Can someone help me

A marble is dropped off the top of a building. Find it's velocity and how far below to its dropping point for the first 10 secon

Yuri [45]

consider the motion in y-direction

v₀ = initial velocity = 0 m/s

a = acceleration = g = - 9.8 m/s²

t = time

v = final velocity at any time "t"

velocity at any time is given as

v = v₀ + at

v = 0 + (- 9.8) t

v = (- 9.8) t

so at t = 1 , v = (- 9.8) (1) = - 9.8 m/s

at t = 2 , v = (- 9.8) (2) = - 19.6 m/s

at t = 3 , v = (- 9.8) (3) = - 29.4 m/s and so on

Y₀ = initial position where the marble was dropped from = 0 m

Y = final position at any time "t"

final position at any time "t" is given as

Y = Y₀ + v₀ t + (0.5) a t²

Y = 0 + (0) t + (0.5) (-9.8) t²

Y = - 4.9 t²

at t = 1 , Y = - 4.9 (1)² = - 4.9 m

at t = 2 , Y = - 4.9 (2)² = - 19.6 m

at t = 3 , Y = - 4.9 (3)² = - 44.1 m

and So on

6 0

3 years ago

What are the two major topics studied by a physicist

Alexxx [7]

Math and the study of motion

6 0

3 years ago

Which of Earths layers is made up of liquid metal

wolverine [178]

The outer core is made out of molten lava also known as liquid metal because of all the iron and other metals in it

7 0

3 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.23 2.23 times a second. A tack is stuck in the t

elena-14-01-66 [18.8K]

Explanation:

The given data is as follows.

Angular velocity ( $\omega$ ) = 2.23 rps

Distance from the center (R) = 0.379 m

First, we will convert revolutions per second into radian per second as follows.

= 2.23 revolutions per second

= $2.23 \times 2 \times 3.14 rad/s$

= 14.01 rad/s

Now, tangential speed will be calculated as follows.

Tangential speed, v = $R \times \omega$

= 0.379 x 14.01

= 5.31 m/s

Thus, we can conclude that the tack's tangential speed is 5.31 m/s.

8 0

3 years ago

Okay guys I need serious help!!

PilotLPTM [1.2K]

The traditional forecast process employed by most NMHSs involves forecasters producing text-based, sensible, weather-element forecast products (e.g. maximum/minimum temperature, cloud cover) using numerical weather prediction (NWP) output as guidance. The process is typically schedule-driven, product-oriented and labour-intensive. Over the last decade, technological advances and scientific breakthroughs have allowed NMHSs’ hydrometeorological forecasts and warnings to become much more specific and accurate.

As computer technology and high-speed dissemination systems evolved (e.g. Internet), National Weather Service (NWS) customers/partners were demanding detailed forecasts in gridded, digital and graphic formats. Traditional NWS text forecast products limit the amount of additional information that can be conveyed to the user community. The concept of digital database forecasting provides the capability to meet customer/partner demands for more accurate, detailed hydrometeorological forecasts. Digital database forecasting also offers one of the most exciting opportunities to integrate PWS forecast dissemination and service delivery, which most effectively serves the user community.

Both the US National Oceanic and Atmospheric Administration(NOAA)/National Weather Service and Environment Canada are currently using digital database forecasting technology to produce routine forecasts. The Australian Bureau of Meteorology is in the process of evaluating and developing an implementation plan for database forecasting using the NOAA/NWS National Digital Forecast Database approach.

3 0

3 years ago