The waste product of photosynthesis is

Answer:

Kinetic Energy

Explanation:

Answer:

In a series circuit, how does the voltage supplied by the battery compare to the voltage on each load? The voltage of the battery is equal to the voltage of each load added together. ... The voltage across the two resistors must both have the same voltage of the battery.

Explanation:

mark me as BRAINLIEST

follow me

carry on learning

Answer:

The voltage of the battery is equal to the voltage of each load added together. The voltage across the two resistors must both have the same voltage of the battery.

>3

Answer

225 meters.

Explanation:

x=x0+30t-(1/2)(1.5)t^2

x=0+30(10)-(1/2)(1.5)(10)^2

x=300-75

x=225

Answer:

Explanation:

1. DECOY PRICING

This occurs when customers make a purchase they must often choose between products with different prices and attributes.

This method of pricing is meant to influence the consumer's purchasing decision and maximise the sales of one particular product. The seller will offer at least three products; two of the products will have a similar or equal price. The two products with the similar prices should be the most expensive ones, and one of the two should be less attractive than the other.

2. LOSS LEADER

This is when a product is sold at a low price (often without profit) in order to stimulate other profitable sales or to attract new customers.

The main is that it will help the business to expand their market share as a whole. It's common practice when first entering a market as it introduces new customers to a service or product in the hope of building a customer base and securing future

3.ODD PRICING

This is a method of psychological pricing a product. Prices ending in 9, 95, 97, 99 are sometimes called “charm prices” and in this type of pricing, the seller fixes a price where the last digits are odd numbers. This is intended to give the buyer no room for manœuvering or for bargaining as the price appears to be less - a product priced at £9.99 will seems much cheaper than one priced at £10.00

4. PRICE DISCRIMINATION

The purpose of price discrimination is to capture the market's consumer surplus and generate the most revenue possible for a product. Identical goods or services are sold at different prices from the same provider to different segments of the market. Industries that commonly use price discrimination include the travel industry, pharmaceuticals and textbook publishers.

5. PRODUCT BUNDLE PRICING

Using this method, sellers will combine several products in the same package. It also serves to move old stock. Blu-ray and videogames are often sold using the bundle approach once they reach the end of their product life cycle. This technique is used at auctions where one attractive item may be included in a lot with a box of less interesting things. Buyers must bid for the entire lot. It’s a good way of moving slow selling products, and in a way is another form of promotional pricing.

Answer:

1=8 ohms 2=0.5 Amps

Explanation:

Answer: Load carrying heavy vehicles have large number of wheels in order to reduce pressure on the contact patch on road. With large number of wheels, it becomes easy to distribute the entire pressure of loads in an even manner.

Explanation:

Question: A ship anchored at sea is rocked by waves that have crests 100 m apart the waves travel at 70m/S, at what frequency do the waves reach the ship?

Answer:

0.7 Hz

Explanation:

Applying,

v = λf............... Equation 1

Where v = velocity of the wave, f = frequency fo the wave, λ = wavelength of the wave

make f the subject of the equation

f = v/λ................. Equation 2

From the question,

Given: v = 70 m/s, λ = 100 m ( distance between successive crest)

Substitute these values into equation 2

f = 70/100

f = 0.7 Hz

Hence the frequency at which the wave reach the ship is 0.7 Hz

Explanation:

u=40

v=?

h=?

v²-u²=2gs

0²-40²=2×10×s

160=20s

s=160/20

=80m/s

total distance= upward distance ×downward distance

=80+80

=160m

total displacement=0 because u and v is the same.

Answer:

The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

Explanation:

Final velocity v = 0

Initial velocity u = 40m/s

We know that,

Using equation of motion

[tex]v^{2} =u^{2} +2gh[/tex]

[tex]0-40^{2} =2[/tex] × [tex]10[/tex] × [tex]h[/tex]

The maximum height is:

[tex]h=80[/tex] [tex]m[/tex]

The  stone will reach at the top and will come down

Therefore, the total distance will be:

[tex]s=h_{1} +h_{2}[/tex]

[tex]s=80m-80m=160m[/tex]

The net displacement is:

[tex]D=h_{1} -h_{2}[/tex]

[tex]D=80m-80m=0[/tex]

Hence, The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

hope this helps.....

Answer:

A cyclotron is a type of particle accelerator.

Explanation:

Cyclotron frequency is the frequency of a charged particle moving perpendicular to the direction of a uniform magnetic field B, since that motion is always circular, the cyclotron frequency is given by equality of centripetal force and magnetic Lorentz force.

Answer: C or B

Explanation:

The EMF induced in the wire moving perpendicularly through a magnetic field is 0.025V. The correct option is C.

The EMF is the electro motive force which causes the current to induce in the object moving in the magnetic field.

Given is the length of wire L =15cm =0.15m, magnetic field strength B = 1.4T and velocity of wire V =0.12 m/s

EMF is related to the length of wire, magnetic field strength  and velocity of wire proportionally.

ε = B x L x V

Plug the values, we get

ε = 1.4 x 0.15 x 0.12

ε = 0.025 Volts

Thus, the correct option is C.

Learn more about EMF.

https://brainly.com/question/15121836

#SPJ2

Answer:

Friction Opposes Motion of an Object.

Now

To get the Net force that Moves an Object and causes acceleration....You subtract the Frictional force

Net force = Pushing Force - Frictional Force

Recall

Net Force; F=Ma

Ma = P - Fr

Now the question asked for How Much force Must be applied to Maintain a Constant velocity.

In a Constant Velocity Motion... Acceleration do not change... Its Zero

So Putting this into the formula above

M(0) = P - Fr

0=P - Fr

Fr = P.

This means

That The force needed to keep this object Moving at Constant Velocity Must be equal to its Frictional Force

Since Frictional Force; Fr =223N

The Applied Force(Pushing Force) Must be equal to 223N too.

Answer:

0.16 h

Explanation:

Speed: 25km per hour

Speed=distance/time

25=4/t

t=4/25

t=0.16 hour

t=9.6 minuites

Brainliest please~~

Answer:

The mass of the dog food added is 9.03 kg

Explanation:

Given;

mass of the shopping cart, m₁ = 14.5 kg

let the mass of the bag added = m₂

the force applied, F = 12 N

initial velocity of the cart-bag system, u = 0

distance traveled by the system, d = 2.29 m

time of motion of the system, t = 3.0 s

The acceleration of the system is calculated as;

[tex]d = ut + \frac{1}{2} at^2\\\\2.29 = 0 + (\frac{1}{2} \times 3^2)a\\\\2.29 = 4.5 a\\\\a = \frac{2.29}{4.5} \\\\a = 0.51 \ m/s^2[/tex]

The total mass of the system (M) is calculated as follows;

F = Ma

M = F/a

M = (12)/(0.51)

M = 23.53 kg

The mass of the dog food added is calculated as;

m₂ = M - m₁

m₂ = 23.53 kg - 14.5 kg

m₂ = 9.03 kg

Answer:

[tex]I=1.9A[/tex]

Explanation:

From the question we are told that:

Meter distance [tex]d=0.0525[/tex]

Magnetic field [tex]B=7.14*19^{-6}T[/tex]

Generally the equation for Magnetic field B is mathematically given by

 [tex]B =\frac{\mu *I }{2* \pi r}[/tex]

 [tex]7.14*19^{-6}T =\frac{ 1.26 *10^{-6} *I }{2* 3.142*0.0525}[/tex]

Where

Constant[tex]\mu=1.26 *10^{-6}[/tex]

 [tex]I=1.9A[/tex]

Therefore

The Current through the wire is

 [tex]I=1.9A[/tex]

Answer: 1.87

Explanation:

acellus

Answer:

The magnitude of the force between the charges is approximately -2,858.77 N

Explanation:

The given details are;

The magnitude of the given charges;

q₁ = -0.00125 C, q₂ = +0.00333 C

The distance between the charges, r = 3.62 m

The magnitude of the electric force between the charges, 'F', between two charged particles or objects is given as follows;

[tex]F = \dfrac{k \cdot q_1 \cdot q_2}{r^2}[/tex]

Where;

k = Constant = 9 × 10⁹ N·m²/C²

q₁ = The magnitude of charge on the first charged object = -0.00125 C

q₂ = The magnitude of charge on the second charged object = +0.00333 C

r = The distance between the charges = 3.62 m

Plugging in the values, gives;

[tex]F = \dfrac{9 \times 10^9 \ \dfrac{N \cdot m^2}{C^2} \times (-0.00125 \, C) \times (+0.00333 \, C)}{\left( 3.62 \, m \right)^2} \approx -2,858.77 \, N[/tex]

The magnitude of the force between the charges, F ≈ -2,858.77 N

Answer:

t = 180 / 1.4 = 129 sec   (time to swim horizontally across river)

S = 129 sec * V     where V is speed of current and S is the distance he will be carried downstream

The problem does not specify V the speed of the river

Answer:

Explanation:

From the given information:

a) the distance(D) showing how dar downstream he will land can be computed as follows:

Assuming the current of the river = 0.2 m/s

[tex]D = \dfrac{180 \ m \times 0.2 \ m/s}{1.4 \ m/s}[/tex]

D = 36 m ÷ 1.4

D = 25.71 m

The required time (t) to reach the other side is:

time (t) = 180 m/ 1.4 m/s

time (t) = 128.57 seconds

Answer:

4

Explanation:

4-3

Answer:

como assim qual a tarefa que educação física? se você me explicar melhor eu consigo te responder !!

Explanation:

Answer:

True

Explanation:

I guess you made a mistake on question.

but I understood what you wanted to say.

Hope this helps... :)

Answer:

a.

[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]

b.

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]

c.

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]

d. The force of gravity acting on the satellite is approximately 1.144 × 10¹³ N

Explanation:

a. The formula for finding the force of gravity, F, acting object on an object is given as follows;

[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]

Where;

F = The force acting between the Earth and the object

G = The gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M = The mass of the Earth = 5.97 × 10²⁴ kg

m = The mass of the object

r = The distance between the center of the Earth and the object

b. Finding the gravitational force, 'F', between the Earth and the scientific satellite, we have;

The given mass of the satellite, m = 1,300 kg

The distance between the center of the Earth and the center of the satellite,  r = The length of the radius of the Earth + The height of orbit of the satellite

The given height of orbit of the satellite, h = 200 km

∴ r = R + h = 6,378 km + 200 km = 6,578 m

Therefore, by plugging in the values, we get;

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]

c. Solving the above equation gives;

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]

d. The force of gravity acting on the satellite, F ≈ 1.144 × 10¹³ Newton

Answer:

the positive charge is the period number

Explanation:

I might be wrong

Answer:

The positive charge is the group number.

Explanation:

Answer:

The answer is Speed=2m/s

Explanation:

S=D/T

S=10m/5s

S=2m/s

Solution :

It is given that the lamp glows brightly for a shorter period of time when the switch is closed on it the switch is put on. But after the some time the lamp goes off and it stops working.

This is because as soon as we on the switch, the current start flowing to the lamp which makes the filament of the lamp to glow, but due to some issue, the current stop flowing even when the switch is on and this stops the lamp from glowing and hence the lamp does not work.

Answer: The equations in column A is matched with gas laws in column B as follows:

21. PV = nRT : (g) Ideal gas law

22. [tex]V_{1}n_{2} = V_{2}n_{1}[/tex] : (f) Avogadro's law

23. [tex]P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex] : (e) Combined Gas Law

24. [tex]P_{1}T_{2} = P_{2}T_{1}[/tex] : (d) Gay-Lusaac's law

25. [tex]V_{1}T_{2} = V_{2}T_{1}[/tex] : (c) Charles' law

26. [tex]P_{1}V_{1} = P_{2}V_{2}[/tex] : (b) Boyle's law

27. [tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex] : (a) Graham's Law of effusion

Explanation:

(A) Ideal gas law: It states that the product of pressure and volume is directly proportional to the product of number of moles and temperature.

So, PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant

T = temperature

So, [tex]P_{1}V_{1} = P_{2}V_{2}[/tex]

[tex]V \propto T\\\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\V_{1}T_{2} = V_{2}T_{1}[/tex]

So,  [tex]P_{1}T_{2} = P_{2}T_{1}[/tex]

So, [tex]V_{1}n_{2} = V_{2}n_{1}[/tex]

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\or, P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex]

[tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex]

Thus, we can conclude that equation in column A is matched with gas laws in column B as follows:

21. PV = nRT : (g) Ideal gas law

22. [tex]V_{1}n_{2} = V_{2}n_{1}[/tex] : (f) Avogadro's law

23. [tex]P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex] : (e) Combined Gas Law

24. [tex]P_{1}T_{2} = P_{2}T_{1}[/tex] : (d) Gay-Lusaac's law

25. [tex]V_{1}T_{2} = V_{2}T_{1}[/tex] : (c) Charles' law

26. [tex]P_{1}V_{1} = P_{2}V_{2}[/tex] : (b) Boyle's law

27. [tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex] : (a) Graham's Law of effusion

Answer:

A...................................

The force of the wall on the car and the car on the wall are equal is true about Newton’s Third Law. Option A is the correct answer.

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that if the car hits the wall, there will be a force exerted by the car on the wall, and an equal and opposite force exerted by the wall on the car. Option A is the correct answer.

The forces involved in the interaction between the car and the wall are equal in magnitude but opposite in direction, as dictated by Newton's Third Law. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object exerts a force of equal magnitude but in the opposite direction on the first object.

Learn more about Newton here:

https://brainly.com/question/28171613

#SPJ2

The complete question is, "In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?

a. The force of the wall on the car and the car on the wall are equal

b. The force of the wall on the car is greatest

c. The force of the car on the wall is greatest

d. There is not enough information to tell"

Explanation:

answer is in the picture above