Can acceleration and velocity be equal vectors?

Answers

Answer 1

Answer:

Velocity is the rate of change of position with respect to time, whereas acceleration is the rate of change of velocity. Both are vector quantities (and so also have a specified direction), but the units of velocity are meters per second while the units of acceleration are meters per second squared.


Related Questions

Which of the following is a vector quantity? i. Force ii. Velocity iii. Acceleration iv. All of these 5771​

Answers

All of these

Option ( iv ) is the correct answer.

Definition

A vector quantity the physical quantity that has both direction as well as magnitude.

A magnet falls through a loop of wire with the south pole entering first. After it has fallen all the way through the wire loop so that the north pole end is moving away from the loop, what is the direction of the induced current in the loop (as viewed from above, looking down on the magnet)

Answers

Answer:

The induced current direction as viewed is clockwise

Explanation:

Lenz's Law states that the induced e. m. f. causes current to be driven in the loop of wire in such a way as to generate magnetic field that are oppose the magnetic flux change which is the source of the induced current

Therefore, as the magnet approaches the coil with the south pole,  the coil produces current equivalent to the upward movement of the south pole of a permanent magnet through it which according to Flemings Right Hand Rule is clockwise

Therefore;

The direction of the induced current in the loop (as viewed from above, looking down the magnet) is clockwise

If the final ​velocity is 0. third equation of motion will be

Answers

Answer:

vf²=vi²+2a∆x

Explanation:

The third equation of motion gives the final velocity of an object under uniform acceleration given the distance traveled and an initial velocity: v 2 = v 0 2 + 2 a d . v^2=v_0^2+2ad. v2=v02+2ad. The graph of the motion of the object.

Phương trình chuyển động thẳng đều của một chất điểm có dạng: x = 2t – 10. (x: km, t: h). Quãng đường đi được của chất điểm sau 2h là bao nhiêu?

Answers

Answer:

Distance cover in 2  hour = 6 kilometer

Explanation:

Given equation:

x = 2t - 10

where

x = kilometer

t = hour

Find:

Distance cover in 2  hour

Computation:

T = 2

So,

x = 2t - 10

x = 2(2) - 10

x = 4 - 10

x = -6

Distance cover in 2  hour = 6 kilometer

Answer:

The distance is 6 km.

Explanation:

The equation of uniform linear motion of a particle has the form: x = 2t – 10. (x: km, t: h). What is the distance traveled by the particle after 2 hours?

x = 2t - 10

distance traveled after t = 2 hours

Substitute t = 2 in the given expression

x = 2 x 2 - 10

x = 4 - 10

x = - 6 km

So, the distance is 6 km.

The refractive index of water is 1.33
and that of diamond is 2.42. Draw a labelled diagram to show how a light ray bends when it travels from water
into diamond.

Answers

Explanation:

light travel slower in daimond

state the term used to describe the turning force exerted by the man pushing down on a lever to lift one end of a heavy log​

Answers

A force called the effort force is applied at one point on the lever in order to move an object, known as the resistance force, located at some other point on the lever.

The way levers work is by multiplying the effort exerted by the user. Specifically, to lift and balance an object, the effort force the user applies multiplied by its distance to the fulcrum must equal the load force multiplied by its distance to the fulcrum. Consequently, the greater the distance between the effort force and the fulcrum, the heavier a load can be lifted with the same effort force.

The term used to describe the turning force exerted when pushing an object is effort.

What is effort?

The is the input force or force applied at one point of a lever in order to overcome a load.

The relationship between effort, load and mechanical advantage of a lever is given as;

M.A = L/E

where;

M.A is mechanical advantageL is the loadE is the applied force or effort

Thus, the term used to describe the turning force exerted when pushing an object is effort.

Learn more about effort here: https://brainly.com/question/24237657

A low-power laser used in a physics lab might have a power of 0.50 mW and a beam diameter of 3.0 mm. Calculate:a.The average light intensity of the laser beam.b. The intensity of a lightbulb producing 100-W light viewed from 2.0 m.c.Compare the intensity of the laser to the intensity of the lightbulb. Is it advisable to look directly at a laser

Answers

Answer:

A) I_laser = 70.74 W/m²

B) I_bulb = 1.989 W/m²

C) it is not advisable to look at the laser beam directly.

Explanation:

We are given;

Power; P = 0.50 mW = 0.5 × 10^(-3) W

Diameter; d = 3 mm = 0.003 m

Radius; r = d/2 = 0.003/2 = 0.0015 m

A) Area of beam; A = πr²

A = 0.0015²π

Now, formula for average intensity is;

I = P/A

I = (0.5 × 10^(-3))/0.0015²π

I = 70.74 W/m²

B) We are told to find the intensity of a lightbulb producing 100-W.

Thus, P = 100 W

A light bulb is spherical in shape. Thus;

Area; A = 4πr²

We are told it's 2 m away.

Thus; r = 2 m

A = 4π(2)²

A = 16π

Thus, I = P/A = 100/16π

I = 1.989 W/m²

C) The intensity of the laser beam is far greater than that of the light bulb. Thus, it is not advisable to look at the laser beam directly.

describe the forest ecosystem with examples​

Answers

A forest ecosystem describes the community of plants, animals, microbes and all other organisms in interaction with the chemical and physical features of their environment: Specifically, a terrestrial environment dominated by trees growing in a closed canopy — a forest, in other words.

Hope this helps you buddy! :)

Write a balanced nuclear reaction for one complete cycle

Answers

Answer:

H + H ------> He + energy. He + He -----> H + H + He.

Explanation:

Hydrogen having one proton and no neutron fuse with hydrogen having one proton and one neutron forming helium atom with the release of photon. After that two helium atoms combine together forming two hydrogen atoms having one proton each whereas one helium atom having two protons and two neutrons present in their nucleus so the end product of this reaction is hydrogen atoms and helium.

A golf club hits a 0.04551 kg golf ball off a golf tee. The club is in contact with the ball for 0.020 s, and the force applied by the club is 115 N. What is the speed of the ball as it leaves the tee

Answers

Answer:

v = 50.5 m/s

Explanation:

F = (m)(^v/^t)

115N = (0.04551kg)(v/(0.020s))

2,526.917161 m/s² = v/(0.020s)

v = 50.53834322 m/s

v = 50.5 m/s

A car travels first 10 km in 20 minutes and another 10 km in 30 minutes. What is the average speed of the car in m/s?

Answers

Total distance = 10 km + 10 km = 20 km

1 km = 1000 m

20km x 1000 = 20,000 m

Total time = 20 min. + 30 min. = 50 minutes

Average speed = Distance /  time

Average speed = 20,000/50 min

Average speed = 400 m/s

; (b) A uniform beam 150cm long weighs 3.5kg and
supported on knife-edges at its ends. The beam
supports a weight 7kg at a distance 30cm from
one end. Find the reactions of the supports.

Answers

Explanation:

(b) A uniform beam 150cm long weighs 3.5kg and

supported on knife-edges at its ends. The beam

supports a weight 7kg at a distance 30cm from

one end. Find the reactions of the supports.

A solenoid passing by a current of 5.0 A generates a magnetic field at its diameter of 50 μT. Thus the number of spirals per length scale is:

A. 5.0 / π Spear / m

B. 10 / π Spear / m

C. 20 / π Spear / m

D. 25 / π Spear / m

Answers

Answer:

D. 25 / π Spiral / m

Explanation:

Given;

current, I = 5 A

magnetic field strength, B = 50 μT = 50 x 10⁻⁶ T

The magnetic field strength is given as;

[tex]B = \mu_0 nI\\\\where;\\\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi \times 10^{-7} T/A.m\\\\n \ is \ the \ number \ of \ spirals \ per \ length\\\\n = \frac{B}{\mu_0 I} = \frac{50 \times 10^{-6}}{5\times 4\pi \times 10^{-7}} = \frac{25}{\pi } \ spirals /m \\\\[/tex]

Therefore, the correct option is D. 25 / π Spiral / m

PLS PLS PLS HELP ME! IT IS DUE IN TEN MINUTES
For his next stunt, Super Dave (80 kg) starts running at 2.00 m/s, then slides down a frictionless slide which is elevated 12.0 m above the ground, then slides across the rough asphalt ground to slow him to a stop (ouch!) He slides across the asphalt for 8.00 m before coming to a stop. What is the coefficient of friction between Super Dave and the ground?

Answers

Answer:

Preface to Students: Welcome to Physics! iv. 1 Motion. 1. 1-1 Speed, Velocity, and. Acceleration. 1. 1-2 Free Fall. 8. 2 Vectors and Projectiles.

257 pages·3 MB

Explanation: I hope this helpssss!!!

Water falls down as a stream from a tap. Why does it not scatter?​

Answers

Answer:

as it hits the ground it's dispersed and this causes it to move in different directions

Please help! Will mark Brainliest.

Answers

Answer:

18 Nm

Explanation:

if the correct answer

1.
"An estimate of the future value of some variable" is the definition of
A. procurement
B. logistics.
C. a forecast.
D. capacity

Answers

The answer is A. procurement.



Define electric current and drift velocity.

Answers

Answer:

Explanation: The voltage or potential difference between two points is defined to be the change in ... I = qnAv relates the drift velocity to the current

A 5.41 kg ball is attached to the top of a vertical pole with a 2.37 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.75 m/s in a horizontal circle with the string remaining taut. Calculate the angle θ, between 0∘ and 90∘, that the string makes with the pole. Use g=9.81 m/s2.

Answers

Answer:

θ = 66º

Explanation:

This exercise of Newton's second law must be solved in part, let's start by finding the slowing down acceleration of the ball

           a = v² / r

the radius of the circle is

          sin θ = r / L

          r = L sin θ

           

we substitute

          a = v² /L sin θ

now let's write Newton's second law

vertical axis

            T_y -W = 0

             T_y = W

radial axis

            Tₓ = m a                 (1)

let's use trigonometry for the components of the string tension

             cos θ = T_y / T

             sin θ = Tₓ / T

             Tₓ = T sin θ

we substitute in 1

            T sin θ = [tex]\frac{m \ v^2}{L \ sin \theta}[/tex]

             T L sin² θ = m v²

we write our system of equations

             T cos θ  = m g

             T L sin ² tea = m v²

we divide the two equations

             L [tex]\frac{sin^2 \theta}{cos \theta}[/tex] = v² / g

             (1 -cos²)/ cos θ  = [tex]\frac{v^2 }{g \ L}[/tex]

             1 - cos² θ  =  [tex]\frac{4.75^2}{9.81 \ 2.37}[/tex]   cos θ

             cos² θ + 0.97044 cos θ -1   = 0

we change variable    cos  θ = x

             x² + 0.97044 x - 1 =0

             x= [tex]\frac{-0.97 \pm \sqrt{0.97^2 - 4 1} }{2}[/tex]

           since the square root is imaginary there is no real solution to the problem, suppose that the radius is 1 m r = 1 m

           T sin θ = [tex]\frac{m \ v^2}{ r}[/tex]

           T cos θ  = m g

resolved

           tan θ =  [tex]\frac{v^2}{ r g}[/tex]

           θ = tan⁻¹ ( 4.75²/ 1 9.81)

           θ = 66º

What is the purpose of the lab the importance of the topic and the question you are trying to answer?

Answers

perpose of lab is to store apparatus and do some experiment

Helps someone to know the exert lengh of something

A 6 kg object's Ug increases by 150 J. What was its change in height?
Please help I don’t understand this and fast please

Answers

Answer:

2.5 m

Explanation:

Potential energy is the energy stored in an object as a result of its position relative to other objects

The change in potential energy is given by:

ΔPE = mgh;

where ΔPE is the change in potential energy, m is the mass if the object, g is the acceleration due to gravity and h is the change in height of the object.

Hence given that g = 10 m/s², ΔPE = 150 J, m = 6 kg, hence:

ΔPE = mgh

150 = 6 * 10 * h

150 = 60h

h = 2.5 m

Hence the change in height is 2.5 m

Observe the given figure and find the the gravitational force between m1 and m2.​

Answers

Answer:

The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N

Explanation:

The details of the given masses having gravitational attractive force between them are;

m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m

The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;

[tex]F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}[/tex]

Where;

F = The gravitational force between m₁ and m₂

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

r₂ = 0.1 m + 0.15 m = 0.25 m

Therefore, we have;

[tex]F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N[/tex]

The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.25 ms2.25 ms from an initial speed of 4.25 m/s4.25 m/s . What is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.75 kg1.75 kg

Answers

Answer:

The force is 3305.6 N.

Explanation:

Final velocity, v = 0

time, t = 2.25 ms

initial velocity, u = 4.25 m/s

mass, m = 1.75 kg

Let the acceleration is a.

Use first equation of motion.

v = u + a t

0 = 4.25 + a x 0.00225

a = - 1888.9 m/s^2

The force is

F = ma

F = 1.75 x 1888.9

F = 3305.6 N

A spring has a spring constant of 5 N/m and is stretched 10 m. What its U,?
a. 50 J
b. 70 J
c. 90 J
d. 250 J

Answers

Answer

I am not sure but it is may be 50J

I hope that's, this answer is fine.

matter can enter and exit which of the following systems?
A. isolated only
B. open only
C. both open and isolated
D. Both closed and isolated

Answers

D will be the correct answer
If I am right
The answer is B open only

Derive the following equations. :a=u-v by t​

Answers

here is the answer!

A 117 kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 62.5 kg person stands on the platform at a distance of 1.05 m from the center, and a 28.3 kg dog sits on the platform near the person 1.43 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.

Answers

Answer:

I_syst = 278.41477 kg.m²

Explanation:

Mass of platform; m1 = 117 kg

Radius; r = 1.61 m

Moment of inertia here is;

I1 = m1•r²/2

I1 = 117 × 1.61²/2

I1 = 151.63785 kg.m²

Mass of person; m2 = 62.5 kg

Distance of person from centre; r = 1.05 m

Moment of inertia here is;

I2 = m2•r²

I2 = 62.5 × 1.05²

I2 = 68.90625 kg.m²

Mass of dog; m3 = 28.3 kg

Distance of Dog from centre; r = 1.43 m

I3 = 28.3 × 1.43²

I3 = 57.87067 kg.m²

Thus,moment of inertia of the system;

I_syst = I1 + I2 + I3

I_syst = 151.63785 + 68.90625 + 57.87067

I_syst = 278.41477 kg.m²

Describe the laws of liquid pressure and explain the term fluid. (Please try to give the correct answer 'coz it's urgent!)​

Answers

Answer:

The laws of liquid pressure are: ... The liquid pressure at a point inside the liquid is same in all directions. The liquid pressure at a point inside the liquid is directly proportional to the depth of that point from the free surface of liquid. The pressure of liquid at same depth is different for different liquids

Answer:

The laws of liquid pressure:

The pressure that is inside the liquid rises with the intensity from the surface of a liquid.

The term fluid:

Fluid is basically liquid and a substance that is known to have no certain shape. The opposite of the term fluid would be solid.

You are pulling a sled using a horizontal rèpe, as shown in the diagram. The rope pulls the sled. exerting a force of 50 N to the right. The snow exerts a friction force of 30 N on the sled to the left. The mass of the sled is 50 kg.

Find the sun of the force on the self

Determine the acceleration of the sled

If the sled has an initial velocity 2m/s to the right, how fast will it be traveling after 5 seconds?

Answers

Answer:

Part 1

20 N

Part 2

0.4 m/s²

Part 3

4 m/s

Explanation:

The force which pulls the sled right = 50 N

The friction force exterted towards left by the snow = -30 N

The mass of the sled = 50 kg

Part 1

The sum of the forces on the sled, F = 50 N + (-30) N = 20 N

Part 2

The acceleration of the sled is given as follows;

F = m·a

Where;

m = The mass of the sled

a = The accelertion

a = F/m

∴ a = (20 N)/(50 kg) = 0.4 m/s²

The acceleration of the sled, a = 0.4 m/s²

Part 3

The initial velocity of the sled, u = 2 m/s

The kinematic equation of motion to determine the speed of the sled is v = u + a·t

The speed, v, of the sled after t = 5 seconds is therefore;

v = 2 m/s + 0.4 m/s² × 5 s = 4 m/s.

What is concave mirror?​

Answers

Answer:

A concave mirror, or converging mirror, has a reflecting surface that is recessed inward (away from the incident light). Concave mirrors reflect light inward to one focal point. They are used to focus light.

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