A car starts from rest. If its acceleration is 1.5 m/s2 in 1.5 seconds, then calculate the
distance travelled by it.

Answers

Answer 1

Answer:

1.6875 m

Explanation:

Here, we are given that,

Initial velocity (u) = 0 (as it starts from rest)Acceleration (a) = 1.5 m/s²Time taken (t) = 1.5 s

Then,

Distance (s) = ?

By using the second equation of motion,

s = ut + ½at²

s is distanceu is initial velocityt is timea is acceleration

⇒ s = 0(1.5) + ½ × 1.5 × (1.5)²

⇒ s = ½ × 1.5 × 2.25

⇒ s = ½ × 1.5 × 2.25

⇒ s = ½ × 3.375

s = 1.6875 m

Distance travelled by it is 1.6875 m.


Related Questions

A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction

Answers

Answer:

The kinetic energy and potential energy lost to friction is 2,420 J.

Explanation:

Given;

total mass, m = 40 kg

initial velocity of the girl, Vi = 5 m/s

hight of the hill, h = 10 m

length of the hill, L = 100 m

initial kinetic energy of the girl at the top hill:

[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]

initial potential energy of the girl at the top hill:

[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]

Total energy at the top of the hill:

E = 500 J + 3920 J

E = 4,420 J

At the bottom of the hill:

final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s

hight of the hill = 0

final kinetic energy of the girl at the bottom of the hill:

[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]

final potential energy of the girl at the bottom of the hill:

[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]

Based on the principle of conservation of energy;  

the sum of the energy at the top hill = sum of the energy at the bottom hill

The energy at the bottom hill is less due to energy lost to friction.

[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]

Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.

Problem

A charged particle is moving in the presence of uniform magnetic field. The mass of the particle

is m = 10−6 kg its charge is Q = 10−5 C and the magnetic field vector is B~ = (1T, 0, 0). At the

beginning the velocity vector of the particle is ~v0 = (12 m/s, 0, 5 m/s).

a.) How large will the x component of the velocity of the particle be in t = 2 s?

b.) Where will the particle be in t = 3.14 s?

c.) How large will the magnitude of the velocity be in t = 2.5 s?​

Answers

Answer:

Answer is a I checked the work

A disk of charge is placed in the x-y plane, centered at the origin. The electric field along the axis of a positive disk of charge... points towards the disk along the z-axis. points away from the disk along the z-axis. always points in the positive z-direction. none of these choices

Answers

Answer:

Points away from the disk along the z-axis.

Explanation:

Along the axis of the disk, which is the z - axis, the total vertical electric field components of the charged disk sum up while the horizontal components cancel out. Thus, leaving only vertical components of electric field along the axis of the disk.

Since the disk is positively charged and electric field lines point away from a positive charge, the electric field along the axis of a positive disk of charge points away from the disk along the z-axis.

PLEASE ANSWR 1ST AND I WILL MARK U BRAINLIEST
Two statements are given- one labeled Assertion (A) and the other labeled Reason ®. Select
the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
a. Both A and R are true, and R is correct explanation of the assertion.
b. Both A and R are true, but R is not the correct explanation of the assertion.
c. A is true, but R is false.
d. A is false, but R is true.
Assertion: An object has a negative acceleration.
Reason: The velocity of an object decreases in the same direction.

Answers

Answer:

Where is the R statement?

A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a constant force of 2850 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s

Answers

Answer:

0.1 s

Explanation:

The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log

So F - f = ma

F - μmg = ma

F/m - μg = a

So, substituting the values of the variables into the equation, we have

a = F/m - μg

a = 2850 N/300 kg - 0.45 × 9.8 m/s²

a = 9.5 m/s² - 4.41 m/s²

a = 5.09 m/s²

Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.

So, making t subject of the formula, we have

t = (v - u)/a

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

t = (v - u)/a

t = (0.5 m/s - 0 m/s)/5.09 m/s²

t = 0.5 m/s ÷ 5.09 m/s²

t = 0.098 s

t ≅ 0.1 s

differentiate between step up and step down transformer​

Answers

Answer:

The main difference between the step-up and step-down transformer is that the step-up transformer increases the output voltage, while the step-down transformer reduces the output voltage.

You swing a bat and hit a heavy box with a force of 1273 N. The force the box exerts on the bat is Group of answer choices less than 1273 N if the box moves. exactly 1273 N whether or not the box moves. None of the above choices are correct. exactly 1273 N only if the box does not move. greater than 1273 N if the bat bounces back. greater than 1273 N if the box moves.

Answers

Answer:

exactly 1273 N whether or not the box moves.

Explanation:

In the case when the bat is swing and it is hitted to a heavy box having a force of 1273 N so here the force of the box that exert on the box should be accurately 1273 N even if the box is moved or not. As the third law of the newton should be equivalent & the opposite reaction

Therefore as per the given situation, the above represent the answer

If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?

Answers

Answer:

72  is the premimum of the insurance.

Explanation:

Below is the given values:

The loading = 0.4

Coinsurance rate = 0.2

Number of units = 100

Total number of units = 100 * 0.4 = 40

Remaining units = 60 * 0.2 = 12

Add the 60 and 12 values = 60 + 12 = 72

Thus, 72  is the premimum of the insurance.

You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 1.8 m/s as they reach the gate. For safety, the skiers should have a speed of no more than 28.0 m/s when they reach the bottom of the ramp. You determine that for a 75kg skier with good form, friction and air resistance will do total work of magnitude 3500 J on him during his run down the slope. What is the maximum height (h) for which the maximum safe speed will not be exceeded?

Answers

Answer:

44.6 m

Explanation:

From the law of conservation of energy, the total energy at the top of the ramp, E equals the total energy at the bottom of the ramp.

E = E'

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at top of ramp = mgh where = height of ramp, K₁ = kinetic energy at top of ramp = 1/2mv₁² where v₁ = speed at top of ramp = 1.8 m/s, W₁ = work done by friction and air resistance at top of ramp = 0 J, U₂ = potential energy at bottom of ramp = 0 J(since the skier is at ground level h = 0), K₂ = kinetic energy at bottom of ramp = 1/2mv₂² where v₂ = speed at bottom of ramp = 28.0 m/s, W₁ = work done by friction and air resistance at bottom of ramp = 3500 J

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

U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh + 1/2mv₁² + W₁ = U₂ + 1/2mv₂² + W₂

mgh + 1/2m(1.8 m/s)² + 0 J = 0 J + 1/2m(28 m/s)² + 3500 J

9.8 m/s² × 75 kg h + 1/2 × 75 kg (3.24 m²/s²) + 0 J = 0 J + 1/2 × 75 kg (784 m²/s²) + 3500 J

(735 kgm/s²)h + 75  kg(1.62 m²/s²) = 75 kg(392m²/s²) + 3500 J

(735 kgm/s²)h + 121.5  kgm²/s² = 29400 kgm²/s² + 3500 J

(735 kgm/s²)h + 121.5 J = 29400 J + 3500 J

(735 kgm/s²)h + 121.5 J = 32900 J

(735 kgm/s²)h = 32900 J - 121.5 J

(735 kgm/s²)h = 32778.5 J

h = 32778.5 J/735 kgm/s²

h = 44.6 m

So, the maximum height of the ramp for which the maximum safe speed will not be exceeded is 44.6 m.

which team won the champions league in 2020 2021​

Answers

Answer:

Chelsea F.C

Explanation:

Chelsea F.C

Soccer

A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .

Required:
How far from the second lens is the final image of an object infinitely far from the first lens?

Answers

Answer:

the required distance is 6 cm

Explanation:

Given the data in the question;

f₁ = 15 cm

f₂ = 5.0 cm

d = 45 cm

Now, for first lens object distance s = ∝

1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'

Now, image distance of first lens s' = 15cm  

object distance of second lens s₂ will be;

s₂ = 45 - 15 = 30 cm

so

1/f₂ = 1/s₂ + 1/s'₂

1/5 = 1/30 + 1/s'₂

1/s'₂ = 1/5 - 1/30  

1/s'₂ = 1 / 6

s'₂ = 6 cm

Hence, the required distance is 6 cm

 

The distance of the final image from the first lens will be is 6 cm.

What is mirror equation?

The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).

The given data in the problem is;

f₁ is the focal length of lens 1= 15 cm

f₂ s the focal length of lens 2= 5.0 cm

d is the distance between the lenses = 45 cm

From the mirror equation;

[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]

If f₁ is the focal length of lens 1 is 15 cm then;

[tex]s'=15 cm[/tex]

f₂ s the focal length of lens 2= 5.0 cm

s₂ = 45 - 15 = 30 cm

From the mirror equation;

[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]

Hence the distance of the final image from the first lens will be is 6 cm.

To learn more about the mirror equation refer to the link;

https://brainly.com/question/3229491

What is the relationship between organ systems and organs? organs are made from one type of organ system organ systems are made from one type of organ organs are made from different types of organ systems organ systems are made from different types of organs

Answers

Organs are made up of different types of organs.

Two point charges exert a 6.10 N force on each other. What will the force become if the distance between them is increased by a factor of 8

Answers

Answer:

0.0953125 N

Explanation:

Applying,

F = kq'q/r²................. Equation 1

Where F = electrostatic force, k = coulomb's constant, q' and q = first and second charge respectively, r = distance between the charge.

From the equation,

If both charges remain constant,

Therefore,

F = C/r²

C = Constant =  product of the two charge(q' and q) and k

Fr² = F'r'²................ Equation 2

From the question,

Given: F = 6.10 N

Assume: r = x m, r' = 8x

Substitute these value into equation 2

6.1(x²) = F'(8x)²

F' = 6.1/64

F' = 0.0953125 N

Hence the new force will become 0.0953125 N

g Calculate the final speed of a solid cylinder that rolls down a 5.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.

Answers

Answer:

[tex]V=8.08m/s[/tex]

Explanation:

From the question we are told that:

Height[tex]h=5.00m[/tex]

Mass [tex]m=0.750kg[/tex]

Radius [tex]r=4.00cm=>0.04m[/tex]

Generally the equation for Total energy is mathematically given by

  [tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]

Therefore

 [tex]V=\sqrt{\frac{4gh}{3}}[/tex]

 [tex]V=\sqrt{\frac{4*9.8*5}{3}}[/tex]

 [tex]V=8.08m/s[/tex]

I need help with physics question.

Answers

(D)

Explanation:

Assuming that the charge q is moving perpendicular to the magnetic field B, the magnitude of the force experienced by the charge is

F = qvB = (2.9×10^-17 C)(4.0×10^5 m/s)(1.7T)

= 2.0×10^-11 N

Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?

Answers

Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.

Explanation:

It is known that formula for path difference is as follows.

[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex]    ... (1)

where, n = 0, 1, 2, and so on

As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.

[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]

Hence, path difference is as follows.

[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]

For lowest frequency, the value of n = 0.

[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]

[tex]\lambda = 4 \Delta L[/tex]

where,

[tex]\lambda[/tex] = wavelength

The relation between wavelength, speed and frequency is as follows.

[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]

where,

[tex]\nu[/tex] = speed

f = frequency

Substitute the values into above formula as follows.

[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]

Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.

A locomotive pulls 11 identical freight cars. The force between the locomotive and the first car is 150.0 kN, and the acceleration of the train is 2 m/s2. There is no friction to consider. 1) Find the force between the tenth and eleventh cars. (Express your answer to two significant figures.)

Answers

Answer:

The force between the 10 th car and the 11 th car is 13636.4 N.

Explanation:

Force, F = 150 kN

acceleration, a = 2 m/s^2

Let the mass of each car is m. \Total numbers of cars = 11

F = n m a

150000 = 11 x m x 2

m = 6818.18 kg

The force between the 10 th and 11 th car is

T = ma = 6818.18 x 2 = 13636.4 N

Please help I need this done within 30 mins

Answers

It may be thinner and more dense? I’m not too experienced in the study of Earth’s crust. However, I know enough to remember that the earths crust is thin.

Given that o.2i+bj+o.4k is a unit vector,what is the value of b?

Answers

Answer:

b = 0.89

Explanation:

The given vector is, [tex]A=0.2i+bj+0.4k[/tex]

A is a unit vector

We need to find the value of b.

For a unit vector, |A| = 1

So,

[tex]0.2^2+b^2+0.4^2=1\\\\0.04+b^2+0.16=1\\\\0.2+b^2=1\\\\b^2=1-0.2\\\\b=0.89[/tex]

So, th value of b is 0.89.

what units of measurement measures both velocity and speed

Answers

Answer:

[tex]metre \: per \: second[/tex]

Explanation:

Velocity is a derived quantity and the S.I unit is metre per second.Speed is also a derived quantity which is has the S.I unit to be metre per second.

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time

Answers

Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

initial velocity of the car, u = 0

final velocity of the car, v = 9.41 m/s

time of motion of the car, t = 4.24 s

mass of the passenger in the car, m = 67 kg

The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

           = 67(9.41 - 0)

           = 67 x 9.41

           = 630.47 kg

Therefore, the impulse experienced by the passenger is 630.47 kg

assuming a filament in a 120W light bulb acts like a prefect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.4×10^_5m^2. The stefan- boltzmann constant is 5.67×10^-8W/(m2.k2) A. 12OOk B. 2400K C. 2100K​

Answers

Answer:

T = 2398 K

Explanation:

To calculate the emission of the light bulb we use the law is Stefan

           P = σ A e T⁴

as they indicate that the filament is a black body, the emissivity is equal to 1 (e = 1)

           T = [tex]\sqrt[4]{\frac{P}{ \sigma A} }[/tex]

    let's calculate

           T =[tex]\sqrt[4]{\frac{120}{5.67 \ 10^{-8} \ 6.4 \ 10^{-5}} }[/tex]

           T = [tex]\sqrt[4]{33.06878 \ 10^{12} }[/tex]

           T = 2,398 10³ K

           T = 2398 K

what is the frequency of a wave related to​

Answers

Answer:

Frequency is the number of complete oscillations or cycles or revolutions made in one second.

A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ​

Answers

Answer:

144 meters

Explanation:

it takes 6 seconds to hit the ground right and the ball lays off 24 m per second .

so by the time the ball hits the ground 6 seconds passed. so that means the cliff is 6.0×24=144

How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *

Answers

Answer:

you drove 50km

Explanation:

10×5 hope this helps

Answer:

50 Km

Explanation:

This is how far you have got on your journey if traveling like this.

Please Mark as Brainliest

Hope this Helps

Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Force and power. Part A How many crates would you have to load onto the truck in one minute for the average power output you use to lift the crates to equal 0.50 hphp

Answers

Answer:

The number of crates is 84580.

Explanation:

mass, m = 30 kg

height, h = 0.9 mm  

Power, P = 0.5 hp = 0.5 x 746 W = 373 W

time, t = 1 minute = 60 s

Let the number of crates is n.

Power is given by the rate of doing work.

[tex]P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580[/tex]

Define Potential Energy
Begin by defining potential energy in your own words within one concise eight word sentence

Answers

Answer:

potential energy is a type of energy an object has because of it's position

16. The sum of kinetic energies in an object.
17. The essential device in power plants that convert mechanical
energy to electricity.
18. The device that converts electricity back to mechanical energy
19. The only EM wave that is seen by naked eye.
20. A device that converts light to electricity.​

Answers

Yes I also need help on this

Help!!
A table is pushed across the floor for a distance of 32 m with a force of 320 N in 150 seconds. How much power was used?
A.70.2W
B.68.3W
C.56.7W
D.49.8W

Answers

Compute the work done on the table:

W = Fd = (320 N) (32 m) = 10,240 J

Divide this by the given time duration to get the power output:

P = W/∆t = (10,240 J) / (150 s) ≈ 63.3 W

Which conclusion can be made based on the information in the table?
Wave speed and wavelengths can vary inversely to produce the same frequency.
O Frequency and wave speed can vary directly to produce the same wavelength.
O Wavelengths and frequency can vary inversely to produce the same wave speed.
O Frequency and wavelengths can vary directly to produce the same wave speed.
Mark this and return
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Previous Activity

Answers

Answer:

The correct option is (b).

Explanation:

The relation between the wavelength and frequency is given by :

[tex]\lambda=\dfrac{v}{f}[/tex]

Where

v is the wave speed

f is the frequency of a wave

It is clear from the above equation that the wavelengths and frequency can vary inversely to produce the same wave speed.

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