Classify the statement below as either Speed or Velocity?

A jeep was driving 40 mph
heading to the grocery store.

Answer:

30 m/s.

Explanation:

The kinetic energy of a toy truck with a mass of 0. 75 kg and a velocity of 4m/s is 6J.

HOW TO CALCULATE KINETIC ENERGY:

The kinetic energy of a moving body can be calculated by using the following formula:

K.E = ½mv²

Where;

According to this question, a toy truck has a mass of 0.75 kg and a velocity of 4 m/s. The kinetic energy is calculated as follows:

K.E = ½ × 0.75 × 4²

K.E = 8 × 0.75

K.E = 6J

Therefore, the kinetic energy of a toy truck with a mass of 0. 75 kg and a velocity of 4m/s is 6J.

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Answer:

-4,200,000 J

Explanation:

Answer:

Approximately [tex]4.18 \times 10^{16}\; \rm N[/tex].

Explanation:

Consider two objects of mass [tex]m_{1}[/tex] and [tex]m_{2}[/tex]. Let [tex]r[/tex] denote the distance between the center of mass of each object. Let [tex]G[/tex] denote the gravitational constant. ([tex]G \approx 6.67 \times 10^{-11}\; {\rm m^{3}\cdot kg^{-1}\cdot s^{-2}}[/tex].)

By Newton's Law of Universal Gravitation, the size of gravitational attraction between these two objects would be:

[tex]\begin{aligned}F &= \frac{G\, m_{1}\, m_{2}}{r^{2}}\end{aligned}[/tex].

In this question, [tex]m_{1} = 5.98\times 10^{24}\; {\rm kg}[/tex] and [tex]m_{2} = 6.24 \times 10^{23}\; {\rm kg}[/tex] are the mass of the two planets. The distance between the two planets is [tex]r = 7.83 \times 10^{10}\; \rm m[/tex] (approximately the same as the distance between the center of mass of planet Earth and the center of mass of Mars.)

Apply Newton's Law of Universal Gravitation to find the size of gravitational attraction between the two planets:

[tex]\begin{aligned}F &= \frac{G\, m_{1}\, m_{2}}{r^{2}} \\ &= \frac{1}{(7.83 \times 10^{10}\; {\rm m})^{2}} \\ &\quad \times (6.67 \times 10^{11}\; {\rm m^{3} \cdot kg^{-1} \cdot s^{-2}}) \\ &\quad \times (5.98 \times 10^{24}\; {\rm kg}) \\ &\quad \times (6.42 \times 10^{23}\; {\rm kg}) \\ &\approx 4.18 \times 10^{16}\; {\rm kg \cdot m \cdot s^{-2}} \end{aligned}[/tex].

Since  [tex]1\; {\rm kg \cdot m \cdot s^{-2}} = 1\; {\rm N}[/tex], the size of gravitational attraction between the two planets would be approximately [tex]4.18 \times 10^{16}\; {\rm N}[/tex].

Answer:

[tex]\\ \sf\longmapsto KE=\dfrac{1}{2}mv^2[/tex]

[tex]\\ \sf\longmapsto \dfrac{1}{2}(0.02)v^2=400[/tex]

[tex]\\ \sf\longmapsto 0.01v^2=400[/tex]

[tex]\\ \sf\longmapsto v^2=40000[/tex]

[tex]\\ \sf\longmapsto v=200m/s[/tex]

Answer:

-8

5.092×10

Explanation:

given,

distance (d) =15m

mass (M1) =350kg

mass(M2) = 492kg

Gravitational constant (G)=6.67×10^-11

we know,

gravitational force=(Gm1m2)÷d^2

=(6.67×10^-11×350×492)÷15^2

= (1.14857×10^-5)÷225

= 5.092×10^-8

Answer:

This is what the Ecological Footprint does: It measures the biologically productive area needed to provide for everything that people demand from nature: fruits and vegetables, meat, fish, wood, cotton and other fibres, as well as absorption of carbon dioxide from fossil fuel burning and space for buildings and roads.

Why it's useful:

Ecological footprint (EF), measure of the demands made by a person or group of people on global natural resources. It has become one of the most widely used measures of humanity’s effect upon the environment and has been used to highlight both the apparent unsustainability of current practices and the inequalities in resource consumption between and within countries.

Answer:

I believe its A: Sports biomechanics.

Answer:

397 j

Explanation:

Because 5.0kg yuh

The kinetic energy of the block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s is 396.9 J.

A moving item or particle has a certain kind of power called kinetic energy. When an item exerts a net force to perform work, which involves the transfer of energy, the object accelerates and as a result, gains kinetic energy.

The amount of kinetic energy that a moving object or particle possesses is determined by both its mass and rate of motion. Any combination of vibration, axis rotation, translation (or movement along a path from one location to another), and translation are all possible motion types.

Given:

The mass of the block, m = 5 kg,

The horizontal force, F = 80 N,

The velocity of the block, v = 12.6 m / s,

Calculate the kinetic energy by the formula given below,

[tex]KE = 1/2mv^2[/tex]

KE = 1 / 2 × 5 × 12.6²

KE =  1 / 2 × 5 × 158.6

KE = 396.9 J

Therefore, the kinetic energy of the block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s is 396.9 J.

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The complete question is A block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s. ​Find its kinetic energy?

Answer:

orange it is lol

Explanation:

Answer:

HELP ORANGE DOESNT LOOK LIKE A WORD ANYMORE

Explanation:

Answer:

The reaction force is the water pushing the swimmer in the rightward direction.

Hope you could get an idea from here.

Doubt clarification - use comment section.

Answer:

It is direct proportionality. The greater the mass, the greater is the gravitational potential energy. The equation for GPE is : GPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height above the ground. As you can see GPE is directly proportional to mass, and height. KT.

Explanation:

Gravitational potential energy is a function of both the mass of your system and the mass of the thing generating the gravity field around your system.

The relationship is linear, which means that if you multiply or divide one of the masses by some number but leave everything else the same, you multiply or divide the potential energy by the same number. A 3kg mass has three times the gravitation potential energy of a 1kg mass, if placed in the same location.

Answer:

Friction

Explanation:

As the skateboarder goes down the ramp, potential energy is converted to kinetic energy. Because of friction, some of the energy in the system is converted to heat energy. Once the kinetic energy is converted to heat, the energy cannot be converted back to the potential or kinetic energy in the system.

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

The net force on the system of the two blocks is calculated as follows;

[tex]m_2 g - T = m_1 a[/tex]

where;

[tex]m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2[/tex]

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

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This question involves the concepts of the law of conservation of momentum and velocity.

The velocity of the eight ball is "5.7 m/s".

According to the law of conservation of momentum:

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

where,

m₁ = mass of number three ball = 5 g

m₂ = mass of the eight ball = 6 g

u₁ = velocity of the number three ball = 3 m/s

u₂ = velocity of the eight ball = - 1 m/s (negative sign due to opposite direction)

v₁ = final velocity of the three number ball = - 5 m/s

v₂ = final velocity of the eight ball = ?

Therefore,

(5 g)(3 m/s) + (6 g)(- 1 m/s) = (5 g)(- 5 m/s) + (6 g)(v₂)

[tex]v_2=\frac{34\ g.m/s}{6\ g}\\\\[/tex]

v₂ = 5.7 m/s

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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.

NOTICE

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

consumables are employed in conjunction with LECO instruments. If purchaser defaults in making payment for

any parts or equipment, this warranty shall be void and shall not apply to such parts and equipment. No late

payment or cure of default in payment shall extend the warranty period provided herein.

LECO Corporation is not responsible for damage to any associated instruments, equipment or apparatus nor wil

LECO be held liable for loss of profit or other special damages resulting from abuse, neglect, or use in disregard

of instructions. The Buyer, their employees, agents and successors in interest assume all risks and liabilities for

the operation, use and/or misuse of the product(s) described herein and agree to indemnify, hold harmless and

defend the seller from any and all claims and actions arising from any cause whatsoever, including seller's

negligence for personal injury incurred in connection with the use of said product(s) and any and all damages

proximately resulting therefrom.

CAUTION

The instrument should be operated only by technically qualified individuals who have fully read and understand

these instructions. The instrument should be operated only in accordance with these instructions.

The operator should follow all ,of the warnings and cautions set forth in the manual and the operator should follow

and employ all applicable standard laboratory safety procedures.

LECO'" is a registered trademark of the LECO Corporatio

As the spring is compressed, it performs

-1/2 (252 N/m) (0.118 m)² ≈ -1.75 J

of work on the clay, while gravity does

(0.263 kg) (9.80 m/s²) (0.118 m) ≈ 0.304 J

on it. Then the total work W performed on the clay ball is approximately -1.45 J.

By the work-energy theorem, W is equal to the change in the clay ball's kinetic energy ∆K. If v is the speed with which it hits the spring, then

W = ∆K

-1.45 J = 0 - 1/2 mv²

Solve for v :

-1.45 J = -1/2 (0.263 kg) v²

v² ≈ 11.0 m²/s²

v ≈ 3.32 m/s

Answer:

When an object is balanced, about a pivot, the total clockwise moment must be equal to the total anticlockwise moment about that pivot.

Hope that helps.

When an object is balanced about a pivot, the total clockwise moment must be equal to the total anticlockwise moment.

Clockwise involves turning to the right while going in the direction of a clock's hands. The moment is referred to as a clockwise moment if the force acting on a body rotates the body clockwise with regard to the axis of rotation.

On the other hand, the moment is referred to as an anti-clockwise moment if the force rotates the body in the opposite direction. The clock face rule can be used to determine this magnet's polarity.

The face of the loop will display the North Pole if the current is flowing counterclockwise. On the other hand, if the current is moving counterclockwise, the South Pole is visible on the loop's face.

Therefore, the total clockwise and total counterclockwise moments must match when an object is balanced about a pivot.

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Answer:

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

Answer:

you should change that name

Answer:

geosphere and hydrosphere

Explanation:

Answer:60mph means it runs every minute.

Explanation: because it makes sense.

Answer:

  2640 ft, or 1/2 mile

Explanation:

Recognizing that 30 s = 0.5 min, we can convert the units like this.

  distance = speed × time

  distance = (60 mi/h) × (1 h/(60 min)) ×(0.5 min) = 0.5 mi

In feet, that is ...

  (5280 ft/mi)(0.5 mi) = 2640 ft

A cheetah runs 1/2 mile, or 2640 feet in 30 s.

Answer:

Friction, normal force, and weight

Explanation:

If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.

Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)

Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.

The force might make the object move or bring the object to rest or change in direction of motion

The magnitude of the magnetic field 0.01 m from the same wire is 0.2 T.

Given to us:

Magnetic field, [tex]B_1 = 0.1\ T[/tex]

Radius of wire, [tex]R_1 = 0.02\ m[/tex]

To find out the magnitude of the magnetic field 0. 01 m from the same wire, we need to find out current first. we will use the formula,

[tex]B = \dfrac{\mu_oI }{2\pi R},\\\rn\\where,\\B= magnetic\ field\\\mu_o = 4\pi\times 10^{-7} m\cdot kg\cdot s^{-2} A^{-2}\ is\ the\ magnetic\ constant\\I= current\\R= radius\ of\ the\ wire[/tex]

Putting the values,

[tex]B_1 = \dfrac{\mu_oI }{2\pi R_1},\\\rn\\\\0.1= \dfrac{4\times \pi \times 10^{-7}\times I}{2\times \pi\times0.02}\\\\I=10,000\ A[/tex]

Now, for [tex]B_2[/tex]

[tex]B_2 = \dfrac{\mu_oI }{2\pi R_2},\\\rn\\\\B_2= \dfrac{4\times \pi \times 10^{-7}\times 10,000}{2\times \pi\times0.01}\\\\B_2= 0.2\ T[/tex]

Hence, the magnitude of the magnetic field 0. 01 m from the same wire is 0.2 T.

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Answer: 0.22

Explanation: Subtract 0.01kg from 0.23kg

The centripetal acceleration which always  act toward the center of the circle. The vectors are perpendicular. The correct answer for this question, is option D

The speed of an object is always constant in a  circular motion while the object keeps changing direction regularly. The velocity will keep changing because of the change in direction.

Whenever there is a change in velocity, there must be acceleration.

When the object is moving in a circular path with a constant speed, the directions of an object's velocity and acceleration vectors will be perpendicular to each other

The reason is because acceleration is centripetal acceleration which always  act toward the center of the circle.

Therefore, the correct answer for this question, is option D

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So acceleration of the bike which started from rest, reaches at speed of 7.10 m/s by travelling distance 35.4 m will be 0.712 m/ sec^2.