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3 years ago

A car has a mass of 1.20 × 103 kilograms and a momentum of 2.00 × 104 kilogram meters/second. What is the velocity of the car?

2 answers:

8 0

momentem = mass * velocity also, be sure to use the carrot for exponents 10^3 and 10^4. 20000/1200 = 16.7,This is your answer,That was hard lol.

Hope the helps.

Vera_Pavlovna [14]3 years ago

3 0

We know, Momentum = Mass * velocity
2 * 10⁴ = (1.20 * 10³) (v)
v = 2 * 10⁴ / 1.20 * 10³
v = 1.67 * 10
v = 16.7 m/s

In short, Your Answer would be Option C

Hope this helps!

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In a physics lab experiment, a spring clamped to the table shoots a 22g ball horizontally. When the spring is compressed 21cm ,

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a square and a triangle I think

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3 years ago

Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range

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Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

X: Always attractive or attractive only

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Explanation:

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3 years ago

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2 years ago

Chapter 36, Problem 007 Light of wavelength 586 nm is incident on a narrow slit. The angle between the first diffraction minimum

svp [43]

Answer:

$d =4.77\times 10^{-5}\ m$

Explanation:

given

wavelength of light λ = 479 nm

= 479 x 10⁻⁹ m

the angle

θ = 1.15 / 2 = 0.575°

using

condition for diffraction minimum ,

d sinθ = m λ

for first minimum m = 1

d sinθ = λ

therefore ,

slit width

$d =\dfrac{\lambda}{sin\theta}$

$d =\dfrac{479\times 10^{-9}}{sin 0.575^0}$

$d =\dfrac{479\times 10^{-9}}{0.01}$

$d =4.77\times 10^{-5}\ m$

hence, the width of the slit is equal to $d =4.77\times 10^{-5}\ m$

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2 years ago