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tia_tia [17]
3 years ago
12

A certain car traveling in a straight line path has a velocity of +10.0 m/s at some instant. after 3.00 s, its velocity is +6.00

m/s. what is the average acceleration of the car during this time interval?
Physics
1 answer:
makvit [3.9K]3 years ago
3 0
The formula is:
v = v o + a t
6 = 10 + 3 * a
3 a = 10 - 6
a = 4 : 3
a =  - 1.33 m/s² ( because the car slows down )
Answer: The average acceleration of the car is - 1.33 m/s²
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shtirl [24]

Answer:

its B

Explanation: i searched it up...

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4 years ago
Please Help!!! It's for a quiz!!
Sphinxa [80]

Answer:

38.8 m/s

Explanation:

Force F(x) = 6 - 2x + 6x²

work

W=\displaystyle\int_{0}^{13.9}F(x)dx=\displaystyle\int_{0}^{13.9}(6-2x+6x^2)dx

=6x-x^2+2x^3|_{0}^{13.9}\\=5261 J

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3 0
2 years ago
Can you access an instance variable from a static method? explain why or why not.
marusya05 [52]
<span>The reason a static method can't access instance variable is because static references the class not a specific instance of the class so there is no instance variable to access.</span>
5 0
4 years ago
What are the basic si units for the wavelength of light?
noname [10]

Answer:

Meter (m)

Explanation:

The wavelenght of a light wave is a measure of the distance between two successive crests (or two successive troughs) of a light wave.

Since the SI units for the distance is the meter (m), then the SI unit for the wavelength is also the meter (m).

Wavelength is related to the frequency of the light wave by:

\lambda=\frac{c}{f}

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7 0
3 years ago
?a wire is stretched 30% what is the percentage change in resistance ​
Marat540 [252]

Answer:

The percentage change in resistance of the wire is 69%.

Explanation:

Resistance of a wire can be determined by,

R = (ρl) ÷ A

Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.

When the wire is stretched, its length and area changes but its volume and resistivity remains constant.

l_{o} = 1.3l, and A_{o} = \frac{A}{1.3}

So that;

R_{o} = (ρl_{o}) ÷ A_{o} = (ρ × 1.3l) ÷ (\frac{A}{1.3})

    = (1.3lρ) ÷ (\frac{A}{1.3})

    = (1.3)^{2} × [(ρl) ÷ A]

   = 1.69R               (∵ R = (ρl) ÷ A)

R_{o} = 1.69R

Where R_{o} is the new resistance, l_{o} is the new length, and A_{o} is the new area after stretching the wire.

The change in resistance of the wire = R_{o} - R

                                      = 1.69R  - 1R

                                      = 0.69R

The percentage change in resistance = \frac{0.69R}{R} × 100

                                                               = 0.69 × 100

                                                              = 69%

The percentage change in resistance of the wire is 69%.

3 0
3 years ago
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