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ahrayia [7]
4 years ago
13

When two sticks are laid end-to-end they cover a length of 8.32 feet. One stick is 2.93 ft longer than the other. What is the le

ngth of the shorter stick in feet?
Physics
1 answer:
77julia77 [94]4 years ago
7 0

To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as

x+2.93

Both cover a magnitude of 8.32 ft, therefore

x +(x+2.97) = 8.32

Now solving for x we have,

x + (x + 2.93) = 8.32

2x + 2.93 = 8.32

2x = 8.32 - 2.93

x = \frac{ 8.32 - 2.93}{2}

x = 2.695 ft

Therefore the shorter stick is 2.695ft long.

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A car moves with a speed 72km/h, the driver uses the brakes, the car stops after 8 seconds, calculate its speed after 10 seconds
vampirchik [111]
The acceleration of the car is solved by subtracting the initial speed from the final speed then dividing the result by the elapsed time.

initial speed = 72 km/hr = 20 m/s

final speed = 0 m/s

elapsed time = 5 seconds

acceleration = (0 m/s – 20 m/s) / 5 s

acceleration = - 20m/s / 5 s

acceleration = -4 m/s^2
8 0
3 years ago
one thing that stringed, wind, and percussion instruments have in common in regard to the sounds they produce is that a. the ins
mario62 [17]

The answer is:

d) the sound originates from a vibration.

The explanation:

The sound waves are generated by a sound source, such as the vibrating diaphragm of a stereo speaker. The sound source creates vibrations in the surrounding medium. As the source continues to vibrate the medium, the vibrations propagate away from the source at the speed of sound, thus forming the sound wave.

3 0
3 years ago
Read 2 more answers
1. The planet Jupiter completes a revolution of the sun in 11.5 years. Express it in seconds. Given that one year= 3.154 × 10^7
xenn [34]

Answer:

The planet Jupiter completes one revolution of the sun in 362710000 seconds. Long time, right?

Explanation:

3.154x10^7=3.154x10000000=31540000

11.5x31540000=362710000

7 0
2 years ago
Coders play an important role in
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3 0
3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
3 years ago
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