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ExtremeBDS [4]
3 years ago
9

How long is the racetrack if it takes a racecar 3.4 a traveling at 75 m/s to finish? ​

Physics
1 answer:
gregori [183]3 years ago
4 0

Answer:

255 metres

Explanation:

The answer to the question is actually quite simple. Since the car goes for 3.4 seconds, and it goes 75 metres every second, the answer is just 3.4 multiplied by 75 Metres.

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If railroad tracks were built over the boundary of two plates what would happen to the railroad tracks as the plates moved
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The railroad tracks will move with the plate boundaries
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A solar reflector is made using 31 identical triangular-shaped mirrors, each having sides . What is the total surface area of th
Stels [109]

Complete Question

Question 18 (3 points) Solve the problem. (3 points) A solar reflector is made using 31 identical triangular-shaped mirrors, each having sides 2.4m, 2. 3m, 1.5 m. What is the total surface area of the reflector?

A) 33 m2  

B) 86 m2

C) 52 m2

D)  34 m2

Answer:

The value is  Area  =2.78 \   m^2

Explanation:

From the question we are told that

   The  sides are  a =  2.4 m

                             b  =  2.3 m

                             c  =  1.5 m

Generally the semi perimeter is mathematically represented as

           s = \frac { a + b  + c  }{2 }

=>       s = \frac { 2.4  + 2.3  + 1.5   }{2 }

=>       s =3.1  \  m

Generally the using Heron's  formula we have that the  surface are a is  mathematically represented as

            Area  =  \sqrt{S (S -  a) (S - b )(S - c ) }

=>         Area  =  \sqrt{3.1  (3.1  -   2.4) (3.1  -  2.3 )(3.1  -  1.5 ) }

=>        Area  =2.78 \   m^2

6 0
2 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

8 0
3 years ago
I need help on this question
raketka [301]

Answer:

there are 5 hydrogen atoms in that formula

3 0
2 years ago
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