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2 years ago

Why does Usain bolt run?

Physics

1 answer:

matrenka [14]2 years ago

5 0

Answer:

because he's fast and speedy

Explanation:

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Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilo

777dan777 [17]

Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert 42 m/s ⇒ km/h

1 km =1000 m

1 h = 36000 sec

$42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\$

$0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}$

Step 2

find kilometers traveled after 4 hours

$V=\frac{s}{t}\\ \\$

V,velocity

s, distance traveled

t. time

now, isolating s

$V=\frac{s}{t} \\s=V * t\\$

and replacing

$s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\$

S=604.8 Km

Have a great day

4 0

3 years ago

Whats the Independent and Dependent Variables

astra-53 [7]

The independent variable is the type of fuel used and the dependent variable is the speed of the race car. The independent variable could be changed through the experimental process to see its relation with the dependent variable<span>. The dependent variable is the result of the independent variable changes.</span>

5 0

3 years ago

A woman does 236 J of work

GREYUIT [131]

Answer:

<h2>workdone = force × distance </h2><h2>236J = 18.9cos(o) × 24.4</h2><h2>236/24.4 = 18.9cos(o)</h2><h2>(0.5117)cos^-1 = (o)</h2><h2><u>59.21°</u></h2>

3 0

3 years ago

Read 2 more answers

The apple is on the tree and it is not moving. What are the two types of forces acting

Alexandra [31]

Answer:

An apple hanging on a tree is acted upon by two basic forces: 1) the downward force exerted by the gravity force and, 2) the upward force exerted by the tree limb. The downward gravity force is also know as "weight".

4 0

3 years ago

a sprinter running a 100m dash leaves the starting block and accelerates to a maximum velocity of 11m/s at 6s into the race. the

mihalych1998 [28]

Answer:

Part(a): The average acceleration is $1.83~m~s^{-2}$ during the first 6s.

Part(b): The average acceleration from 6s to 8s is zero.

Explanation:

Part(a):

The average acceleration is defined as the rate at which velocity is changing with respect to time. So if ' $v_{i}$ ', ' $v_{f}$ ' and ' $a_{av}$ ' represents the initial velocity, final velocity, and average acceleration of a particle, then mathematically

$a_{av} = \dfrac{v_{f} - v_{i}}{t}$

where ' $t$ ' is the time taken by the particle to achieve the velocity ' $v_{f}$ ' starting from initial velocity ' $v_{i}$ '

Given in the problem, $v_{i} = 0, v_{f} = 11~m~s^{-1}~and~t = 6~s$ .

So the average acceleration( $a_{av}$ ) during the first 6 s will be

$a_{av} = \dfrac{11 - 0}{6}~m~s^{-2} = 1.83~m~s^{2}$

Part(b):

During the time between 6 s to 8 s, as mentioned in the problem, the sprinter maintains the constant velocity. So the average acceleration during this time interval will be zero.

7 0

3 years ago