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2 years ago

A car is at x1 = 15m at time t1 = 5 secs & later seen at x2=20m at time t2 = 10secs. find the average velocity of the car.

Physics

1 answer:

Otrada [13]2 years ago

8 0

Answer:

Looks hard wish I could answer it fast

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43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a

olga2289 [7]

Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

$P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J$

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J$

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

$\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N$

The frictional force that acts on the sliding bear is 378.47166 N

5 0

2 years ago

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Finish the sentence

Amiraneli [1.4K]

Answer:

1. Equal and opposite reaction

2. Pairs

3. That is equal

Explanation:

I have tried my best but without knowing your curriculum or the possible answers it is difficult to know exactly what the answer would be... :(

Please mark as branliest ;)

- Hope this helped!

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2 years ago

Considering , will F ever equal zero?

Kamila [148]

No it will not because F will equal 3 not 0

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3 years ago

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What can iron filings sprinkled on a piece of paper demonstrate?

jolli1 [7]

<span>c the pattern of the magnetic fields lines</span>

6 0

2 years ago

A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What

Digiron [165]

Answer:

83.1946504051 m

Explanation:

u = Initial velocity = $60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s$

s = Displacement = $123\ ft=\dfrac{123}{3.281}=37.4885705578\ m$

$\theta$ = Angle = $26^{\circ}$

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2$

Coefficient of friction

$\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561$

$mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m$

The stopping distance is 83.1946504051 m

6 0

3 years ago

A car is at x1 = 15m at time t1 = 5 secs & later seen at x2=20m at time t2 = 10secs. find the average velocity of the car.​

A car is at x1 = 15m at time t1 = 5 secs & later seen at x2=20m at time t2 = 10secs. find the average velocity of the car.