Answers please and thank you

Cheetahs can accelerate to a speed of 19.6 m/s in 2.45 s and can continue to accelerate to reach a top speed of 27.6 m/s . Assum

marta [7]

Answer:

$v_{top} = 61.96 \frac{mi}{h}$

Explanation:

To express the cheetah's top speed in miles per hour, we just need to find the conversion factor.

We know that the top speed is

$v_{top} = 27.7 \frac{m}{s}$

So, we want to obtain miles from meters and hours from seconds.

<h3>miles from meters</h3>

First we write the equivalence:

$1609.34 \ m = 1 \ mi$

Now, we can divide by 1609.34 meters on both sides:

$\frac{1609.34 \ m}{ 1609.34 \ m} = \frac{1 \ mi}{ 1609.34 \ m}$

The left sides equals 1, so

$1 = \frac{1 \ mi}{ 1609.34 \ m}$

And this is our conversion factor from meters to miles. Now, we can multiply our top speed by this conversion factor, as the conversion factor equals one, and is dimensionless, the physical meaning will be the same.

$v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}$

$v_{top} = 0.0172120 \frac{mi}{s}$

This is the top speed in miles per second, now, for obtaining miles per hour:

<h3>hours from seconds</h3>

We can do pretty much the same, first, the equivalence:

$1 \ h = 3600 \ s$

as the seconds are dividing in the velocity, we know divide by 1 hour.

$\frac{1 \ h}{ 1 \ h} = \frac{3600 \ s}{ 1 \ h}$

$1 = \frac{3600 \ s}{ 1 \ h}$

and know we just multiply our top speed by this conversion factor

$v_{top} = 0.0172120 \frac{mi}{s} \frac{3600 \ s}{ 1 \ h}$

$v_{top} = 61.96 \frac{mi}{h}$

8 0

3 years ago

If you cut a piece of wood in half, each half would have *

Anon25 [30]

Answer:

c

Explanation:

let's assume that you are splitting the piece of wood horizontally. This would not change the density of the wood at all, only the length. I rest my case. Ur awesome and have an amazing day.

7 0

3 years ago

I need some help with this!

Sonja [21]

Explanation:

ummm I believe it's frequency

5 0

3 years ago

How does the mass of an object affect its acceleration during free fall

ira [324]

The mass of an object has no effect whatsoever on the object's
acceleration during free-fall. If there is no air resistance to interfere
with the natural effects of gravity, then a feather and a battleship ...
dropped at the same time ... fall together, and hit the ground at the
same time.

8 0

3 years ago

Read 2 more answers

Help me please, need more assistance

Dmitrij [34]

Explanation:

12) q = mCΔT

125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)

T = 82.0°C

13) Solving for ΔT:

ΔT = q / (mC)

a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C

b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C

c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C

d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C

e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C

f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C

14) q = mCΔT

q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)

q = 502,000 J

20) q = mCΔT

q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)

q = 742,000 J

24) q = mCΔT

q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)

q = -0.091 J

6 0

3 years ago

Answers please and thank you​

Answers please and thank you