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Stolb23 [73]
2 years ago
9

Kinetic energy can be calculated using the formula KE = 1/2mv2. What is the kinetic energy of a bicycle that has a mass of 20 kg

and is traveling at a velocity of 10 m/sec?​
Physics
1 answer:
Alenkasestr [34]2 years ago
8 0
<h3>Answer:</h3>

KE = 1000 J

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Physics</u>

<u>Energy</u>

Kinetic Energy Formula: \displaystyle KE = \frac{1}{2}mv^2

  • Energy is in Joules
  • m is mass (in kg)
  • v is velocity (in m/s)
<h3>Explanation:</h3>

<u>Step 1: Define</u>

m = 20 kg

v = 10 m/s

<u>Step 2: Find KE</u>

  1. Substitute [KE]:                    \displaystyle KE = \frac{1}{2}(20)(10)^2
  2. Exponents:                          \displaystyle KE = \frac{1}{2}(20)(100)
  3. Multiply:                               \displaystyle KE = (10)(100)
  4. Multiply:                               \displaystyle KE = 1000
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A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a
Svet_ta [14]

A) The car overtakes the truck after 7.56 s

B) Initial distance between car and truck: 37.1 m

C) Speed of the truck: 15.9 m/s, speed of the car: 25.7 m/s

D) See graph in attachment

Explanation:

A)

The truck starts from rest and has a constant acceleration, so its position at time t can be written as

x_t(t)=d+\frac{1}{2}a_tt^2

where

d is the initial distance between the truck and the car (the truck starts some distance ahead of the car)

a_t=2.10 m/s^2 is the acceleration of the truck

The car position instead it is given by the equation

x_c(t)=\frac{1}{2}a_ct^2

where

a_c=3.40 m/s^2 is the acceleration of the car

The car overtakes the truck when the truck has moved 60.0 m, so when

x_t(t') = d + 60

Therefore, solving the equation, we find the time t when  this occurs:

d+\frac{1}{2}a_t t'^2 = d+60\\\frac{1}{2}a_tt'^2=60\\t'=\sqrt{\frac{2\cdot 60}{a_t}}=\sqrt{\frac{120}{2.1}}=7.56 s

B)

In order to find the initial distance between the car and the truck (d), we have to calculate first the distance covered by the car during these 7.56 s. It is given by:

x_c(t')=\frac{1}{2}a_c t'^2=\frac{1}{2}(3.40)(7.56)^2=97.2 m

This means that after 7.56 s, when the car reaches the truck, the car has covered 97.2 m while the truck has covered 60 m. However, their positions are now equal, so we can write:

x_c(t')=x_t(t')

And by solving the equation, we find the value of d, the initial distance between car and truck:

\frac{1}{2}a_c t'^2 = d + \frac{1}{2}a_t t'^2\\d=\frac{1}{2}(a_c-a_t)t'^2 = \frac{1}{2}(3.40-2.10)(7.56)^2=37.1 m

C)

In order to find the speed of each vehicle, we use the following suvat equation:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time

For the truck, we have:

u = 0

a_t = 2.10 m/s^2

So its speed after t = 7.56 s is

v_t = 0+(2.10)(7.56)=15.9 m/s

For the car, we have

u = 0

a_c=3.40 m/s^2

So its speed after t = 7.56 s is

v_c=0+(3.40)(7.56)=25.7 m/s

D)

Find the graph required in attachment.

On the x-axis, it is represented the time in seconds. On the y-axis, it is represented the position in meters.

Both curves are in the shape of a parabola since the motion of both vehicles is an accelerated motion.

The curve that starts at -37.1 m is the curve representing the car: in fact, the car starts behind the truck by 37.1 m. The curve that starts from x = 0, t= 0 is that of the truck.

The two curves meets when t = 7.56 s: at that time, the two vehicles have reached the same position, and we see that occurs when x = 60 m, which means that this happens when the truck has covered 60 meters.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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Maksim231197 [3]

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

\displaystyle{t=D/v}

and from Hubble's Law:

v =H_0D

Therefore:

\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}

With the given value for the Hubble's constant we have:

H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s

and thus,

t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years

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Which equals 23 kilograms?
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How does the table show that the balloon went downwards ?
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