1. Which among the sports you stated above are you familiar with?​

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

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:

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:

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:

The half-life of the given isotope will be 10 days, if after 30 days only 31.25 grams are remaining from a sample of 250 grams of the sample size taken in the beginning.

The Half-life is the time which is required for a quantity to reduce the content to half of the amount present as its initial value. The term is used in nuclear physics to describe how quickly an unstable atom undergo radioactive decay or how long does stable atoms survive. The term is also used generally to characterize any type of exponential decay.

The half-life of the isotope can be calculated by the formula:

FR = 0.5n

FR = Fraction Remaining = 31.25 g / 250.0 g = 0.1250

n = number of half lives elapsed = ?

0.125 = 0.5n

log 0.125 = n log 0.5

-0.9031 = -0.3010 n

n = 3.000 half lives have elapsed

3 half lives = 30 days

1 half live = 10 days

Therefore, the half-life of the isotope will be 10 days.

Learn more about Half-life here:

https://brainly.com/question/24710827

#SPJ2

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.

The fastest time trial for the first quarter checkpoint is 2.02 s.

The slowest time trial for the first quarter checkpoint is 2.7 s.

The range of the times measured for the checkpoint is 0.68 s.

The given parameters;

The fastest time trial for the first quarter checkpoint is the least measured time value.

fastest time trial = least time measured

fastest time trial =  2.02 s

The slowest time trial for the first quarter checkpoint is the highest measured time value.

slowest time trial = 2.7 s

The range of the times measured for the checkpoint is difference between the fastest time and slowest time.

Range = fastest time - slowest time

Range = 2.7 - 2.02

Range = 0.68 s

Learn more about range of data here: https://brainly.com/question/24326172

Answer:

2.02, 2.15, 0.13

Explanation:

A 2-column table with 5 rows. The first column has entries empty, time of trial number 1 (seconds), time of trial number 2 (seconds), time of trial number 3 (seconds), average time (seconds). The second column labeled one quarter checkpoint has entries 2.15, 2.05, 2.02, 2.07.

Use the table to answer the questions.

What is the fastest time trial for the first quarter checkpoint?

2.02

seconds

What is the slowest time trial for the first quarter checkpoint?

2.15 seconds

What is the range of times measured for this checkpoint?

0.13 seconds

Answer:

60n would be the needed force

The force needed to push a box weighing 30 N up a ramp that has an ideal mechanical advantage of 3 is equal to 10 N.

The mechanical advantage can be described as the ratio of the input force to the output force. The mechanical advantage of any machine can be determined by the ratio of the forces involved to do the work.

The ratio of the resistance force to the effort is called the actual mechanical advantage which will be comparatively less. The efficiency of a machine is always determined by equating the ratio of its output to its input.

The efficiency of the machine is equal to the ratio of the actual mechanical advantage (M.A.) and theoretical mechanical advantage. Mechanical advantage can be defined as the force produced by a machine to the force applied to it.

Given the load = 30 N and the ideal mechanical advantage = 3

Mechanical advantage = Load/ Effort

Input force or effort = Load/ M.A.

Force = 30/3

Input Force = 10 N

Learn more about the Mechanical advantage,  here:

brainly.com/question/16617083

#SPJ2

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].

Answer:

It can rotate in any direction. The sign of the angular acceleration depends on how you set the reference system, it can be both negative or positive.

Answer:

the wave is carrying more energy

Explanation:

trust me broski

Answer:

a

Explanation:

Answer:

20 meters/sec

Explanation:

See attachment.

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

v = 20 m/s

Answer:

papi sus

Explanation:

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

learn more about safe distance

https://brainly.com/question/4144073?referrer=searchResults

#SPJ2

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:

So as two objects are separated from each other, the force of gravitational attraction between them also decreases. If the separation distance between two objects is doubled (increased by a factor of 2), then the force of gravitational attraction is decreased by a factor of 4 (2 raised to the second power).

Explanation:

hope his helps

Answer:

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

Hope that helps. x

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.

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:

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:

Answer:

>400N is needed to balance that lever

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.

[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:

acid rain corrodes a monument. d.

Explanation:it only makes sense because its doing something to a physical object.and the other ones aren't.