A car of mass 5000 kg was initially moving at 100 km/h and stops at a distance of 55 m. Find the
magnitude of the net force (in N) acting to stop the car.

Factors such as more crops, fewer predators, invasive species, and changes in the environment explain a decrease or increase in the birds and rodent populations.

Before 1997, both rodents and birds populations increased due to different factors. Moreover, these factors can be classified as natural or human-made factors. Here are the factors that mainly affected these animals:

Bird population:

Increase:

Decrease in predators that increase birds chances to survive (natural factor)

Increase in crops that improve food access for birds (human factor)

Decrease:

Invasive species or increase of predators (natural factor)

Destruction of natural habitats due to an increase in industry and extraction (human factor)

Rodent population:

Increase:

Decrease:

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Answer: C.

Correct me if I'm wrong.

Answer:

Equal reaction from the pair in every action there's an equal and opposite reaction

The item will keep moving at a consistent speed if the object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.

All mass-bearing objects are attracted by gravitational force. Because it consistently attempts to bring masses together rather than push them apart, the gravitational force is referred to as attractive.

As we know, the gravitational force is given by:

[tex]\rm F = \dfrac{Gm_1m_2}{r^2}[/tex]

Where G is the gravitational constant.

m1 and m2 are masses.

r is the distance between the masses.

It is given that:

An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.

As we know,

An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg and the item will keep moving at a consistent speed.

Thus, the item will keep moving at a consistent speed if the object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.

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Answer:

Explanation:

I'm not 10% sure but i think that if you were to do:

603 + 277 = 880

1000 - 880 = 120

So to keep it from moving, i think you would need to pull with a force of 120N

The force with which it should be pulled to keep the knot from moving is 663.58 N.

Newton's third law states that, for every action, there is an equal and opposite reaction.

Here,

Force exerted by Roberta, F₁ = 277 N

Force exerted by Michael, F₂ = 603 N

Since, Roberta is pulling to the west and Michael is pulling towards the south, the angle between the forces applied by them is 90°.

The force needed to keep the knot from moving is the resultant force of these two forces.

Therefore, the resultant force,

Fₙ = √F₁² + F₂²

Fₙ = √(277)² + (603)²

Fₙ = 663.58 N

Hence,

The force with which it should be pulled to keep the knot from moving is 663.58 N.

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Answer:

Explanation:

All

Explanation:

I think this is it, give it a try

Answer:

F-F(gr) = ma

a= {F-F(gr)}/m =

=(15-10)/15=0.33 m/s² (upward)

Hi there!

(A)

A grinding wheel is the same as a disk, having moment of inertia of:

[tex]I = \frac{1}{2}MR^2[/tex]

Plug in the given mass and radius (REMEMBER TO CONVERT) to find the moment of inertia:

[tex]I = \frac{1}{2}(0.380)(0.085)^2 = 0.00137 kgm^2[/tex]

(B)

We can use the rotational equivalent of Newton's Second Law to calculate the needed torque:

Στ = Iα = τ₁ - τ₂

Begin by solving for the angular acceleration. Convert rpm to rad/sec:

[tex]\frac{1750r}{min} * \frac{1 min}{60 s} * \frac{2\pi rad}{1 r} = 183.26 rad/sec[/tex]

Now, we can use the following equation:

ωf = wi + αt (wi = 0 rad/sec, from rest)

183.26/5 = α = 36.65 rad/sec²

τ = Iα = 0.0503 Nm

Since there is a counter-acting torque on the system, we must begin by finding that acceleration:

[tex]\frac{1500r}{min} * \frac{1 min}{60 s} * \frac{2\pi rad}{1 r} = 157.08 rad/sec[/tex]

ωf = wi + αt

-157.08/55 = α = -2.856 rad/sec²

τ₂ = Iα = 0.0039 Nm

Now, calculate the appropriate torque using the above equation:

[tex]\Sigma\tau = \tau_1 - \tau_2[/tex]

[tex]\Sigma\tau + \tau_2 = \tau_1[/tex]

[tex]0.0503 + 0.0039 = \large\boxed{0.054 Nm}[/tex]

Answer:

F=15N

Explanation:

F=m.a

m=1500g ÷1,000 = 1.5kg

a= 10m/s/s

F =1.5 × 10

F = 15 Newton

Answer:

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Explanation:

The behavior of an object that has been charged is dependent upon whether the object is made of a conductive or a nonconductive material. Conductors are materials that permit electrons to flow freely from particle to particle. An object made of a conducting material will permit charge to be transferred across the entire surface of the object. If charge is transferred to the object at a given location, that charge is quickly distributed across the entire surface of the object. The distribution of charge is the result of electron movement. Since conductors allow for electrons to be transported from particle to particle, a charged object will always distribute its charge until the overall repulsive forces between excess electrons is minimized. If a charged conductor is touched to another object, the conductor can even transfer its charge to that object. The transfer of charge between objects occurs more readily if the second object is made of a conducting material. Conductors allow for charge transfer through the free movement of electrons

Answer:

Explanation:

95.0 km/hr = 26.39 m/s

65 km/hr = 18.06 m/s

Circumference of a tire is 0.9π m

77 revolutions is a distance of

77(0.9π) = 69.3π m

v² = u² + 2as

a = (v² - u²) / 2s

a = (18.06² - 26.39²) / (2(69.3π))

a = -0.85 m/s²

s = (v² - u²) / 2a

s = (0² - 26.39²) / 2(-0.85)

s = 409 m

Answer: Even though the genetic susceptibility plays a crucial role on the IQ of the individual, various modifiable environmental factors like education, premature birth, nutrition, pollution, drug and alcohol abuse, mental illnesses, and diseases can have an influence on an individual's IQ.

[tex]\\ \rm\rightarrowtail Avg\: velocity=\dfrac{Total\: Displacement }{Total\: time}[/tex]

[tex]\\ \rm\rightarrowtail Avg\: velocity=\dfrac{4}{2}=2m/s[/tex]

Answer:

Kinetic energy = (1/2) (mass) (speed²)

First object: (1/2) (2 kg) (2 m/s)² = 4 joules .

Second object: (1/2) (4 kg) (3 m/s)² = 18 joules .

The second object had more kinetic energy than the first one had.

Explanation:

Answer:

Kinetic energy = (1/2) (mass) (speed²)

First object: (1/2) (2 kg) (2 m/s)² = 4 joules .

Second object: (1/2) (4 kg) (3 m/s)² = 18 joules .

The second object had more kinetic energy than the first one had.

The magnitude of the force that must be applied to push the book across the desk is 3.97 N.

The given parameters;

The acceleration of the book across the desk is calculated as follows;

a = μg

where;

a = 0.3 x 9.8

a = 2.94 m/s²

The magnitude of the force that must be applied is calculated as follows;

F = ma

F = 1.35 x 2.94

F = 3.97 N

Thus, the magnitude of the force that must be applied to push the book across the desk is 3.97 N.

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The total distance covered is 24 Km and Peter was not moving between the points marked 10 mins and 30 mins on the graph. His displacement according to the graph is zero.

The distance time graph shows the distance covered plotted on the vertical axis against the time taken plotted on the horizontal axis. Using this graph, the total distance covered can easily be obtained.

The total distance covered is 12 km + 12 km since equal distance was covered to and fro. Hence the total distance covered is 24 km. Perter was not moving between the points marked 10 mins and 30 mins on the graph.

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Answer:

138

Explanation:

(since there's a blank next to "9x y j", I'm assuming the y of F(x, y) is 9x[tex]\sqrt{y}[/tex] j)

1) find the partial derivative of each:

[tex](6y^{\frac{3}{2} })i + (9x\sqrt{y} )j[/tex]  

 [tex]f_{x} =(6y^{\frac{3}{2} })_{x} = \int\limits{6y^{\frac{3}{2} }} \, dx = 6xy^{\frac{3}{2} } +c \\\\f_{y} = (9x\sqrt{y} )_{y} = \int\limits{9xy^{\frac{1}{2} } } \, dy = 9x(\frac{2}{3} )y^{\frac{3}{2} } +c = 6xy^{\frac{3}{2} } +c[/tex]

2) use partial integrals to make gradient of f:

take whatever you got from partial integral and add them together (if they repeat, just use it once)

[tex]F =[/tex] Vf (V = gradient of)

[tex]F(x, y) = 6xy^{\frac{3}{2} }[/tex]

3) Evaluate the integrals with given points:

Integral of F dotted with dr = F(point B) - F(point A) = F(3, 4) - F(1, 1)

[tex]F(point B) = 6(3)(4)^{\frac{3}{2} }\\F(Point A) = 6(1)(1)^{\frac{3}{2}}\\F(point B) - F(pointA) = 6(3)(4)^{\frac{3}{2} }-(6(1)(1)^{\frac{3}{2}})[/tex]

= 144 - 6 = 138 units of work

Work done by the force field F in moving an object from A to B = 138 J

Given data :

Force field F(x,y) = [tex]6y^{\frac{3}{2} }i + (9x\sqrt{y} ) j[/tex]  

Step 1 : determine the partial derivatives of the vector quantity

Fx = ∫ [tex]6y^{\frac{3}{2} }i = 6xy^{\frac{3}{2} } + c[/tex]

Fy = ∫ [tex](9x \sqrt{y})_{y} = 9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex]

Equating the partial derivatives :  

[tex]9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex]  = [tex]6xy^{\frac{3}{2} } + c[/tex]

therefore the gradient of F  i.e. F = vF  = F( x,y ) = [tex]6xy^{\frac{3}{2} }[/tex]

Next step : Determine the work done

Work done ( F.dr ) = [ F(point b ) = F( 3,4 ) ]  - [ F(point A) = F( 1,1 ) ]

F(3,4 ) = 6(3)(4)[tex]^{\frac{3}{2} }[/tex]  = 144

F( 1,1 )  = 6(1)(1)[tex]^{\frac{3}{2} }[/tex]    = 6

Therefore the work done by the force field = 144 - 6 = 138 J

Hence we can conclude that the work done by the force field F is = 138 J

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Answer:

OK sure why not!!!!!!!!!!!!

Answer:

Newton's second law states that .

Formulically

F=Force

m=mass

a=acceleration

The best example is hitting a tennis ball.

Answer:

Critical angle is a angle of incidence, for which refraction is 90 degree, this is a least angle of incidence at which total reflection takes place, please mark me Brainliest

Good morning dear...

Have a beautiful and joyful day ahead.

Answer

An Achievement test is an assessment of developed knowledge or skill. ... Achievement tests are developed to measure skills and knowledge learned in a given grade level, usually through planned instruction, such as training or classroom instruction. Achievement tests are often contrasted with aptitude tests.

Explanation:

Answer:

i think its D or B

Explanation:

i just think

On Earth,in atmosphere the hydrogen is found in the less quantity. Option C is correct.

The chemical element hydrogen has the atomic number 1 and the symbol H. The smallest element is hydrogen.

Hydrogen is found in large numbers on Earth in combination with other elements, such as water and hydrocarbons, yet it is only 0.00005 percent present in Earth's atmosphere.

On Earth,in atmosphere the hydrogen is found in the less quantity.

Option C is correct.

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Answer:

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lutfsn...

We have that for the Question "(a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor"  

Answer:

From the question we are told

In an oscillating LC circuit, when 86.6% of the total energy is stored in the inductor's magnetic field

A) When 86.6\% energy is stored in inductor

[tex]\%[/tex]of energy stored in electric field = [tex]1 - 0.866 = 13.4\%[/tex]

[tex]\frac{V_E}{V} = \frac{\frac{q^2}{2c}}{\frac{Q^2}{2c}} = 0.134\\\\\frac{q}{Q} = \sqrt0.134\\\\\frac{q}{Q} = 0.366\\\\q = 0.366Q_{max[/tex]

B)

[tex]\frac{V_B}{V} = \frac{\frac{Li^2}{2}}{\frac{LI^2}{2}} = 0.866\\\\\frac{i}{I} = \sqrt0.866\\\\\frac{i}{I} = 0.931\\\\i = 0.931I_{max[/tex]

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Answer:

30.3 meters, 172 degrees

Explanation:

To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (east-west) displacements and the sum of all the vertical (north-south) displacements.

Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West

Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North

The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A non-scaled sketch is useful for visualizing the situation.

Applying the Pythagorean theorem leads to the magnitude of the resultant (R).

R2 = (30.0 m)2 + (4.0 m)2 = 916 m2

R = Sqrt(916 m2)

R = 30.3 meters

The angle theta in the diagram above can be found using the tangent function.

tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m)

tangent(theta) = 0.1333

theta = invtan(0.1333)

theta = 7.59 degrees

This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.

The magnetic field in the gap is [tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]

In a circuit, a magnetic flux is circulated or followed via a confined environment or passage called a magnetic field gap. The narrow air gap is a non-magnetic component of a magnetic circuit that is normally connected to the remainder of the circuit magnetically in series. This enables a significant amount of magnetic flux to pass via the gap.

The magnetic field in the gap at a distance s < a can be computed by using the formula:

[tex]\mathbf{ \oint Bdl = \mu_oI_{enclosed}}[/tex]

where;

[tex]\mathbf{B( 2 \pi d) = \mu _o \oint _s J_d da }[/tex]

where;

[tex]\mathbf{B( 2 \pi d) = \mu _o J_d \oint _sda }[/tex]

[tex]\mathbf{B( 2 \pi d) = \mu _o J_d (\pi d^2) }[/tex]

Making the magnetic flux density the subject, we have:

[tex]\mathbf{B =\dfrac{ \mu _o J_d (\pi d^2) }{( 2 \pi d)}}[/tex]

[tex]\mathbf{B =\dfrac{ \mu _o J_dd}{ 2 }}[/tex]

Recall that, the drift current density [tex]\mathbf{J_d = \dfrac{I}{\pi a^2}}[/tex]

[tex]\mathbf{B = \dfrac{\mu_o d}{2}(\dfrac{I}{\pi a^2})}[/tex]

Recall that distance in question is said to be (s);

∴

[tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]

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Answer:

watts is for real power

volt amperes reactive (VA) for reactive power

volt amperes (VA) for apparent power

Answer:

Explanation:

The work of the buoyancy force will convert to gravity potential energy

Fresh Water has a density of 1000 kg/m³

The ball has a density of 370 kg/m³

assume the ball is 1 m³

weight of the ball is 370g

The buoyancy force is 1000g

Assume the buoyancy force drops suddenly to zero when the center of the ball clears the water level.

The Work done on the ball is

W = Fd = 1000g(0.590)

the change in potential energy is

PE = mgh = 370g(0.590 + y)

where y is the height above the water level.

1000g(0.590) = 370g(0.590 + y)

1000(0.590) = 370(0.590 + y)

y = (0.590)(1000 - 370) / 370

y = 1.004594... ≈ 1.00 m