Differentiate between Scalar quantity and Vector quantity and give two examples each​

Answers

Answer 1
Answer:

A scalar quantity is a physical quantity which only has magnitude. Scalars can be added algebraically.A vector quantity is a physical quantity which has both magnitude and direction. Vectors should be added vectorially.


Related Questions

find the upward force in Newton when each of these is under water(density of 1g/cm3) a lump of iron of volume 2000cm3​

Answers

Answer:

Upthrust = 19.6 N

Explanation:

When an object is immersed under water, the upward force it experience is called an upthrust. An upthrust is a force which is applied on any object in a fluid which acts in an opposite direction to the direction of the weight of the object.

Upthrust = density of liquid x gravitational force x volume of object

i.e U = ρ x g x vol

Given: ρ = 1g/[tex]cm^{3}[/tex] (1000 kg/[tex]m^{3}[/tex]), volume = 2000 c[tex]m^{3}[/tex] (0.002 [tex]m^{3}[/tex]) and g = 9.8 m/[tex]s^{2}[/tex]

So that;

U = 1000 x 9.8 x 0.002 (kg/[tex]m^{3}[/tex] x [tex]m^{3}[/tex] x m/[tex]s^{2}[/tex])

   = 19.6 Kg m/[tex]s^{2}[/tex]

U = 19.6 Newtons

The upthrust on the iron is 19.6 N.

Which one of the following physical quantities has its S.I. unit m/s?
(i) Acceleration
(ii) Velocity
(iii) Force
(iv) Density​

Answers

Answer:

velocity is the answer of this question.

Answer:

Velocity is the right answer ok

Under normal circumstances: _________
a. Fetal Hb binds to oxygen more tightly than Mb binds.
b. Fetal Hb binds oxygen more tightly in the absence of 2,3-BPG.
c. Fetal Hb does not bind to oxygen.
d. Adult Hb has the lowest affinity for oxygen of the 3.
e. More than one of these statements is correct.

Answers

Answer:

Fetal Hb binds oxygen more tightly than adult Hb (not option a)

a 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its kinetic energy when it's 2.00 m above the ground

Answers

Answer:

KE_2 = 3.48J

Explanation:

Conservation of Energy

E_1 = E_2

PE_1+KE_1 = PE_2+KE_2

m*g*h+(1/2)m*v² = m*g*h+(1/2)m*v²

(0.0780kg)*(9.81m/s²)*(5.36m)+(.5)*(0.0780kg)*(4.84m/s)² = (0.0780kg)*(9.81m/s²)*(2m)+KE_2

4.10J+0.914J = 1.53J + KE_2

5.01J = 1.53J + KE_2

KE_2 = 3.48J

When the lightbulbs were used as the resistors, you observed only a flash of light, as opposed to a continuous glow. Explain why that behavior is expected. After all, the light bulb is directly connected to the power supply.

Answers

Solution :

Whenever the lightbulbs are used as resistors, we throw the switch to the left. This allows the current to flow through the circuit which causes the bulb to glow and also the capacitor gets charged. When the capacitor gets fully charged, the electric field becomes constant between its two plates. Now there is no displacement current induced in the plates of the capacitor. The capacitor works as an open switch and the bulb gets switched off.

And thus the bulb flashes for the moment as opposed to continuous glow.

find the equivalent resistance of this circuit

Answers

Answer:

Req = 564 Ω

Explanation:

The equivalent resistance between R1 and R2:

1/R =1/R1 + 1/R2

1/R =1/960 + 1/640

1/R = 1/384

R = 384

Now, the equivalent resistance between R and R3:

Req = 384 + 180

Req = 564 Ω

if 6000j of energy is supplid to a machine to lift a load of 300N through a vvertical height of 1M calculatework out put​

Answers

Answer:

300J

Explanation:

Work done = Force x the distance travelled in the direction of the force

=300 x 1

=300J

Express 6revolutions to radians

Answers

Answer:

About 37.70 radians.

Explanation:

1 revolution = 2[tex]\pi[/tex] radians

∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)

6 revolutions = 37.6991 or ≈ 37.70 radians

2- A student ran 135 meters in 15 seconds. What was the student's velocity?
*
7.5 m/s
9 m/s
12 m/s
15 m/s

Answers

Answer:

9 Brainly hahaha ............huh

How do you know that a liquid exerts pressure?​

Answers

Answer:

The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.

How much amount of water can be decomposed
through electrolysis by passing 2 F charge?

Answers

Answer:

So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.

If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.

Answers

Answer:

W = 641.52 J

Explanation:

The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:

[tex]Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\[/tex]

where,

W = Work Done = ?

m = mass = 6 kg

v = speed = 4.2 m/s

g = acceleration dueto gravity = 9.81 m/s²

h = height = 10 m

Therefore,

[tex]W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)[/tex]

W = 52.92 J + 588.6 J

W = 641.52 J

What Are the type's of Tidal turbines?

Answers

Answer:

Types of tidal turbines

Axial turbines.

Crossflow turbines.

Flow augmented turbines.

Oscillating devices.

Venturi effect.

Tidal kite turbines.

Turbine power.

Resource assessment.

Answer:

Axial turbines

Crossflow turbines

flow augmented turbines

I need help with this please!!!!

Answers

Answer:

1.84 hours

I hope it's helps you

Someone help me with these questions please!

Answers

Answer:

a 25 and b 25

2. 26

60n

can you guys pls also solve for average speed.

Answers

Answer:

d_t = 3.05km

v_a = 4.3km/h

Explanation:

42mins*(2/3) = 28mins

42mins-28mins = 14mins

d = v*t

d_1 = (4km/h)*(1h/60mins)*(28mins)

d_1 = 1.87km

d_2 = (5km/h)*(1h/60mins)*(14mins)

d_2 = 1.17km

d_t = d_1+d_2

d_t = 1.87km+1.17km

d_t = 3.05km

v_a = (v_1+v_2)/2

v_a = [(2*4km/h)+5km/h)]/3

v_a = 4.3km/h

state the laws of reflection​

Answers

Answer:

Explanation:

The law of reflection says that the reflected angle (measured from a vertical line to the surface  called the normal) is equal to the reflected angle measured from the same normal line.

All other properties of reflection flow from this one statement.

an aluminum atom has an atomic number of 13 and a mass number of 27,how many
a)protons
b) electrons

pls write the formula too ​

Answers

Element is

[tex]\boxed{\sf {}^{27}Al_{13}}[/tex]

Atomic number=13Mass number=27

[tex]\\ \sf\longmapsto No\:of\:Protons=Atomic \:Number=13[/tex]

And

[tex]\\ \sf\longmapsto No\:of\:Neutrons=Mass\:number-Atomic\:Number[/tex]

[tex]\\ \sf\longmapsto No\:of\:Neutrons=27-13[/tex]

[tex]\\ \sf\longmapsto No\:of\:Neutrons=14[/tex]

And

[tex]\\ \sf\longmapsto No\:of\:electrons=No\:of\:Protons=13[/tex]

In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.

Answers

Answer:

λ₂ = 357.3 nm

Explanation:

The expression for double-slit interference is

          d sin θ = m λ                 constructive interference

          d sin θ = (m + ½) λ        destructive interference.

The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits

         d sin θ = m λ₁

       

now ask for destructive interference for m = 4

        d sin θ = (m + ½) λ₂

we match these two expressions

         m λ₁ = (m + ½) λ₂

         λ₂ = ( m / m + ½) λλ₁  

let's calculate

         λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]

        λ₂ = 357.3 nm

the specific heat capacity of a substance is 500J/kg/oC. Find the heat required to rise the temperature of 10 quintial of the substance by 3 degree celcius

Answers

Heat capacity=c=5000J/kg°CMass=10quintal=1000kg=mTemeperature=T=3°CHeat=Q

[tex]\boxed{\sf Q=mc\Delta T}[/tex]

[tex]\\ \sf\longmapsto Q=1000(5000)(3)[/tex]

[tex]\\ \sf\longmapsto Q=15000000J[/tex]

[tex]\\ \sf\longmapsto Q=1.5\times 10^7J[/tex]

what is the dimensional formula of young modulas​

Answers

Answer:

The dimensional formula of Young's modulus is [ML^-1T^-2]

Answer:

G.oogle : The dimensional formula for Young’s modulus is:

A. [ML−1T−2]A. [ML−1T−2]

B. [M0LT−2]B. [M0LT−2]

C. [MLT−2]C. [MLT−2]

D. [ML2T−2]

explain what would happen if the cell was disconnected from the circuit ( please help me)​

Answers

Most of the time, "cell" means a battery, and if it gets disconnected, all voltage and current in the circuit goes away.  

I know this is a lame answer, but it kinda depends on what the cell was doing in the circuit, and what else is in the circuit besides the cell, and you haven't told us anything about these details, so that's really all we can guess.

Although your question is incomplete a general answer within the concept of your question is provided:    when the cell is disconnected from the Circuit the flow of voltage across the circuit will be halted.

A cell/battery is often used as a voltage source in electrical circuits, since the cell is a power source, when the cell is connected to a circuit the cell discharges some of its voltage to the components of the circuit ( such as capacitors ) to keep the circuit functional. so when the cell is disconnected from the circuit the flow of voltage is halted

Hence we can conclude that without a voltage source ( cell ) in a circuit there will be no flow of voltage across the components of the circuit.

learn more : https://brainly.com/question/16598952

Parallel Wires: Two long, parallel wires carry currents of different magnitudes. If the current in one of the wires is doubled and the current in the other wire is halved, what happens to the magnitude of the magnetic force that each wire exerts on the other?

Answers

Answer:

Explanation:

Given force between 2 currents carrying

wires = F₀

Magnetic force between the2 wires =F₀= (μ₀/4π) x ( 2 (μ₀/4π) x ( 2I₁I₂ / μ) x L

where I₁=Current in wire 1

           I₂= Current in wire 2

           L= Length of the wire

when one current is doubled and the other is halved

I₁= 2 I₁

I₂=    I₂/2

F₀ = (μ₀/4π) x ( 2× (2I₁) (I₂/2) / μ) x L

Where does a body have more weight the poor at the eqator of the earth.​

Answers

At the North Pole or South Pole but ur body itself doesn’t not change it is the force of gravity that changes as u approach the pole

Answer:

Explanation:

Your body weighs more at the pole for two important reasons. Both have to do to the spin of the earth on its axis.

Because of its spin the earth is thicker around the equator than it is through the poles. This means that when you stand on the equator, you are farther away from the center of earth than you would be at the poles. As gravity decreases with the inverse of the square of distance, gravity will be weaker at the equator.

As you are also spinning with the earth, you will have a required centripetal acceleration and force to keep you attached to the ground, This force decreases the effect of gravity so again, you would weigh less at the equator.

1 Poin Question 4 A 85-kg man stands in an elevator that has a downward acceleration of 2 m/s2. The force exerted by him on the floor is about: (Assume g = 9.8 m/s2) А ON B 663 N C) 833 N D) 1003 N​

Answers

Answer:

D) 1003 N​

Explanation:

Given the following data;

Mass of man = 85 kg

Acceleration of elevator = 2 m/s²

Acceleration due to gravity, g = 9.8 m/s²

To find the force exerted by the man on the floor;

Force = mg + ma

If you dive underwater, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum has an area of about 70 mm217 * 10-5 m22, and it can sustain a force of about 7 N without rupturing. If your body had no means of balancing the extra pressure (which, in reality, it does), what would be the maximum depth you could dive without rupturing your eardrum

Answers

Answer:

[tex]h=10m[/tex]

Explanation:

From the question we are told that:

Area [tex]a=70 x 10^{-6}[/tex]

Force [tex]F=7N[/tex]

Generally the equation for Pressure is mathematically given by

Pressure = Force/Area

[tex]P=\frac{F}{A}[/tex]

[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]

[tex]P= 1*10^{5} Pa[/tex]

Generally the equation for Pressure is also mathematically given by

[tex]P=hpg[/tex]

Therefore

[tex]h=\frac{P}{hg}[/tex]

[tex]h=\frac{10000}{1000*9.8}[/tex]

[tex]h=10m[/tex]

Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.189 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.39 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, part (a) being the one with the greater (and positive) value and part (b) being the other value?

Answers

Answer:

The charges are + 74.3 μC and - 74.3 μC

Explanation:

Let the charges be q and q'.

Since the charges initially attract each other with a force of 1.39 N, the force of attraction is given by

F = kqq'/r² where k = 9 × 10⁹ Nm²/C² and r = distance between the charges = 0.189 m

When the charges are brought together, they share their charge equally and have a net charge of (q + q')/2 each.

They now repel each other.

So, the magnitude of the force of repulsion is given by

F' = k[(q + q')/2][(q + q')/2]/r²

F' = k[(q + q')²/4r²

Since the magnitude of the force of attraction and repulsion are the same, we have that

F = F'

kqq'/r² = k[(q + q')²/4r²

qq' = (q + q')²/4

(q + q')² = 4qq'

q² + 2qq' + q'² = 4qq'

q² + 2qq' - 4qq' + q'² = 0

q² - 2qq' + q'² = 0

(q - q')² = 0

q - q' = 0

q = q'

Substituting q = q' into F, we have

F = kqq'/r²

F = kq²/r²

making q subject of the formula, we have

q² = Fr²/k

q = √(Fr²/k)

q = r√(F/k)

Substituting the values of the variables into the equation, we have

q = 0.189 m√(1.39 N/9 × 10⁹ Nm²/C²)

q = 0.189 m√(0.15444 × 10⁻⁹ Nm²/C²)

q = 0.189 m(0.3923 × 10⁻³ C/m)

q = 0.0743 × 10⁻³ C

q = 74.3 × 10⁻³ × 10⁻³ C

q = 74.3 × 10⁻⁶ C

q = 74.3 μC

Since q and q' initially attract, it implies that they initially had opposite charges.

So, q = 74.3 μC and q' = -74.3 μC

So, the charges are + 74.3 μC and - 74.3 μC

Assume the speed of sound is 343 m/s. You are sitting 150 m away from home plate at a baseball game. How much time in seconds elapses between the batter hitting a home run and the moment you actually hear the batter hitting the ball

Answers

Answer:

  t = 0.437 s

Explanation:

Sound is a wave so its speed is constant

         v = x / t

         t = x / v

         

indicates that the distance is x = 150 m

         t = 150/343

         t = 0.437 s

this is the time it takes to hear the hit

To see the blow it is almost instantaneous since the speed of light is much greater c = 3 10⁸ m / s

A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially

Answers

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s

A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m vertically above its starting point. a) what is the vertical component of its initial velocity? b) what is the horizontal component of velocity?​

Answers

Explanation:

Given:

t = 20 seconds

x = 3000 m

y = 450 m

a) To find the vertical component of the initial velocity [tex]v_{0y}[/tex], we can use the equation

[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]

Solving for [tex]v_{0y}[/tex],

[tex]v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}[/tex]

[tex]\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}[/tex]

[tex]\:\:\:\:\:\:\:=120.5\:\text{m/s}[/tex]

b) We can solve for the horizontal component of the velocity [tex]v_{0x}[/tex] as

[tex]x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}[/tex]

or

[tex]v_{0x} = 150\:\text{m/s}[/tex]

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