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3 years ago

HALP me!! This is a physics question.

Physics

2 answers:

zepelin [54]3 years ago

5 0

Distance 6.9 m

Work done

345 kJ = 345000

Force 5 x 10 4 = 50000 N

emmasim [6.3K]3 years ago

4 0

<h2>Answer:</h2>

<u>Distance covered is 6.9 meters</u>

<h2>Explanation:</h2>

Data given:

Work Done = 345 kJ = 345000 J

Force = 5 x 10 ^ 4 = 50000 N

Distance = ?

Solution:

As we know that

Work Done = Force applied x Distance covered

By arranging the equation we get

Work / Force = Distance covered

By putting the values

345000 / 50000 = 6.9

So distance covered is 6.9 meters

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What evidence can you cite to support the claim that the frequency of light does not change upon reflection?

sergiy2304 [10]

Answer: The color of an image is identical to the color of the object forming the image. When you look at yourself in a mirror, the color of your eyes doesn’t change. The fact that the color is the same is evidence that the frequency of light doesn’t change upon reflection. 2.

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5 0

3 years ago

A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget

mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, $m_{1}$ = 4 kg

Speed, $v_{1}$ = 5 m/s

$m_{2}$ = 1 kg

$v_{2}$ = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

$\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2$

By plugging in the values we get,

$KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\$

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

$KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)$

$KE = 40J + KE(lost)$

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0

3 years ago

In a compound, the atom does or does not take on a new set of properties

disa [49]

It does take on new set of proerties

7 0

3 years ago

Hi i do not have a question i just want to show u my cute doggo :)) ik everyone on here is probably stressed so hope this makes

Arisa [49]

Answer:

wooowwwwww

Explanation:

it's so so cuteeeeeeee

I love dogs tho:))

5 0

3 years ago

Read 2 more answers

A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

malfutka [58]

Answer:

Acceleration, $a=-2.48\ m/s^2$

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}$

$a=-2.48\ m/s^2$

So, the acceleration on the rough ice $-2.48\ m/s^2$ and negative sign shows deceleration.

8 0

3 years ago