Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1. 2 m high. Student 2 pushes an identical box up a 5 m ramp from the floor to the top of the same table. Which student did the MOST work?.

Answer:

20 meters/sec

Explanation:

See attachment.

2m/s^2 = (v - 0)/10s

v = 20 m/s

Answer:

This is because some kinetic energy had been transferred to something else.

Explanation:

An inelastic collision is a collision in which there is a loss of kinetic energy. While momentum of the system is conserved in an inelastic collision, kinetic energy is not.

Answer:

>400N is needed to balance that lever

The synonym of a word( observe, infer, repetition, replication and data) means another word that is nearest in meaning to the given word.

The synonym of the words given from the question is given below:

Therefore, synonym of a word means another word that is nearest in meaning to the given word.

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Hi there!

a.

The equation for centripetal acceleration is as follows:

[tex]\large\boxed{a_c = \frac{v^2}{r}}}[/tex]

Plug in the given values to solve:

[tex]\large\boxed{a_c = \frac{(65)^2}{75} = 56.33 m/s^2}[/tex]

b.

According to Newton's Second Law:

[tex]\large\boxed{\Sigma F = ma}}[/tex]

The acceleration is v²/r, so the net force is:

[tex]\large\boxed{\Sigma F = m(\frac{v^2}{r})}}[/tex]

Multiply by the given mass:

[tex]\large\boxed{\Sigma F = 700(56.33) = 39433.33 N}}[/tex]

c.

There is NO net force in the vertical direction since the object is NOT accelerating in the vertical direction (normal force and weight cancel).

Thus, the ONLY net force experienced by the object is in the horizontal direction and is EQUAL to the centripetal force.

Answer:

its true since i already did this last year, if i am wrong i apologize

Explanation:

Answer:

The wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.

Given data in the question;

Speed of wave;

Frequency of wave;

wavelength;

To determine the wavelength of the radio wave, we use the expression for the relations between wavelength, frequency and speed.

Where  is wavelength, f is frequency and c is the speed.

We substitute our given values into the equation

Therefore, the wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.

Explanation:

Photovoltaic panels are used to produce alternating current using sunlight. It uses solar cells which are photoactive and produce a photoelectric effect and thereby a potential is created.

When photons from the light energy is hit on a metallic surface the photon collides with the electrons of the metals and electrons are ejected from the metal surface. This phenomenon is called photoelectric effect.

In a solar photovoltaic panel, photovoltaic cells layers of semi-conducting materials, such as silicon, are sandwiched photovoltaic cells. When photons from sunshine strike each layer, its unique electronic properties energise and produce an electric field.

The current required to generate electricity is produced through what is known as the photoelectric effect. Electricity produced by solar panels is in the direct current form. This is then transformed into an alternating current by passing via an inverter.

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Answer:

the orbital in which the electron is located relative to the atom's nucleus.

Explanation:

Answer:

the principal energy level of an electron refers to the shell or orbital in which the electron is located relative to the atom's nucleus.

Answer:

Explanation:

Equipment manufactured by LECO(8 Corporation, St. Joseph, Michigan is warranted free from defect in material

and workmanship for a period of six months from the date of purchase. Equipment not manufactured by LECO is

covered to the extent of warranty provided by the original manufacturer and this warranty does not cover any

equipment, new or used, purchased from anyone other than the LECO Corporation. All replacement parts shall

be covered under warranty for a period of thirt days from date of purchase. LECO MAKES NO OTHER

REPRESENTATION OR WARRANTY OF ANY OTHER KIND, EXPRESSED OR IMPLIED, WITH RESPECT TO

THE GOODS SOLD HEREUNDER, WHETHER AS TO MERCHANTABILITY, FITNESS FOR PURPOSE, OR

OTHERWISE.

Expendable items such as crucibles, combustion tubes, chemicals and items of like nature are not covered by

this warranty.

LECO's sole obligation under this warranty shall be to repair or replace any part or parts which, to our

satisfaction, prove to be defective upon return prepaid to LECO Corporation, St. Joseph, Michigan. This

obligation does not include labor to install replacement parts, nor does it cover any failure due to accident, abuse,

neglect, or use in disregard of instructions furnished by LECO. In no event shall damages for defective goods

exceed the purchase price of the goods, and LECO SHALL NOT BE LIABLE FOR INCIDENTAL OR

CONSEQUENTIAL DAMAGES WHATSOEVER.

All claims in regard to the parts or equipment must be made within ten (10) days after Purchaser learns of the

facts upon which the claim is based. Authorization must be obtained from LECO prior to returning any other

parts. This warranty is voided by failure to comply with these notice requirements.

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The warranty on LECO equipment remains valid only when genuine LECO replacernent parts are employed.

Since LECO has no control over the quality or purity of consumable products not manufactured by LECO, the

specifications for accuracy of results using LECO instruments are not guaranteed unless genuine LECO

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LECO'" is a registered trademark of the LECO Corporatio

[tex]\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }} [/tex]

[tex]\qquad[/tex]______________________________

Then –

[tex]\qquad[/tex] [tex]\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]

[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }[/tex]

[tex]\qquad[/tex][tex] \longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }[/tex]

[tex]\qquad[/tex] [tex]\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }[/tex]

[tex]\qquad[/tex] [tex] \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }[/tex]

Therefore, the vertical component of velocity of projectile at this height will be–

☀️[tex]\qquad[/tex][tex] \pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }[/tex]

Answer:

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is v sintheta / √2

Answer:

papi sus

Explanation:

Range be R and height be h

[tex]\boxed{\sf R=\dfrac{u^2sin2\theta}{g}}[/tex]

[tex]\boxed{\sf h=\dfrac{u^2sin^2\theta}{2g}}[/tex]

u=initial velocity

theta is angle of projection.

g=acceleration due to gravity

ATQ

[tex]\\ \sf\longmapsto R=2h[/tex]

[tex]\\ \sf\longmapsto \dfrac{u^2sin2\theta}{g}=\dfrac{2u^2sin^2\theta}{2g}[/tex]

Cancelling required ones

[tex]\\ \sf\longmapsto sin^2\theta=sin2\theta[/tex]

sin2O=2sinOcosO

[tex]\\ \sf\longmapsto sin^2\theta=2sin\theta cos\theta [/tex]

[tex]\\ \sf\longmapsto \dfrac{sin^2\theta}{sin\theta cos\theta=2[/tex]

[tex]\\ \sf\longmapsto \dfrac{sin\theta}{cos\theta}=2[/tex]

[tex]\\ \sf\longmapsto tan\theta=2[/tex]

[tex]\\ \sf\longmapsto \theta=tan^{-1}(2)[/tex]

[tex]\\ \sf\longmapsto \theta=63.4°[/tex]

Done

Option B is correct

Answer:

C

Explanation:

All of the other options are affecting the gravitational force and opposing force that pushed upwards, the diagram represents an object that is not moving up or down.

Answer:

Because in correspondence to the same distance from a mass, the gravitational acceleration is the same for all the bodies. It doesn't depend on the mass of the objects.

Answer:

the wave is carrying more energy

Explanation:

trust me broski

Answer:

a

Explanation:

Answer:

a).  The truck's distance covered for the trip is

                      (60) + (60 - 15)  =  105 kilometers .

b).  Its displacement for the whole trip is the distance

and direction from the start-point to the end-point.

                         15 kilometers east .

Explanation:

Answer:

The force of the car is 15000N.

Explanation:

The unit of force is Newtons (N), so based on the question, the force is 15000 Newtons.

Answer:

[tex]0.6\; \rm m\cdot s^{-1}[/tex], assuming that drag of the water on the log is negligible.

Explanation:

The momentum [tex]p[/tex] of an object is equal to the product of mass [tex]m[/tex] and velocity [tex]v[/tex]. That is, [tex]p = m\, v[/tex].

If the drag of water on the log is negligible, momentum of the beaver and the log, combined, would be preserved.

The momentum of the beaver and the log, combined, was initially [tex]0\; \rm kg\cdot m \cdot s^{-1}[/tex].

The momentum of the beaver right after the dive would be [tex]7.5 \; {\rm kg} \times 4\; {\rm m \cdot s^{-1}} = 30\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

The sum of the momentum of the beaver and the log is conserved and should continue to be [tex]0\; {\rm kg\cdot m \cdot s^{-1}}[/tex] even after the dive. Since the momentum of the beaver is [tex]30\; {\rm kg \cdot m \cdot s^{-1}[/tex] after the dive, the momentum of the log should become [tex](-30)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Since the mass of the log is [tex]50\; {\rm kg}[/tex], the new velocity of this log would be:

[tex]\begin{aligned}v &= \frac{p}{m} \\ &= \frac{(-30)\; {\rm kg \cdot m \cdot s^{-1}}}{50\; {\rm kg}} \\ &= (-0.6)\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].

(The new velocity of the log is negative because the log would be moving away from the beaver.)

The speed of an object is the magnitude of velocity. For this log, a velocity of [tex](-0.6)\; {\rm m \cdot s^{-1}}[/tex] would correspond to a speed of [tex]|(-0.6)\; {\rm m \cdot s^{-1}| = 0.6\; {\rm m \cdot s^{-1}[/tex].

The same formula of work can be applied to all the questions. The answers are:

E1. 100J

E2. 40N

E3. 5m

E4. a.) 360 J           b.) 240 J               c.) 120 J

El. If a horizontally directed force of 40 N is used to pull a box a distance of 2.5 m across a tabletop. The formula to get the much work that will be done by the 40 - N force will be

Work done = force x distance

Work done = 40 x 2.5

Work done = 100 J

E2. If a woman does 160 J of work to move a table 4 m across the floor. We will use the same formula to calculate the magnitude of the force that the woman applied to the table assuming the force is applied in the horizontal direction.

Work done = force x distance

160 = 4F

F = 160/4

Force F = 40 N

E3. Given that a force of 60 N used to push a chair across a room does 300 J of work. Same formula to get how far the chair move in this process.

Work done = force x distance

300 = 60 x distance

distance = 300/60

Distance = 5 m

Therefore, the chair moved 5m away.

E4. Given that a rope applies a horizontal force of 180 N to pull a crate a distance of 2 m across the floor. And a frictional force of 120 N opposes this motion.

a. The work done by the force applied by the rope can  be found by

W = F x S

W = 180 x 2

W = 360 J

b. What is the work done by the frictional force?

W = [tex]F_{r}[/tex] x s

W = 120 x 2

W = 240 J

c. What is the total work done on the crate?

W = (F - [tex]F_{r}[/tex]) x distance

Where  [tex]F_{r}[/tex]  = frictional force

Substitute all the parameters

W = (180 - 120) x 2

W = 60 x 2

W = 120 J

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Answer:

Nuclear energy is the energy stored in atoms that can produce electricity.

Hope that helps. x

When traveling with average speed 40 km/h, the car would cover a distance of 80 km in time t such that

40 km/h = (80 km) / t   ⇒   t = 2 h

so the total travel time is 2 h + 3 h = 5 h.

Average speed is defined as the total distance traveled divided by the time it took to cover that distance. So if the overall average speed was 30 km/h, then

30 km/h = (80 km + x) / (5 h)   ⇒   x = 70 km

where x is the distance traveled in the last 3 h of the trip.

Then the total displacement of the car is 80 km + 70 km = 150 km.

The velocity of the tennis ball when it hits the ground is calculated as 4.43m/s. Details about velocity can be found below.

The velocity of a moving object can be calculated using the equation of motion as follows:

v² = u² + 2as

Where;

v² = 0² + 2(9.8) (1)

v² = 19.6

v = ✓19.6

v = 4.43m/s

Therefore, the velocity of the tennis ball when it hits the ground is calculated as 4.43m/s.

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First, you need to find out the speed of 6x10^10. Then find out the value of 3x10^30 after you do that, Make this visual,

3x10^30. 5x10 to the 19 power

_______.

6x10^10

The answer is c

If the spaceship travels a distance of 6 × 10¹⁰ kilometers with an average speed the time taken to travel 3 × 10³⁰ kilometers is 5 × 10¹⁹ years. Thus, option a is correct.

Speed is a physical quantity associated with a moving body. It is the measure of distance travelled per unit  time . Thus, mathematically it is the ratio of distance to the time.

If we have the time and speed known we can find the distance travelled and vice versa. As the speed increases, the distance covered per unit time will be higher.

Here the spaceship travels  6 × 10¹⁰ kilometers . Thus the time taken to travel 3 × 10³⁰ kilometers can be calculated as follows:

1 year   -    6 × 10¹⁰ km

  ?           -  3 × 10³⁰ km

Cross multiply them we get, 3 × 10³⁰ km/  6 × 10¹⁰ km = 5 × 10¹⁹ years.Hence, option a is correct.

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Answer:

The first one.

If the road becomes wet or crowded, you should ___ slow down and increase your following distance_

The safe distance between vehicles traveling in column specified by the command in light of safety requirements

The slower speed will help you save gas and avoid potential accidents. You can easily eliminate chances of rear end collision by maintaining a safe distance between your car and the vehicles ahead.

hence , a) slow down and increase your following distance is a correct option

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