What is the velocity of a dropped object after it has fallen for 3.0 s?

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:

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.

Answer:

20 meters/sec

Explanation:

See attachment.

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

v = 20 m/s

Answer:

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

Hope that helps. x

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 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

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

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.

To find more about photoelectric effect, refer the link ;

https://brainly.com/question/26465043

#SPJ5

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:

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 wave is carrying more energy

Explanation:

trust me broski

Answer:

a

Explanation:

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:

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.

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.

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:

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

Explanation:

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

Learn more about work here: https://brainly.com/question/8119756

Answer:

papi sus

Explanation:

Answer:

>400N is needed to balance that lever

[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:

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:

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:

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