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aleksandr82 [10.1K]
3 years ago
7

Here is a graph of the movement of a car. What was the acceleration of this vehicle between 15 and 20 seconds? Show your work an

d your answers.

Physics
1 answer:
tekilochka [14]3 years ago
6 0
<span>-2.4 m/s^2 At 15 seconds, the velocity was 12 m/s, at 20 seconds, the velocity was 0 m/s. The graph shows a straight line between those 2 points, so the acceleration was uniform. So we have a velocity change of 0 m/s - 12 m/s = -12 m/s And a time i</span>
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A laser produces light at 5.32x10-7 m.
Blababa [14]

Answer:

D

Explanation:

speed = frequency x wavelength

speed of light in vacuum is 3.0 x 10^8

wavelength = 5.32 x10 ^-7

3.0 x 10 ^ 8 = 5.32 x 10^-7 x frequency

frequency = 5.63909 x 10^14

round it off = 5.64 x 10^14 Hz

thus the answer is D

hope this helps please mark it

6 0
3 years ago
PLSLPSLLPSLPSLPS HELP I ONLY HAVE 5 MINS PLEASE
UkoKoshka [18]

Answer:

0.19m/s²

Explanation:

Initial velocity(u) = 50×1000/60×60

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a= -0.19m/s²

-a = 0.19m/s²

The magnitude of retar dation is 0.19m/s²

5 0
2 years ago
Read 2 more answers
he calculation of kinetic energy uses the formula, KE = 1 2 mv2. Identify the unit that is required for mass and velocity, respe
Umnica [9.8K]
Mass is measured in kg
Velocity is measured in ms^-1
Hope this is what you were looking for
5 0
3 years ago
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240
Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

3 0
3 years ago
What occurs when a swimmer pushes through water to swim
Triss [41]
When we swim we apply force and push the water backward with the help of our hands. In response, The water pushes us forward with an equal force. Thus, in order to move forward and swim, the swimmer lushes the water backward. Newton's 3rd law of motion
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