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3 years ago

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A yellow and green car traveled 400 miles to Dayton, OH. The green car made the trip in 10 hours. The yellow arrived in 8 hours.

Calculate the speed of each car and tell which car was faster. (s=d/t)
A)The green car was faster. Green traveled at a speed of 50 mph while yellow was traveling at an average of 40 mph.

B)The green car was faster since it arrived in 10 hours.

C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

D)The yellow car was faster since it was traveling at 40 mph

2 answers:

elixir [45]3 years ago

4 0

Answer: C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

Explanation:

The speed of each car is defined as:

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

where d is the distance traveled by the car and t is the time taken.

For the yellow car, d=400 mi and t=8 h, so its speed is

$v=\frac{400 mi}{8 h}=50 mph$

For the green car, d=400 mi and t=10 h, so its speed is

$v=\frac{400 mi}{10 h}=40 mph$

So, the correct choice is

C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

forsale [732]3 years ago

4 0

Answer:

C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

Explanation:

The speed is defined as ratio of distance covered with time:

$v = \frac{distance}{time}$

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

here

d = the distance traveled by the car and

t = the time taken.

For the yellow car,

d=400 mi

t=8 hours

so speed of yellow car is given as

$v_{yellow} = \frac{400}{8} = 50 mph$

For the green car,

d=400 mi

t=10 hours,

so speed of green car is

$v_{green} = \frac{400}{10} = 40 mph$

So, the correct choice is

C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

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konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

$K_f = \frac{1}{2}mv^2$ is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

$K_i = \frac{1}{2}mu^2$ is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

$W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J$

Learn more about work and kinetic energy:

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A ball is dropped from a window and takes two seconds to reach the ground .it starts from rest and reaches a final speed of 20m/

Akimi4 [234]

The ball can't reach the speed of 20 m/s in two seconds, unless you THROW it down from the window with a little bit of initial speed. If you just drop it, then the highest speed it can have after two seconds is 19.6 m/s .

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(Gravity on Earth is only 9.8 m/s^2.)

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Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

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liubo4ka [24]

Answer:

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A 3.3 kg ball sits on the ground and is kicked with a FAPP of 36N

jeyben [28]

a) 32.3 N

The force of gravity (also called weight) on an object is given by

W = mg

where

m is the mass of the object

g is the acceleration of gravity

For the ball in the problem,

m = 3.3 kg

g = 9.8 m/s^2

Substituting, we find the force of gravity on the ball:

$W=(3.3)(9.8)=32.3 N$

b) 48.3 N

The force applied

$F_{app} = 36 N$

The ball is kicked with this force, so we can assume that the kick is horizontal.

This means that the applied force and the weight are perpendicular to each other. Therefore, we can find the net force by using Pythagorean's theorem:

$F=\sqrt{W^2+F_{app}^2}$

And substituting

W = 32.3 N

Fapp = 36 N

We find

$F=\sqrt{32.3^2+36^2}=48.3 N$

c) $14.6 m/s^2$

The ball's acceleration can be found by using Newton's second law, which states that

F = ma

where

F is the net force on an object

m is its mass

a is its acceleration

For the ball in this problem,

m = 3.3 kg

F = 48.3 N

Solving the equation for a, we find

$a=\frac{F}{m}=\frac{48.3}{3.3}=14.6 m/s^2$

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