When an object absorbs light, what usually happens to its temperature?

Stan is driving north on his scooter at 8m/s, accelerates 11m/s (North) in 4s, drives a constant velocity for the next 15s, and

kow [346]

A) Acceleration: $a_1 = 0.75 m/s^2, a_2 = 0, a_3 = -1.57 m/s^2$

B) The total displacement is 209.5 m north

C) The average velocity is 8.06 m/s north

Explanation:

A)

Acceleration is defined as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

Here we have:

- In the first segment,

u = 8 m/s north

v = 11 m/s north

t = 4 s

So the acceleration is

$a_1 = \frac{11-8}{4}=0.75 m/s^2$ (north)

- In the second segment, Stan drives at a constant velocity: so the final velocity is equal to the initial velocity,

u = v

Therefore, the acceleration is zero: $a_2 = 0$

- In the third segment,

u = 11 m/s (north)

v = 0 (he comes to a stop)

t = 7 s

So the acceleration is

$a=\frac{0-11}{7}=-1.57 m/s^2$

And the negative sign means the acceleration is south, opposite to the direction of motion.

B)

In a uniformly accelerated motion, the displacement can be calculated as:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

t is the time

- For the first segment, we have

$u = 0\\a = 0.75 m/s^2\\t=4 s$

So the displacement is

$s_1 = 0+\frac{1}{2}(0.75)(4)^2=6 m$

- For the second segment, we have

$u = 11 m/s\\a = 0\\t=15 s$

So the displacement is

$s_2 = (11)(15)+0=165 m$

- For the third segment, we have

$u = 11\\a = -1.57 m/s^2\\t=7 s$

So the displacement is

$s_3 = (11)(7)+\frac{1}{2}(-1.57)(7)^2=38.5 m$

So the total displacement is:

s = 6 m + 165 m + 38.5 m = 209.5 m

In the north direction (positive direction)

C)

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the total displacement

t is the total time

Here we have:

d = 209.5 m

t = 26 s

Therefore, the average velocity is

$v=\frac{209.5}{26}=8.06 m/s$ (north)

Learn more about accelerated motion:

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#LearnwithBrainly

7 0

3 years ago

A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

marin [14]

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

4 0

3 years ago

HEEEEELLLLLLLLLLLLLLLLPPPPPPPPPPPPPPP i need to get unstuck

pochemuha

Geyshbdgsggefsgahevayagvdvdgavd

8 0

3 years ago

Read 2 more answers

why tidal waves are not produced by gravitational pull of Sun? Although the gravitational pull of Sun is far more Stronger than

Elina [12.6K]

First of all, don't forget that the sun is 400 times farther from us than the moon is. That fact alone tells us that anything on the earth is attracted to each kilogram of the moon with a force that's 160,000 times stronger than the force that attracts it to each kilogram of the Sun.
But more to your point ... The tides ARE greatly influenced by the sun. That's why tides are considerably higher at New Moon, when the sun and moon are both pulling in the same direction.

5 0

3 years ago

After combining components of two vectors to be added, it was found that the resultant vector has an x-component of -177 cm and

3241004551 [841]

The angle measured counterclockwise from the positive x-axis is θ = 50.4°

<h3>How to get the angle correspondent to a vector?</h3>

Here we know that the vector is:

V = < -177 cm, -214 cm>

To get the correspondent angle for this vector, we can think that this is the hypotenuse of a right triangle, such that the y-component and x-component are the cathetus.

Then, to get the angle (measured counterclockwise from the positive x-axis) is given by:

Tan(θ) = (opposite cathetus)/(adjacent cathetus)

Tan(θ) = (-214cm)/(-177 cm)

Using the inverse tangent function we get:

Atan(Tan(θ)) = Atan((-214cm)/(-177 cm))

θ = 50.4°

So the angle is 50.4°

If you want to learn more about vectors, you can read:

brainly.com/question/3184914

8 0

2 years ago