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Lelechka [254]
3 years ago
6

Whoever answers correctly gets brainlist!

Physics
2 answers:
Elenna [48]3 years ago
7 0

Answer: police use infred waves,

radio waves have largest wave length, the

only part of the spectrum seen by humans is visible waves

gamma waves have shortest wave length

Explanation:

UNO [17]3 years ago
5 0

Infared = used by police

gamma = short wavelength

radio = largest wavelength

visible = only ones we can see

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A 5000 g toy car starts from rest and moves a distance of 300 cm in 3 s under the action of a single constant force. Determine t
sveticcg [70]

Answer:

3.33 N

Explanation:

First, find the acceleration.

Given:

Δx = 3 m

v₀ = 0 m/s

t = 3 s

Find: a

Δx = v₀ t + ½ at²

3 m = (0 m/s) (3 s) + ½ a (3 s)²

a = ⅔ m/s²

Use Newton's second law to find the force.

F = ma

F = (5 kg) (⅔ m/s²)

F ≈ 3.33 N

4 0
3 years ago
What comes down but never goes up
White raven [17]

Answer:

Explanation:

rain and your age

5 0
3 years ago
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Troyanec [42]

Answer:

2

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5 0
2 years ago
A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal
Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, a=g=-9.8\ m/s^2(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

S=ut+\frac{1}{2}at^2

Plug in the given values and solve for 'u'. This gives,

0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0
3 years ago
A small grinding wheel has a moment of inertia of 4.0*10-5kgm2. What net torque must be applied to the wheel for its angular acc
kvv77 [185]

Hi there!

We can use the rotational equivalent of Newton's Second Law:

\huge\boxed{\Sigma \tau = I \alpha}

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I = Moment of inertia (kgm²)

α = Angular acceleration (rad/sec²)

We can plug in the given values to solve.

\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}

4 0
3 years ago
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