What is the distance traveled by an object that moves with an average velocity of 8.0 m/s for a total of 4.0 seconds

According to conservation of energy, the energy of interacting bodies in a closed system remains constant. The total energy of an isolated system remains constant; it is said to be conserved over a period of time.

Elastic energy is the mechanical implicit energy stored in the configuration of a material or physical system as it's subjected to elastic distortion by work performed upon it. Elastic energy occurs when objects are impermanently compressed, stretched or generally misshaped in any manner.

The mechanical energy is never lost forever , rather it gets converted to thermal energy because of the friction .

To know more about Law of conservation of energy ,visit:

https://brainly.com/question/29775341

#SPJ1

When you push a block with your hand into the wall to hold it stationary, the direction of the normal force and friction force respectively on the block are as follows: Direction of normal force: It is the force that is exerted perpendicular to the surface of contact between the block and the wall.

In this case, the normal force acts in the upward direction against the weight of the block. It is responsible for balancing the weight of the block and preventing it from sinking into the wall.

Direction of friction force:

It is the force that opposes the motion of the block and acts parallel to the surface of contact between the block and the wall.

The friction force acts in the backward direction opposite to the force applied by the hand on the block.

It is responsible for holding the block stationary and preventing it from sliding down the wall.

Know more about direction   here:

https://brainly.com/question/28108225

#SPJ11

\(\\ \sf\longmapsto m1v1+m2v2=(m1+m2)v3\)

\(\\ \sf\longmapsto 7(17)+8(-14)=(7+8)v3\)

\(\\ \sf\longmapsto 119-112=15v3\)

\(\\ \sf\longmapsto 15v3=7\)

\(\\ \sf\longmapsto v3=7/15\)

\(\\ \sf\longmapsto v3=2.1m/s\)

The force on the glove is 12.8 N.

Solving for the  force on the glove:

The force on the glove is equal to the impulse (change in momentum) experienced by the ball divided by the time over which the force was applied.

Impulse = force x time

              = 5.0 N x (1/8.0 m/s)

Impulse = 0.625 Ns

The change in momentum of the ball ΔP is

ΔP =  m x v

where,

m = is  the mass of the ball

v= its velocity

Momentum = mass x velocity

                   = 0.2 kg x 8.0 m/s

Momentum= 1.6 kg m/s

The force on the glove is equal to the change in momentum divided by the time over which the force was applied:

Force = change in momentum/time

           = 1.6 kg m/s

               (1/8.0 m/s)

Force = 12.8 N

Learn more about momentum and force on

https://brainly.com/question/29021338

#SPJ1

The catch in this one is:  We don't know how much force the student used to push the bike.  

It wasn't necessarily the 100N.  That's just the weight of the bike. But you know that you can push a car, a wagon, or a bicycle hard, you can push it not so hard, you can give it a little push, you can give it a big push, you can push it strong, you can push it weak, you can push it medium.  The harder you push, the more it'll accelerate, but it's completely up to you how hard you want to push.  That's what's so great about wheels !  That's why they were such a great invention ! This is where I made my biggest mistake. This guy came into my store one day and said he's got this great invention, it's definitely going to take off, it'll be a winner for sure, he called it a "wheel".  I looked at it, I turned it over and I looked on all sides. I thought it was too simple.  I didn't know then it was elegant. I threw him out.  I was so dumb.  I could have invested money in that guy, today I would have probably more than a hundred dollars.

Anyway, can we figure out how much force the student used to push with ?  Stay tuned:

-- The bike covered 15 meters in 5 seconds.  Its average speed during the whole push was (15m/5s) = 3 meters/sec.

-- If the bike started out with no speed, and its average speed was 3 m/s, then it must have been moving at 6 m/s at the end of the push.

-- If its speed increased from zero to 6 m/s in 5 seconds, then its acceleration was (6m/s / 5 sec) = 1.2 m/s²

-- The bike's weight is 100N.  

(mass) x (gravity) = 100N

Bikemass = (100N) / (9.8 m/s²)

Bikemass = 10.2 kilograms

-- F = m A

Force = (mass) x (acceleration)

Force = (10.2 kg) x (1.2 m/s²)

Force = 12.24 N

-- Work = (force) x (distance)

Work = (12.24 N) x (15 m)

Work = 183.67 Joules

-- Power = (work done) / (time to do the work)

Power = (183.67 joules) / (5 seconds)

Power = 36.73 watts

Answer:

Explanation:

lll

Answer:

Unbalanced

Explanation:

Hussein is pushing 2 N harder.

A "screen" or even just a set of parallel bars are highly reflective to electromagnetic waves as long as the open spaces are small compared to the wavelengths.

"Grid" dishes work fine ... with less weight and less wind resistance ... for frequencies below about 3 GHz. (Wavelengths of at least 10 cm.)

(I even worked on a microwave system in South America where huge grid dishes were used on a 90-mile link.)

(i) The hydraulic gradient is 0.45.

(ii) The flow rate is approximately 0.00000167 cubic meters per second.

(iii) The hydraulic conductivity is approximately 0.000037 meters per second.

(i) Hydraulic gradient:

The hydraulic gradient (i) can be calculated by dividing the head difference (h) by the distance (L) between the manometer tappings:

i = h / L

Given:

Head difference (h) = 45 mm

Distance between manometer tappings (L) = 100 mm

Converting the units to meters:

h = 45 mm / 1000 = 0.045 m

L = 100 mm / 1000 = 0.1 m

Substituting the values into the formula:

i = 0.045 m / 0.1 m = 0.45

(ii) Flow rate:

The flow rate (Q) can be calculated using the equation:

Q = (V / t) / A

Where V is the quantity of water collected, t is the duration of the test, and A is the cross-sectional area of the test sample.

Given:

Quantity of water collected (V) = 500 ml = 0.5 L

Duration of test (t) = 300 seconds

Diameter of test sample (d) = 100 mm

Converting the units to meters:

V = 0.5 L = 0.5 / 1000 = 0.0005 m³

t = 300 seconds

d = 100 mm / 1000 = 0.1 m

Calculating the cross-sectional area (A) using the formula for the area of a circle:

A = π * (d/2)^2

Substituting the values:

A = π * (0.1/2)^2 = π * 0.005^2 = 0.00007854 m²

Substituting the values into the formula for flow rate:

Q = (0.0005 m³ / 300 s) / 0.00007854 m²

Calculating the flow rate:

Q = 0.00000167 m³/s

(iii) Hydraulic conductivity:

The hydraulic conductivity (K) can be calculated using Darcy's Law:

K = Q / (A * i)

Given the values we calculated:

Q = 0.00000167 m³/s

A = 0.00007854 m²

i = 0.45

Substituting the values into the formula:

K = 0.00000167 m³/s / (0.00007854 m² * 0.45)

Calculating the hydraulic conductivity:

K ≈ 0.000037 m/s

To know more about Hydraulic visit-

https://brainly.com/question/31453487

#SPJ11

The exercise tells us that the electric field is given by the following equation:

And it also gives us, A and Q. Thus, our electric field inside the capacitor is:

As we know, the electric force can be written as:

The charge of an electron is a constant, which is q=1.6*10^(-19) C.

Finally, our force can be written as:

Our final answer is 0.00001149 micro Newtons

Answer:

kinetic energy to electrical energy.

The rate of change of the distance between the two particles is 10 units per second.

Let's call the distance between particle A and particle B "d". To find the rate of change of "d", we need to take the derivative of "d" with respect to time. By using the Pythagorean theorem to relate the distance between the two particles to their x and y coordinates:

d^2 = x^2 + y^2

Taking the derivative of both sides with respect to time, we get:

2dd/dt = 2x(dx/dt) + 2y(dy/dt)

where dx/dt is the rate of change of particle A's x-coordinate (which is equal to 2), and dy/dt is the rate of change of particle B's y-coordinate (which is equal to 3).

We need to find x and y in terms of time. Particle A is moving at a constant rate of 2 per second along the positive x-axis, so its x-coordinate is given by:

x = 2t

where t is time in seconds.

Similarly, particle B is moving at a constant rate of 3 per second along the positive y-axis, so its y-coordinate is given by:

y = 3t

Substituting these expressions for x and y into our equation for the rate of change of "d", we get:

2dd/dt = 2(2)(1) + 2(3)(1) = 10

So the rate of change of the distance between the two particles is 10 units per second.

To know more about Distance:

https://brainly.com/question/15172156

#SPJ4

The surface gravity of the new planet is approximately 1/32 times the surface gravity of Earth.

To determine the average distance of an asteroid from the Sun based on its period, we can use Kepler's Third Law of planetary motion, which states that the square of the orbital period is proportional to the cube of the semi-major axis (average distance) of the orbit.

Let's calculate the semi-major axis (a) using the given period (P) of the asteroid's orbit.

P = 1.283 years

The formula for Kepler's Third Law is:

P² = a³

Substituting the values:

(1.283)² = a³

1.646289 = a³

Taking the cube root of both sides:

a ≈ 1.188 AU

Therefore, the average distance of the asteroid from the Sun is approximately 1.188 astronomical units (AU).

Now let's move on to the second question about the surface gravity of a newly discovered planet.

The surface gravity of a planet can be determined using the formula:

g = (G * M) / R²

where:

g = surface gravity

G = gravitational constant (approximately 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²)

M = mass of the planet

R = radius of the planet

We know that the new planet has half the density but four times the radius of Earth. Density is defined as mass divided by volume, so if the density is half, the mass (M) will also be half compared to Earth.

Let's assume the mass of the Earth is Me and the radius of the Earth is Re. Therefore, the mass (M) and radius (R) of the new planet would be:

M = (1/2) * Me

R = 4 * Re

Substituting these values into the formula for surface gravity:

g = (G * (1/2) * Me) / (4 * Re)²

g = (G * (1/2) * Me) / (16 * Re²)

Since the question asks for the surface gravity of the new planet relative to Earth, we can simplify further:

g = (1/32) * (G * Me / Re²)

The surface gravity of the new planet is approximately 1/32 times the surface gravity of Earth.

Learn more about surface gravity from the link given below.

https://brainly.com/question/12611262

#SPJ4

Answer:

Circular motion requires a"centripetal" force. Without a centripetal force, an object cannot travel in circular motion. 

Answer:

1.27seconds

Explanation:

Speed = 324 m/s , Distance = 411.48m

Time = Distance/Speed

Time = 411.48/324 seconds

Time = 1.27seconds ( Ans)

Answer:

yes

Explanation:

The atom is the smallest and most fundamental unit of matter. All living things are made of cells the cell itself is the smallest fundamental unit of structure and function in living organisms.

have great day!!

Answer:

yes

Explanation:

Atoms are the smallest particles of matter that cannot break down further.

Answer:

electromagnetic waves can travel through a vacuum where as mechanical waves have to have a medium to travel

Explanation:

Answer:

1 IS CORRECT

Explanation:

Answer:

Explanation:The Hertzsprung-Russell Diagram is a graphical tool that astronomers use to classify stars according to their luminosity, spectral type, color, temperature and evolutionary stage. Stars in the stable phase of hydrogen burning lie along the Main Sequence according to their mass.

hoped this helped

Apply hooke's law

\(\\ \rm\rightarrowtail F=kx\)

\(\\ \rm\rightarrowtail 1176=k(0.0075)\)

\(\\ \rm\rightarrowtail k=156800N/m\)

#b

We need mass again

\(\\ \rm\rightarrowtail F=kx\)

\(\\ \rm\rightarrowtail mg=kx\)

\(\\ \rm\rightarrowtail 9.8m=156800(0.0048)\)

\(\\ \rm\rightarrowtail 9.8m=752.64\)

\(\\ \rm\rightarrowtail m=76.8kg\)

Answer:

In legal terminology, the assured clear distance ahead (ACDA) is the distance ahead of any terrestrial locomotive device such as a land vehicle, typically an automobile, or watercraft, within which they should be able to bring the device to a halt.[1] It is one of the most fundamental principles governing ordinary care and the duty of care for all methods of conveyance, and is frequently used to determine if a driver is in proper control and is a nearly universally implicit consideration in vehicular accident liability.[2][3][4] The rule is a precautionary trivial burden required to avert the great probable gravity of precious life loss and momentous damage.[5][6][7] Satisfying the ACDA rule is necessary but not sufficient to comply with the more generalized basic speed law, and accordingly, it may be used as both a layman's criterion and judicial test for courts to use in determining if a particular speed is negligent, but not to prove it is safe.[8] As a spatial standard of care, it also serves as required explicit and fair notice of prohibited conduct so unsafe speed laws are not void for vagueness.[9][10][11] The concept has transcended into accident reconstruction and engineering.[12]

This distance is typically both determined and constrained by the proximate edge of clear visibility, but it may be attenuated to a margin of which beyond hazards may reasonably be expected to spontaneously appear. The rule is the specific spatial case of the common law basic speed rule,[13] and an application of volenti non fit injuria. The two-second rule may be the limiting factor governing the ACDA, when the speed of forward traffic is what limits the basic safe speed, and a primary hazard of collision could result from following any closer.[2][3]

As the original common law driving rule preceding statutized traffic law,[13] it is an ever important foundational rule in today's complex driving environment. Because there are now protected classes of roadway users–such as a school bus, mail carrier, emergency vehicle, horse-drawn vehicle, agricultural machinery, street sweeper, disabled vehicle,[14] cyclist, and pedestrian–as well as natural hazards which may occupy or obstruct the roadway beyond the edge of visibility,[14] negligence may not depend ex post facto on what a driver happened to hit, could not have known, but had a concurrent duty to avoid.[13][15] Furthermore, modern knowledge of human factors has revealed physiological limitations–such as the subtended angular velocity detection threshold (SAVT)–which may make it difficult, and in some circumstance impossible, for other drivers to always comply with right-of-way statutes by staying clear of roadway.[16][17]

As common law rule or statute

Determining the ACDA

ACDA rule-specific case generalized to the Basic Speed Law

Safe speed

"Assurance" beyond proximate edge of clear visibility as transference of liability

Derivations

Tables of reference constants and safe speeds

See also

Notes

References

pa follow at pa brainlest na den

Answer:if you cannot clearly discern time and distance

Explanation:

The time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.

Based on the given information,

• The air-track glider connected with a spring oscillates between the 10 cm mark and the 60 cm mark.  

• In 33 seconds, the glider completes 10 oscillations.  

Here we have to find the period, frequency,amplitude and maximum speed of glider.

The time period of one oscillation is 33s/10 = 3.3s

The frequency f is the reciprocal of the time period.

f= 1/T =1/3.3 S = 0.303 Hz

The amplitude A is

A = 1/2 (60cm - 10cm)

= 25cm

Maximum speed of glider is

V max = 2π/t (0.25m)

V max = 0.47575m/s

Thus,  time period is 3.3 s, frequency is 0.303 Hz, amplitude is 0.25 m, and maximum speed is 0.47575 m/s.

To know more about frequency click-

https://brainly.com/question/254161

#SPJ4

Answer:

D

Explanation:

I’m pretty sure it’s correct but I don’t really know. Just trying to pass science

Answer:

The correct answer is D trust me I did the test

When the concave lens moves in it exit by spreading out like water

Answer:

Due to the resistance of air, a drag force acts on a falling body (parachute) to slow down its motion. Without air resistance, or drag, objects would continue to increase speed until they hit the ground. The larger the object, the greater its air resistance. Parachutes use a large canopy to increase air resistance. Also, Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down. Sorry if not helpful.

Both magnitude and direction. Centripetal acceleration always act towards the center of the circle and have constant value.

"When a body moves in a circular path the acceleration acting towards the center of the circle is called centripetal acceleration." Hence the magnitude and direction will not change.

Know more about centripetal acceleration here

https://brainly.com/question/14394290

#SPJ2

Answer:

Acceleration, a = 0.04 m/s²

Explanation:

Given the following data;

Force = 60N

Mass = 1500kg

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

\( F = ma\)

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

Making acceleration (a) the subject, we have;

\(Acceleration (a) = \frac{F}{m}\)

Substituting into the equation;

\(Acceleration (a) = \frac{60}{1500}\)

Acceleration, a = 0.04 m/s²

The initial potential energy of the swing is 0 J. The final potential energy of the swing is 345.35J. The change in potential energy of the swing is 345.35J. And the work done is 345.35J.

The initial potential energy of the swing is given by:

PE_i = mgh_i

PE_i = 35 kg * 9.81 m/s^2 * 0 m

PE_i = 0 J

where m is the mass of the tire, g is the acceleration due to gravity, and h_i is the initial height of the tire.

The final potential energy of the swing is given by:

PE_f = mgh_f

PE_f = 35 kg * 9.81 m/s^2 * 1.0 m

PE_f = 343.35 J

where h_f is the final height of the tire.

The change in potential energy is:

ΔPE = PE_f - PE_i

ΔPE = 343.35 J - 0 J

ΔPE = 343.35 J

Therefore, the work done on the tire is:

W = ΔPE

W = 343.35 J

So it would take 343.35 J of work to move the tire a vertical distance of 1.0 m at a constant speed of 1 m/s.

To know more about potential energy, here

brainly.com/question/15101923

#SPJ4

For a twin-sized air mattress used for camping has dimensions of 100 cm by 200 cm by 15 cm when blown up, the weight is mathematically given as

M=2923.38N

Generally, the equation for the maximum mass   is mathematically given as

M=Vp-m

Therefore

M=100*200*15(1*10^3)-2kg

M=298kg

In conclusion,  the weight

W=298kg*9.81

W=2923.38N

Read more about mass

https://brainly.com/question/15959704

#SPJ1

Gravitational force between two masses m, and m, is represented as F = Gm₂ m₂ / r^2 where r = xi+yj + zkG, m, m₂ are nonzero constants and let's assume that I = 0

a) Calculation:For F = Gm₂ m₂ / r^2.

Using r = xi+yj + zk and let r^2 = x^2 + y^2 + z^2∴ F = Gm₂ m₂ / (x^2 + y^2 + z^2), Where G, m, m₂ are nonzero constants. Divergence of F = ∇ · F= 1/r^2(d/dx(r^2Fx) + d/dy(r^2Fy) + d/dz(r^2Fz))= 1/r^2(d/dx(r^2Gm₂ m₂ x/(x^2+y^2+z^2)^(3/2)) + d/dy(r^2Gm₂ m₂ y/(x^2+y^2+z^2)^(3/2)) + d/dz(r^2Gm₂ m₂ z/(x^2+y^2+z^2)^(3/2)))= 1/r^2(d/dx(r^2Gm₂ m₂ x/(x^2+y^2+z^2)) * (x^2+y^2+z^2)^(3/2) + d/dy(r^2Gm₂ m₂ y/(x^2+y^2+z^2)) * (x^2+y^2+z^2)^(3/2) + d/dz(r^2Gm₂ m₂ z/(x^2+y^2+z^2)) * (x^2+y^2+z^2)^(3/2))= 1/r^2(Gm₂ m₂ [2x(x^2+y^2+z^2)-3x^2]/(x^2+y^2+z^2)^(5/2) + Gm₂ m₂ [2y(x^2+y^2+z^2)-3y^2]/(x^2+y^2+z^2)^(5/2) + Gm₂ m₂ [2z(x^2+y^2+z^2)-3z^2]/(x^2+y^2+z^2)^(5/2))= 1/r^2(Gm₂ m₂ [(2x^2+2y^2+2z^2-3x^2)/(x^2+y^2+z^2)^(3/2)] + [2x^2+2y^2+2z^2-3y^2]/(x^2+y^2+z^2)^(3/2)] + [2x^2+2y^2+2z^2-3z^2]/(x^2+y^2+z^2)^(3/2)])= 1/r^2(Gm₂ m₂ [x^2+y^2+z^2]/(x^2+y^2+z^2)^(3/2))= 0.

Curl of F = ∇ × F= i(d/dy(Fz) - d/dz(Fy)) - j(d/dx(Fz) - d/dz(Fx)) + k(d/dx(Fy) - d/dy(Fx))= i(d/dy(Gm₂ m₂ z/(x^2+y^2+z^2)) - d/dz(Gm₂ m₂ y/(x^2+y^2+z^2))) - j(d/dx(Gm₂ m₂ z/(x^2+y^2+z^2)) - d/dz(Gm₂ m₂ x/(x^2+y^2+z^2))) + k(d/dx(Gm₂ m₂ y/(x^2+y^2+z^2)) - d/dy(Gm₂ m₂ x/(x^2+y^2+z^2)))= i(Gm₂ m₂ [-2xz]/(x^2+y^2+z^2)^(5/2)) - j(Gm₂ m₂ [-2yz]/(x^2+y^2+z^2)^(5/2)) + k(Gm₂ m₂ [(x^2+y^2-2z^2)]/(x^2+y^2+z^2)^(5/2))

b) Calculation:The line integral of F along a curve C can be evaluated by the following formula∫C F.dr = ∫∫ ( ∇ x F) ds, Where r is the position vector of the curve, s is the scalar parameter representing the curve, and the integral is evaluated from the initial point to the final point.

Using the curl of F obtained in part a) and for the surface with ∂S as C∫C F.dr = ∫∫ ( ∇ x F) ds= ∫∫ curl(F) ds= ∫∫ (-2xz i -2yz j + (x^2+y^2-2z^2)k) ds...[1]

Let's consider the surface S as a plane perpendicular to the z-axis of the form ax+by+c=0 and the curve C as the intersection of the plane and the cylinder x^2 + y^2 = a^2.

Let's choose the unit normal to the surface S as k (along the z-axis).

The curl of F is a vector field perpendicular to the plane and along the direction of k.

Thus the integral can be written as∫C F.dr = ∫∫ ( ∇ x F) . k ds= ∫∫ (x^2+y^2-2z^2) ds...[2]

Now let's evaluate the integral over the given plane ax+by+c=0. We can write x = t, y = (c-at)/b and z = 0, where t is the scalar parameter along the line of intersection of the plane and the cylinder (x^2 + y^2 = a^2).

Since the curve C is on the cylinder of radius a, we have x^2+y^2 = a^2 ⇒ t^2+(c-at)^2/b^2 = a^2On solving for t, we have t = (bc±ab √(a^2-b^2-c^2))/[a^2+b^2].

Substituting t in x and y, we get the curve C in the x-y plane as a function of the scalar parameter s asx = (bc±ab √(a^2-b^2-c^2))/[a^2+b^2]y = (c-at)/b= (c-(bc±ab √(a^2-b^2-c^2))/[a^2+b^2])/b.

Now we can evaluate the integral over the curve C, which is along the intersection of the plane and the cylinder.

Integral over C (x^2+y^2-2z^2) ds= ∫t₁^t₂ [(t^2 + [(c-at)^2]/b^2 - 2(0)^2)^(1/2)] dt= ∫t₁^t₂ [(a^2-b^2-c^2)t^2+2bc(c-at)+b^2c^2-a^2b^2]^(1/2) dt.

Now we can choose the value of t₁ and t₂ such that the square root in the integrand is minimized (so that the integral is path-independent).

This can be done by choosing the value of t that gives the minimum value of (a^2-b^2-c^2)t^2+2bc(c-at)+b^2c^2-a^2b^2 over the range of t from t₁ to t₂.

On differentiation with respect to t and equating to 0, we get the value of t = bc/(a^2+b^2).

Substituting this value of t in the integrand, we get the minimum value of the square root in the integrand to be |c| √(a^2+b^2)/|b|.

Thus the integral over C is given by∫C F.dr = ∫∫ (-2xz i -2yz j + (x^2+y^2-2z^2)k) ds= ∫∫ (x^2+y^2-2z^2) ds= ∫t₁^t₂ |c| √(a^2+b^2)/|b| dt= |c| √(a^2+b^2)/|b| (t₂-t₁).

Now we can see that the integral is path-independent as it depends only on the end points t₁ and t₂ and not on the path taken to reach them.

Learn more about position vector here ;

https://brainly.com/question/14552074

#SPJ11