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raketka [301]
3 years ago
12

CAN SOMEONE HELP ME PLEASE!? After chasing its prey, a cougar leaves skid marks that are 236 m in length. Assuming the cougar sk

idded to a complete stop with a constant acceleration of -2.87 m/s^2, identify the speed of the cougar before it began to skid.​
Physics
1 answer:
malfutka [58]3 years ago
5 0

Answer:

u=36.8m/s

Explanation:

because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations

u^2=v^2-2ā*s. where:

u^2 stands for intial velocity

v^2 stands for final velocity

since the cougar skidded to a complete stop the final velocity is zero.

u^2=v^2-2ā*s

u^2=(0)^2 -2(-2.87 m/s^2)*236 m

u^2=0+5.74m/s^2* 236m

u^2=1354.64m^2/s^2

u=√1354.64m^2/s^2

u=36.8m/s (approximate value)

when ever the acceleration is constant you can use one of the following equation to find the required value.

1. v = u + at. (no s)

2. s= 1/2(u+v)t. (no ā)

3. s=ut + 1/2at^2. ( no v)

4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)

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Light refracts when traveling from air into glass because light
ella [17]

Answer:

Speed of light changes

Explanation:

Since glass is denser than air so the speed of light in glass is less than in air so light rays bend i.e. refract.

3 0
3 years ago
Which is hotter, Canopus or Vega? How much brighter?
marusya05 [52]

Temperature determines the hotness or coldness. The surface temperature of Canopus is 7350 K. The surface temperature of Vega is 9602 K. Because the surface temperature of Vega is greater than Canopus, Vega is hotter.

Brightness is determined by the star's magnitude. The apparent magnitude of Canopus is (-0.72) and that of Vega is (0.03). we know that, lesser the magnitude, brighter is the star. Hence, Canopus is brighter.

Vega is hotter but Canopus is brighter. This is because, brightness depends on the star's distance and size of the star as well other than the factor temperature.

4 0
3 years ago
When starting a foot race, a 64 kilogram sprinter exerts an average force of 693 newtons backward on the ground for 0.59 seconds
cluponka [151]

The distance traveled by the sprinter in meters is determined as 1.88 m.

<h3>Acceleration of the sprinter</h3>

The acceleration of the sprinter is the rate of change of velocity of the sprinter with time.

The acceleration of the sprinter is calculated as follows;

Apply Newton's second law of motion as follows;

F = ma

a = F/m

where;

  • F is the applied force by the sprinter
  • m is mass of the sprinter
  • a is acceleration of the sprinter

a = 693 N / 64 kg

a = 10.83 m/s²

<h3>Distance traveled by the sprinter</h3>

The distance traveled by the sprinter is calculated as follows;

s = ut + ¹/₂at²

where;

  • u is initial velocity = 0

s = ¹/₂at²

where;

  • t is time of motion
  • a is acceleration

s = (0.5)(10.83)(0.59²)

s = 1.88 m

Thus, the distance traveled by the sprinter in meters is determined as 1.88 m.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

6 0
1 year ago
Imagine you are on an airplane. Your plane has flown from 30 degrees North to 42 degrees North. How far have you flown?
ycow [4]
The answer to your question is 12 degrees because you subtract 42 degrees minus 30 degrees to get 12 degrees
7 0
3 years ago
A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr
const2013 [10]

Answer:

388 cm^3

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

pV=const.

which can also be rewritten as

p_1 V_1 = p_2 V_2

In our case, we have:

p_1 = 75.9 kPa is the initial pressure

V_1 = 639 cm^3 is the initial volume

p_2 = 125 kPa is the final pressure

Solving for V2, we find the final volume:

v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3

7 0
3 years ago
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