Which event occurred approximately 4.6 billion years ago

A student needs to travel a total of 400 miles to reach his vacation destination . if he drives at an average speed of 50 mph ho

Arte-miy333 [17]

Answer:

The answer is 8 hours.

Explanation:

M : H

50 : 1

400 : X

400 * 1 = 50 x

400 / 50 = 8

8 0

3 years ago

Read 2 more answers

What grade of sprain is a completely torn ligament?

Alex Ar [27]

Grade 1: Stretching or slight tearing of the ligament with mild tenderness, swelling and stiffness. The ankle feels stable and it is usually possible to walk with minimal pain.

Grade 2: A more severe sprain, but incomplete tear with moderate pain, swelling and bruising. Although it feels somewhat stable, the damaged areas are tender to the touch and walking is painful.

Grade 3: This is a complete tear of the affected ligament(s) with severe swelling and bruising. The ankle is unstable and walking is likely not possible because the ankle gives out and there is intense pain.

source - https://www.rushcopley.com/health/physician-articles/varying-degrees-of-ankle-sprains/

6 0

3 years ago

What is the wavelength of a 100 mhz radio signal?

Daniel [21]

100 MHz = 100,000,000 Hz = 10^8<span> Hz

And using basic conversions between frequencies, I've determined that the wavelength is roughly 3 meters.</span>

8 0

3 years ago

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

where the relation between time (t) and period (T) is:

$\mathsf{t = \dfrac{T}{2}}$

T = 2t

T = 2(0.247)

T = 0.494 seconds

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

By making the spring constant k the subject of the formula:

$\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}$

$\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}$

$\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}$

$\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}$

$\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}$

$\mathbf{ k =8.25 \ N/m}$

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

6 0

3 years ago

Compare tidal range to spring tide

makkiz [27]

Spring tides have higher high tides and lower low tides whereas neap tides have lower high tides and higher low tides. Hence, the range is much larger in a spring tide than in a low tide.

7 0

3 years ago