All categories

3 years ago

A runner achieves a velocity of 12.1 m/s 8.2 sec after he begins. What is

Physics

1 answer:

Karo-lina-s [1.5K]3 years ago

7 0

Given - 

Velocity (v) = 12.1 m/s
Time (t) = 8.2 sec

To find -

Acceleration

Solution - 

★ Acceleration = velocity/ time

→ A = 12.1/8.2

→ 1.47.... m/s²

Therefore, the acceleration of the runner will be 1.47 m/s².

You might be interested in

What distance will a car cover while uniformly accelerating from 12m/s to 26m/in 14 seconds

VikaD [51]

Answer:

Distance = 266m

Explanation:

6 0

3 years ago

Answer ?

morpeh [17]

Answer:

2.75 m/s^2

Explanation:

The airplane's acceleration on the runway was 2.75 m/s^2

We can find the acceleration by using the equation: a = (v-u)/t

where a is acceleration, v is final velocity, u is initial velocity, and t is time.

In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1

a = 2.75 m/s^2

5 0

2 years ago

100 point good luck answering this 2+2-4+8+-5+???

Darya [45]

The answer to that is 3

7 0

3 years ago

Velocity is a vector quantity which has both magnitude and direction. ... Net force is also a vector quantity which has both mag

GuDViN [60]

When the object is moving in the elliptical orbit, it means that the direction of its acceleration should be towards the two foci (plural of focus) of the ellipse to keep the elliptical motion. As force according to the Newton's second law: F = ma, the net force must be in the direction of the acceleration. As far as the magnitude of net force is concerned, you can use Newton's gravitational law to find its magnitude.

7 0

3 years ago

Read 2 more answers

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 re

valina [46]

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 = µ n

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

4 0

2 years ago