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3 years ago

Which THREE forms of light are invisible light?

Physics

2 answers:

PtichkaEL [24]3 years ago

3 0

Answer:

infrared light

Explanation:

Nana76 [90]3 years ago

3 0

Answer:

The infrared rays, radio waves and x-rays are forms of light are invisible light. Explanation: We know that, The electromagnetic spectrum shows the wave length of the electromagnetic radiation of visible light.

hope this helps

Explanation:

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Two horizontal forces act on a 1.4 kg chopping block that can slide over a friction-less kitchen counter, which lies in an xy pl

kogti [31]

Answer:

Part a)

$a = (0.64\hat i - 0.5 \hat j)m/s^2$

Part b)

$a = (0.64\hat i + 5.21 \hat j)m/s^2$

Part c)

$a = (4.92\hat i - 0.5 \hat j)m/s^2$

Explanation:

As per Newton's II law we know that

F = ma

so we will have

$a = \frac{F}{m}$

so we will have

$a = \frac{F_1 + F_2}{m}$

Part a)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i - 4\hat j)}{1.4}$

$a = \frac{0.9 \hat i - 0.7 \hat j}{1.4}$

$a = (0.64\hat i - 0.5 \hat j)m/s^2$

Part b)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i + 4\hat j)}{1.4}$

$a = \frac{0.9 \hat i + 7.3 \hat j}{1.4}$

$a = (0.64\hat i + 5.21 \hat j)m/s^2$

Part c)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (3\hat i - 4\hat j)}{1.4}$

$a = \frac{6.9 \hat i - 0.7 \hat j}{1.4}$

$a = (4.92\hat i - 0.5 \hat j)m/s^2$

3 0

3 years ago

How much heat is needed to melt 1.5kg of ice and then to raise the temperature of the

Kipish [7]

Answer:

> The amount of heat required to melt ice and raise the temperature of water T o C T^oC ToC is Q = m L f + m c Δ T Q=mL_f+mc\Delta T Q=mLf+mcΔT Here m = 1.5 k g m=1.5 kg m=1.5kg L f = 3.33 ∗ 1 0 5 J

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3 years ago

A careful photographic survey of Jupiter's moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

Over [174]

The concept used to solve this problem is that given in the kinematic equations of motion. From theory we know that the change in velocities of a body is equivalent to twice the distance traveled by acceleration, in other words:

$v_f^2-v_i^2 = 2ax$

Where,

$v_{f,i} =$ Final and initial velocity

a = Acceleration

x = Displacement

For the given case, the displacement is equivalent to the height (x = h) and the acceleration is the same gravitational acceleration (a = g). In turn we do not have initial speed therefore

$v_f^2 = 2hg$