PHYSICS HIGH SCHOOL

How far does a car travel as it accelerates from rest to 24 m/s in 6 seconds?

Answers

Answer 1

Answer:

Answer:

$\boxed{\sf Distance \ travelled = 72 \ m}$

Given:

Initial velocity (u) = 0 m/s (Starts from rest)

Final velocity (v) = 24 m/s

Time taken (t) = 6 seconds

ToFind:

Distance travelled by car (s)

Explanation:

From equation of motion of object moving with uniform acceleration in straight line we have:

$\boxed{ \bold{s = ((v + u)/(2) )t}}$

By substituting value of v, u & t in the equation we get:

$\sf \implies s = ( (24 + 0)/(2) ) * 6 \n \n \sf \implies s = (24)/(2) * 6 \n \n \sf \implies s = 12 * 6 \n \n \sf \implies s = 72 \: m$

$\therefore$

Distance travelled by car (s) = 72 m

Related Questions

How do streams flow?
Imagine that a car is traveling from your left to your right at a constant velocity. Which two actions could the driver take that may be represented as (a) a velocity vector and an acceleration vector both pointing to the right and then (b) changing so the velocity vector points to the right and the acceleration vector points to the left?
From which of the following do spiral galaxies form?A. High-density proto-galactic clouds B. Rapidly spinning proto-galactic clouds C. Colliding/merging proto-galactic clouds D. Rapidly cooling proto-galactic clouds
How do you calculate the initial speed given time and displacement?The question is: The longest kick in CFL history was 83.2m. If the ball remained in the air for 4.12s, determine its initial speed. I calculated it several times and got the same answer, 40.38m/s, but the textbook says it's 28.6m/s.
Which of these causes summer in the northern hemisphere?The Sun is closer to Earth during the summer.The northern hemisphere receives more direct sunlight during the summer.Earth's northern axis is tilted away from the Sun during the summer.The North Pole is tilted away from the Sun during the summer.

An object is dropped from the edge of a cliff and is moving at 26.5 m/s just before it hits the ground. How high is the cliff?11.0 m

260 m

35.8 m

848 m

Answers

Answer:

35.8 m

Explanation:

An appropriate formula is ...

$V_f^2-V_i^2=2ad$

The initial velocity is 0, and the acceleration due to gravity is 9.8 m/s², so the distance, d, is ...

$d=(26.5^2)/(2\cdot 9.8)\approx 35.8 \quad\text{m/s}$

Why are the vector quantities magnitude and direction

Answers

Because color, cost, temperature, weight, and age
don't tell you enough about the motion.

Because, by definition, a vector has to have magnitude and direction. Otherwise, it's not a vector. If it only has magnitude, then it's a scalar, not a vector. Does that help at all?

A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at rest, how fast is it moving after it travels 3 meters?

Answers

The box has 3 forces acting on it:

• its own weight (magnitude w, pointing downward)

• the normal force of the incline on the box (mag. n, pointing upward perpendicular to the incline)

• friction (mag. f, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ F = -f + w sin(35°) = m a

• net perpendicular force:

∑ F = n - w cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

n = w cos(35°)

n = (15 kg) (9.8 m/s²) cos(35°)

n ≈ 120 N

Solve for the mag. of friction:

f = µn

f = 0.25 (120 N)

f ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) a

a ≈ (54.3157 N) / (15 kg)

a ≈ 3.6 m/s²

Now solve for the block's speed v given that it starts at rest, with v₀ = 0, and slides down the incline a distance of ∆x = 3 m:

v² - v₀² = 2 a ∆x

v² = 2 (3.6 m/s²) (3 m)

v = √(21.7263 m²/s²)

v ≈ 4.7 m/s

You need to raise a heavy block by pulling it with a massless rope. You can either (a) pull the block straight up height h, or (b) pull it up a long, frictionless plane inclined at a 15∘ angle until its height has increased by h. Assume you will move the block at constant speed either way.

Answers

Final answer:

In both scenarios, the work done on the heavy block is the same, as it is determined by the change in the vertical height. However, pulling the block up the inclined plane may require less force because the work is distributed over a larger distance.

Explanation:

The subject of this question is based on the concept of work and energy in physics. When you pull the heavy block straight upwards (scenario a), the work done is equal to the force times the distance, or Work = mg*h, where m is the mass of the block, g is the acceleration due to gravity, and h is the height it needs to rise. For pulling the block up the inclined plane (scenario b), the work done still equals mg*h as the vertical distance it rises is the same.

This is because, according to the principle of work and energy, the work done on an object is equal to the change in its kinetic energy. Since the speed of the block remains constant in both scenarios, the kinetic energy does not change, meaning the work done on the block is the same in both scenarios.

However, pulling the block up the inclined plane may require less force because of the larger distance over which the work is done. But the overall work is the same in both cases.

Learn more about Work and Energy here:

brainly.com/question/16987255

#SPJ12

1. Coherent beams with wavelengths of 400 nm intersect at points on the screen. What will be observed at these points interference maximum or minimum, if the optical path difference of the beam is in the 1st case A1=20 um and in the 2nd - A2=15 um? Find the order of minimum and maximum. 2. An installation for observing Newton's rings in reflected light is illuminated by monochromatic light incident normally to the plate surface. The distance between the second and fourth dark rings is 5 mm. Find the distance between the eighth and sixteenth light rings. 3. Find the angle between the main planes of the polarizer and the analyzer if the intensity of the natural light passing through the polarizer and the analyzer decreased by 5 times.

Answers

1. The observed interference at points will be maximum for $A_1 = 20 \, \mu m$ and minimum for $A_2 = 15 \, \mu m$ , with orders 12th maximum and 12th minimum.
2. The distance between the eighth and sixteenth light rings is approximately 2.583 mm.
3. The angle between the main planes of the polarizer and the analyzer is approximately $63.43^\circ$ when the intensity decreases by 5 times.

1. For coherentbeams with wavelengths of 400 nm, the interference observed at points on the screen will be maximum when the optical path difference (OPD) is a whole number of wavelengths ( $m \lambda$ ), and minimum when the OPD is a half-integer number of wavelengths ( $(m + 0.5) \lambda$ ).

Given

$A_1 = 20 \, \mu m$

$A_2 = 15 \, \mu m$ , the difference in optical path between the two cases is $$\Delta A = A_1 - A_2 = 5 \, \mu m$.$

The number of wavelengths difference can be calculated as $(\Delta A)/(\lambda)$ , which is approximately 12.5 wavelengths.

2. In Newton's ringspattern, the radius of the m dark ring is given by $r_m = √(m \lambda R)$ , where R is the radius of curvature of the lens.

Given the distance between the second and fourth darkrings ( $m = 2\) to \(m = 4$ ) as 5 mm, we can set up an equation:

$$√(4 \lambda R) - √(2 \lambda R) = 5 \, \text{mm}$.$

Solving for R gives $$R \approx 1.904 * 10^(-2) \, \text{m}$.$

Now, the radius of the 16th light ring can be calculated as $r_(16) = √(16 \lambda R)$ , and the radius of the 8th light ring is $r_(8) = √(8 \lambda R)$ . The distance between these two light rings is then given by:

$$d = r_(16) - r_(8) \n\n= √(16 \lambda R) - √(8 \lambda R)$.$

Plugging in the values, we get $d \approx 2.583 \, \text{mm}$ .

3. When the intensity decreases by a factor of 5, the transmission of the polarizer and analyzer combination becomes $(1)/(√(5))$ times the original transmission, as intensity is proportional to the square of the transmission.

The relationship between the intensity and the angle $\theta$ between the main planes of the polarizer and the analyzer is given by Malus's law: $I = I_0 \cos^2 \theta$ , where $I_0$ is the initial intensity.