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3 years ago

A cyclist is traveling 10 m/s with an acceleration of 6 m/s2. How fast is the cyclist traveling at the end of 20 seconds

Physics

1 answer:

krok68 [10]3 years ago

7 0

Answer:

<em>The cyclist is traveling at 130 m/s</em>

Explanation:

<u>Constant Acceleration Motion </u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

$v_f=v_o+at$

The cyclist initially travels at 10 /s and it's accelerating at a=6m/s^2. We need to know the new speed when t= 20 seconds have passed.

Apply the above equation:

$v_f=10+6\cdot 20$

$v_f=10+120$

$v_f=130\ m/s$

The cyclist is traveling at 130 m/s

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3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between

Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

$v^2 -u^2 = 2ad$ (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

$F=ma=-\mu mg$

where

m = 1000 kg is the mass of the car

$\mu = 0.80$ is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

$a=-\mu g$

And we can substitute it into eq.(1) to find d:

$v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m$

7 0

4 years ago

What is the specific enthalpy of benzene vapor at 45 c and 0.7 atm absolute pressure, relative to a reference state of benzene v

vazorg [7]

The specific enthalpy of benzene vapor at 45 c and 0.7 atm absolute pressure, relative to a reference state of benzene vapor at 45 c and 1.27 atm absolute pressure will be 0 kJ/mol.

<h3>What is specific enthalpy and how was it calculated in the question?</h3>

A thermodynamic system has a property called enthalpy (H). It is calculated by the sum of the internal energy (U) of the thermodynamic system and the product of its volume (V) and pressure (p). The SI Unit is Joule (J).

Equation:

H = U+pV

The specific enthalpy of vapor can be defined as the amount of energy spent in order to transform a liquid substance into its vapor or gaseous form. The SI Unit is kJ/mol.

In the above question, the formula to be used is

P1/P2 = (Δ Hvap)/R)(1/T2-1/T1)

T1 & P1 --> the starting temperature & pressure respectively (= 1.27 atm and 45c),

T2 & P2 --> the final temperature & pressure respectively (= 0.7 atm and 45c),

R --> the real gas constant i.e. 8.314kJ/mol and

ΔHvap --> The specific enthalpy of vaporization.

Putting the values in the equation;

1.27/0.7=(ΔHvap/8.314)(1/45-1/45)

Hence as after subtracting the equation becomes 0, our final answer also comes out to be ΔHvap= 0 kJ/mol.

To know more about specific enthalpy, visit:

brainly.com/question/16244647

#SPJ4

6 0

2 years ago

Make a sketch of the radius of gyration of a polymer chain vs. the degree of polymerization N as it would appear on a log-log sc

zvonat [6]

Answer:

hhhhhhhhhhhhhhhhhhhhh

Explanation:

sjdnxjwodj1oeixjwkw9dijwqoisjd1

sjssusidej

4 0

3 years ago

Helppppp plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz!

Fynjy0 [20]

C real,inverted and smaller than the object

6 0

3 years ago

Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to

jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x $10^{-8}$ J.

(b) The wavelength of the photon is 1.2 x $10^{-17}$ m.

Explanation:

Let;

$E_{1}$ = -13.60 ev

$E_{2}$ = -3.40 ev

(a) Energy of the emitted photon can be determined as;

$E_{2}$ - $E_{1}$ = -3.40 - (-13.60)

= -3.40 + 13.60

= 10.20 eV

= 10.20(1.6 x $10^{-9}$ )

$E_{2}$ - $E_{1}$ = 1.632 x $10^{-8}$ Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x $10^{-8}$ Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x $10^{-34}$ Js), c is the speed of light (3 x $10^{8}$ m/s) and λ is the wavelength.