt{5.07 ^2 +4.23 ^ 2}[/tex]

Then the magnitude of the final displacement was:

6.6 km

The displacement of the rock will be the same as the total horizontal distance traveled. Here the rock's horizontal position is given by

[tex]x=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\cos30^\circ\,t[/tex]

so to find the horizontal distance it traversed, we need to know the time it took for the rock to return to the ground. We use the rock's vertical position over time to figure that out:

[tex]y=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\sin30^\circ\,t-\dfrac g2t^2=0[/tex]

where [tex]g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. Then we find that [tex]t\approx1.5\,\mathrm s[/tex], at which point we find [tex]x\approx20\,\mathrm m[/tex].

kangaroo is capable of jumping to a height of 2.62 m

It means it will reach maximum height of 2.62 m when it will jump off with some maximum capable speed from the ground

So here as it will reach to its maximum height the final speed of the kangaroo will be zero

and also we know that during the motion of kangaroo the acceleration of kangaroo is due to gravity which is given by g = 9.8 m/s^2

now we can use kinematics equation to find out take off speed

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we know that

[tex]v_f = 0[/tex]

a = - 9.8 m/s^2

d = 2.62 m

now we will have

[tex]0^2 - v_i^2 = 2 * (-9.8)* 2.62[/tex]

[tex] - v_i^2 = - -51.352 [/tex]

[tex]v_i = \sqrt{51.352}[/tex]

[tex]v_i = 7.17 m/s[/tex]

so the take off speed of the kangaroo will be 7.17 m/s and correct answer is "option d"

Answer:

vi = 7.17 m/s

Explanation:

(4225 m2/s2)/(6 m/s2) = d

d = 704 m

Return to Problem 8

Given:

vi = 22.4 m/s

vf = 0 m/s

t = 2.55 s

Find:

d = ??

d = (vi + vf)/2 *t

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

d = 28.6 m

Return to Problem 9

Given:

a = -9.8 m/s2

vf = 0 m/s

d = 2.62 m

Find:

vi = ??

vf2 = vi2 + 2*a*d

(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)

0 m2/s2 = vi2 - 51.35 m2/s2

51.35 m2/s2 = vi2

vi = 7.17 m/s

Hello!

To find the speed of the train, we have to divide the distance by the time.

Speed = [tex]\frac{distance}{time}[/tex]

600 / 4 = 150

The speed of the train is 150 km per hour.

The x-component of a vector can be calculated using the magnitude of the vector and the angle it forms with the positive x-axis.

In a rectangular coordinate system in a plane, the x-component of a vector is given by the dot product of the vector with the unit vector Î (î). The x-component can be calculated as:

Ex = E * cos(θ)

where E is the magnitude of the vector and θ is the angle the vector forms with the positive x-axis.

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The x-component of a vector describes the part of the vector in the x direction. In the case of vector E⃗, it can be computed based on the angle θ it makes with the x-axis and its magnitude E through the relationship E cosθ.

The x-component of a vector describes the effect of the vector in the 'x' direction. This can be visualized by understanding that any vector can be viewed as a resultant of its components along the various axes available (here, along 'x'). Thus, when a given vector is expressed in rectangular or Cartesian components, the respective projections of the vector on 'x' (and 'y') directions essentially give the x-component (and y-component).

In the case of vector E⃗, the x-component can be computed based on the angle θ (theta) it makes with the 'x' axis and its overall magnitude E, using simple trigonometric principles. The x-component (Ex) is given by E cosθ because cosθ gives the fraction of the magnitude E that lies along the 'x' direction.

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The correct awnser is true i belive

TRUE

Answer

It is false

Explanation:

In old days it was assumed that if we place wooden pallets on concrete. The casing of pallets may break down and their may be leakage of battery. So to  protect the case, people placed the pallets on sand or gravel.

Here they use the concept that if you place a glass jar on concrete with a force it may got a crack and if you place this glass jar with same force on sand it will not break.

But it is not true because pallets have strong casing which cant break on placing it on concrete.

Hence When using wooden pallets to store batteries it can be placed anywhere.

(A) The resistance of the block of aluminum is given by:

[tex]R=\frac{\rho L}{A}[/tex]

where

[tex]\rho[/tex] is the aluminum resistivity

L is the length of the cylinder

A is the cross-sectional area of the cylinder

We already know the aluminum resistivity ([tex]\rho=2.65 \cdot 10^{-8} \Omega m[/tex]) and the length of the cylinder (L=20 m), so we must find the cross-sectional area A, which is given by the difference between the area of the larger cylinder and the area of the radial hole:

[tex]A=\pi (R^2 -r^2)[/tex]

where [tex]R=2.5 cm=0.025 m[/tex] and [tex]r=20 mm=0.02 m[/tex] (assuming that the 20 mm removed radially refers to the radius of the hole).

Therefore, the cross-sectional area is

[tex]A=\pi ((0.025 m)^2-(0.020 m)^2)=7.06 \cdot 10^{-4} m^2[/tex]

Substituting into the initial formula of the resistance, we find:

[tex]R=\frac{\rho L}{A}=\frac{(2.65 \cdot 10^{-8} \Omega m)(20 m)}{7.06 \cdot 10^{-4} m^2}=7.51 \cdot 10^{-4} \Omega[/tex]

(B) The resistivity of the tungsten is [tex]\rho=5.6 \cdot 10^{-8} \Omega m[/tex], so a cylinder of tungsten of the same size would have a resistance of

[tex]R=\frac{\rho L}{A}=\frac{(5.6 \cdot 10^{-8} \Omega m)(20 m)}{7.06 \cdot 10^{-4} m^2}= 1.59 \cdot 10^{-3} \Omega[/tex]

The behaviour of the resistance versus temperature is given by:

[tex]R=R_0 (1 + \alpha \Delta T)[/tex]

where [tex]\alpha[/tex] is a coefficient that for aluminum is equal to [tex]\alpha = 0.004308[/tex], R0 is the resistance of the piece of aluminum we found at point (A), and R is the resistance of the tungsten. Re-arranging the formula and substituting, we find

[tex]\Delta T = \frac{R/R_0 -1}{\alpha}=\frac{\frac{1.59 \cdot 10^{-3} \Omega}{7.51 \cdot 10^{-4} \Omega}-1}{0.004308}=259.3^{\circ}C[/tex]

So, the temperature must increase by 259.3 degrees.

(C) The power dissipated is given by:

[tex]P=I^2 R[/tex]

where I=30 A is the current. Substituting the numbers into the formula, we find

[tex]P=(30 A)^2 (7.51 \cdot 10^{-4} \Omega)=0.68 W[/tex]

Forces are presented as straight arrows pointing in the direction they function  on the body.

The force arrows always points straight arrows.

Scientific law is also known as natural law. It  imply a cause and effect between the observed elements and always apply under the same conditions .

Scientific law is a statement based on repeated experimental observations that describes some aspect of the Universe.

Ummm thats not a question buddy...

Volume=0.09m³

Density=4000kg/m³

Force=?

Density=mass/volume ⇒mass=volume×density

m=0.09×4000=360kg

Force=mass×accelaration

F=360×9.8

F=3528N

Answer

3,528 N

Explanation

density = mass/volume

ρ = m/v

∴ m = ρv

      = 4000 × 0.09

        = 360 Kg

Weight is defined as the force that acts on  a body and is directed towards the center of the earth.

Weight = mass × acceleration due to gravity

W = mg          Where g = 9.8 m/s²

W = 360 × 9.8

  = 3,528 N

• Objects “fall” toward the center of Earth.  (TRUE)

Since the gravitational force between earth and any object is always along the line joining the centers of two so all the objects near the surface of earth will always feel force towards the centre of the earth due to gravity.

• Objects are pulled by Earth but do not pull on Earth.  (FALSE)

Since gravitational force will always follow Newton's III law as per which every action has equal and opposite reaction force so object near the surface of earth will also pull the earth as earth is pulling the object towards the center.

• Objects accelerate at a rate of 9.8 m/s each second.  (TRUE)

Since acceleration due to gravity of Earth is 9.8 m/s^2 so here we can say that all objects near the surface of earth will accelerate towards the center of earth at rate of 9.8 m/s^2 due to gravitational pull.

• Objects travel at a rate of 9.8 m/s to are the center of Earth. (FALSE)

Since earth is pulling all objects near its surface with gravitational force so objects can not move at constant speed but it will move with constant acceleration.

• Objects are pulled by Earth more strongly than they pull on Earth. (FALSE)

Since gravitational force will always follow Newton's III law which say every action has equal and opposite reaction so we can say that force due to all objects on earth will be same as the force of earth on the object.

The velocity of an object dropped and falling for 3.0 seconds under the influence of gravity and ignoring air resistance will be approximately 29.4 m/s downward.

The velocity of an object dropped and falling under the influence of gravity increases with time due to the acceleration caused by gravity. In the absence of air resistance, an object in freefall near the surface of the Earth accelerates downwards at a rate of 9.8 m/s². This means that for every second that passes, the object's downward speed will increase by about 9.8 m/s.

So, if an object has been falling for 3.0 seconds, we can find its velocity using the formula v = gt, where v is velocity, g is the acceleration due to gravity (9.8 m/s² near the surface of the Earth), and t is time. Substituting in our values we get: v = 9.8 m/s² * 3.0 s = 29.4 m/s.

Therefore the velocity of a dropped object after it has fallen for 3.0 seconds would be approximately 29.4 m/s downward.

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efficiency of the heat engine is calculated by the formula

[tex]\eta = 1- \frac{T_c}{T_h}[/tex]

here we know that

[tex]T_c [/tex] = cold temperature

[tex]T_h [/tex] = hot temperature

now we will find the efficiency of all

1) for W

[tex]\eta_1 = 1- \frac{T_c}{T_h}[/tex]

[tex]\eta_1 = 1- \frac{120}{620}[/tex]

[tex]\eta_1 = 0.806[/tex]

2) for X

[tex]\eta_2 = 1- \frac{T_c}{T_h}[/tex]

[tex]\eta_2 = 1- \frac{100}{840}[/tex]

[tex]\eta_2 = 0.88[/tex]

3) For Y

[tex]\eta_3 = 1- \frac{T_c}{T_h}[/tex]

[tex]\eta_3 = 1- \frac{300}{900}[/tex]

[tex]\eta_2 = 0.67[/tex]

4) for Z

[tex]\eta_4 = 1- \frac{T_c}{T_h}[/tex]

[tex]\eta_4 = 1- \frac{25}{500}[/tex]

[tex]\eta_4 = 0.95[/tex]

So here most efficient engine is Z and least is Y

so we can arrange it as

Z > X > W > Y

so 3rd option is correct here

Answer:

C.) Z, X,W, Y

We are standing still inside an elevator

So our speed in the frame of elevator is zero

we can write this relative speed as

[tex]v_{pE} = v_p - v_E[/tex]

also we know that speed of the elevator is 55 mph

[tex]v_E = 55 mph[/tex]

now the velocity in relative frame is zero as he is  still in the moving elevator while in the ground frame or observing from the ground its velocity will be same as velocity of elevator.

So whenever an object is at rest in a moving frame then its relative velocity with respect to that frame must be ZERO

C. a windy and powerful storm system

C. a windy and powerful storm system

ANSWER:

Abseiling rope is tested beyond twice the maximum possible weight when in use because of safety purposes.

EXPLANATION:

Every rope has its elastic limit.Abseiling rope is used basically for climbing or descending purposes so this type of rope must be of quite good quality and  it must bear maximum possible weight. We test it  just to check the elastic limit of the rope so that the rope shouldnot increase its elastic limit and it doesnot break when we use it.

Well, there aren't really any choices on the list you provided that I would consider to be an acceptable answer to this question.  So I just have to make something up:

From the given information, I can conclude:

-- The car's velocity is constant, and points East.

-- The car's acceleration is zero.

-- The forces on the car are balanced.

-- The centripetal force acting on the car is zero.

-- The force delivered from the engine to the wheels is EXACTLY EQUAL to the sum of the forces of air resistance and friction with the road.

Let the salmon jump with speed v at an angle of 33.3 degree

So here we can find out the components of his velocity in x and y direction

[tex]v_x = v cos33.2[/tex]

[tex]v_y = v sin33.2[/tex]

now the horizontal displacement of the salmon is 3.15 m so he cover this horizontal distance with constant speed as there is no acceleration in x direction

here we can say

[tex]v_x * t = 3.15[/tex]

[tex]v cos33.2 * t = 3.15[/tex]

[tex]v*t = 3.76[/tex]

now in Y direction it is an accelerated motion as it is accelerated due to gravity

[tex]y = v_y * t + \frac{1}{2}at^2[/tex]

[tex]-0.45 = v sin33.2* t - \frac{1}{2}*9.8* t^2[/tex]

now we will plug in v*t = 3.76

[tex]-0.45 = 3.76 * sin33.2 - 4.9 * t^2[/tex]

[tex]4.9 t^2 = 2.51[/tex]

[tex]t^2 = 0.512[/tex]

[tex]t = 0.72 s[/tex]

now speed v is given by equation above

[tex]v*t = 3.76[/tex]

[tex]v*0.72 = 3.76[/tex]

[tex]v = 5.25 m/s[/tex]

so here he must have to jump with minimum speed of 5.25 m/s

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

[tex]F_{net} = F_1 - F_2[/tex]

[tex]F_{net} = 60 - 45[/tex]

[tex]F_{net} = 15 N[/tex]

so boat will have net force F = 15 N in forward direction due to both jumps

The 'slope' of the road is (rise between two points) / (hotizontal distance between the same two points).  Or (rise)/(run).  It's also the tangent of the angle between the horizontal and the incline.

In this problem, your 'run' is 2.4 km, and the 'rise' is 160 m.

The slope of the incline is (160) / (2,400) .

That's 0.0667 .

To answer the question, you need to find the ANGLE whose tangent is that same number.

If your calculator has ANY functions on it, then " tan⁻¹ " is the one you want.  Key in a number, and " tan⁻¹ " will give you the angle whose tangent is that number.

The angle of the road above the horizontal is found using trigonometry, specifically the tangent function, which is the ratio of the elevation gain to the distance traveled along the road. By applying the arctan function to 160m/2400m, we can calculate the angle of inclination in degrees.

The question involves finding the angle of the road above the horizontal when the elevation change and the distance traveled along the road are known. This requires the use of trigonometry to determine the angle, given the rise (elevation change) and run (distance traveled along the road).

To find the angle, we can use the tangent function, which is the ratio of the opposite side to the adjacent side in a right triangle. In this scenario, the rise is 160 m and the run is 2400 m (2.4 km). We need to convert the run into meters to match the units of the rise, making the run 2400 m.

Now we compute the angle θ:

Once we calculate the arctan(1/15), we will get the angle of inclination of the road in degrees.

Its average would be 35km in 30 minutes Hope this helps :D