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3 years ago

9

A sidewalk has a length of 75.00m. How many inches is this? (Hint: you need to use two unit conversion fraction. 1 cm equals abo

ut 0.3937 inches)

1 answer:

maria [59]3 years ago

5 0

Length = (75.00 m)

Length = (75 meter) x (3.28084 foot/meter) x (12 inch/foot)

Length = (75 x 3.28084 x 12) (meter-foot-inch / meter-foot)

<em>Length = 2,952.76 inches</em>

You might be interested in

Which quantity can be calculated using Newton’s second law of motion?

DiKsa [7]

Newton's 2nd law of motion:

Net Force = (mass) x (acceleration) .

The law shows the relationship among an object's mass
and acceleration, and the net force acting on it.

If you know any two of the quantities in the formula,
the law can be used to calculate the third one.

3 0

3 years ago

Need help asap pls

ozzi

Detailed Explanation:

1) Rusting of Iron

4Fe + 3O2 + 2H2O -> 2Fe2O32H2O

Reactants :-
Fe = 4
O = 3 * 2 + 2 = 8
H = 2 * 2 = 4

Products :-
Fe = 2 * 2 = 4
O = 2 * 3 + 2 = 8
H = 2 * 2 = 4

2) Fermentation of sucrose…

C12H22O11 + H2O -> 4C2H5OH + 4CO2

Reactants :-
C = 12
H = 22 + 2 = 24
O = 11 + 1 = 12

Products :-
C = 4 * 2 + 4 = 12
H = 4 * 5 + 4 = 24
O = 4 * 2 + 4 = 12

Looking closely at the way I have taken the total number of elements on the reactants and products side, you can solve the rest.

All the Best!

8 0

2 years ago

Read 2 more answers

Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul

Yuki888 [10]

Answer:

$63.4^{\circ}$

Explanation:

When unpolarized light passes through the first polarizer, the intensity of the light is reduced by a factor 1/2, so

$I_1 = \frac{1}{2}I_0$ (1)

where I_0 is the intensity of the initial unpolarized light, while I_1 is the intensity of the polarized light coming out from the first filter. Light that comes out from the first polarizer is also polarized, in the same direction as the axis of the first polarizer.

When the (now polarized) light hits the second polarizer, whose axis of polarization is rotated by an angle $\theta$ with respect to the first one, the intensity of the light coming out is

$I_2 = I_1 cos^2 \theta$ (2)

If we combine (1) and (2) together,

$I_2 = \frac{1}{2}I_0 cos^2 \theta$ (3)

We want the final intensity to be 1/10 the initial intensity, so

$I_2 = \frac{1}{10}I_0$

So we can rewrite (3) as

$\frac{1}{10}I_0 = \frac{1}{2}I_0 cos^2 \theta$

From which we find

$cos^2 \theta = \frac{1}{5}$

$cos \theta = \frac{1}{\sqrt{5}}$

$\theta=cos^{-1}(\frac{1}{\sqrt{5}})=63.4^{\circ}$

6 0

3 years ago

An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec

aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

$x = V*t = -8\frac{m}{s} *3s = -24m$

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

$v^2 = v_0^2 + 2a*d$

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

$d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m$

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

$t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s$

Then, the time of the second phase will be:

$t_2 = 9s - t_1 = 9s - 1.143s = 7.857s$

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

$x = \frac{1}{2}a*t^2 + v_0*t + x_0$

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

$x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m$

So, the total distance covered by this object in meters will be the sum of all the distances we found:

$x_total = 24m + 4.57m + 216.07m = 244.64m$

8 0

3 years ago

The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the

ratelena [41]

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

4 0

3 years ago