Answer:
Geology is the study of the Earth that involves the process at Earth, materials of which it is made, and its history.
<u>Geologists combine both laboratory and field data to illustrate the results of their research. Some observations that can the geologist make by working outdoors instead of in a lab are as follows:</u>
- Understanding and exploring the earth's surface closely using geophysical tools.
- Collecting samples by own and make some interpretations at the same time.
- Observation of the landscapes
- Close observation of outcrops
First, calculate how long the ball is in midair. This will depend only on the vertical displacement; once the ball hits the ground, projectile motion is over. Since the ball is thrown horizontally, it originally has no vertical speed.
t = time vi = initial vertical speed = 0m/s g = gravity = -9.8m/s^2 y = vertical displacement = -45m
y = .5gt^2 [Basically, in this equation we see how long it takes the ball to fall 45m] -45m = .5 (-9.8m/s^2) * t^2 t = 3.03 s
Now we know that the ball is midair for 3.03s. Since horizontal speed is constant we can simply use:
x = horizontal displacement v = horizontal speed = 25m/s t = time = 3.03s
x = v*t x = 25m/s * 3.03s = 75.76 m Thus, the ball goes about 75 or 76 m from the base of the cliff.
The base of the pyramid has the producers and everything else above the base falls under the consumers category i.e the locusts,frogs and the snake. The grass is the producer, the locust is a consumer, the frog is a special type of omnivore, termed the "life-history omnivore" since they eat both plants and animals but at different times in their lives. In this case they are just omnivores and lastly, the snake is a carnivore.
Answer:
The magnetic field will be
, '2d' being the distance the wires.
Explanation:
From Biot-Savart's law, the magnetic field (
) at a distance '
' due to a current carrying conductor carrying current '
' is given by
![\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}](https://tex.z-dn.net/?f=%5Clarge%7B%5Coverrightarrow%7BB%7D%20%3D%20%5Cdfrac%7B%5Cmu_%7B0%7DI%7D%7B4%20%5Cpi%7D%7D%20%5Cint%20%5Cdfrac%7B%5Coverrightarrow%7Bdl%7D%20%5Ctimes%20%5Chat%7Br%7D%7D%7Br%5E%7B2%7D%7D%7D)
where '
' is an elemental length along the direction of the current flow through the conductor.
Using this law, the magnetic field due to straight current carrying conductor having current '
', at a distance '
' is given by
![\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}](https://tex.z-dn.net/?f=%5Clarge%7B%5Coverrightarrow%7BB%7D%7D%20%3D%20%5Cdfrac%7B%5Cmu_%7B0%7DI%7D%7B2%20%5Cpi%20d%7D)
According to the figure if '
' be the current carried by the top wire, '
' be the current carried by the bottom wire and '
' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the
symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by
symbol.
Given
and ![\large{I_{B} = 12.5 A}](https://tex.z-dn.net/?f=%5Clarge%7BI_%7BB%7D%20%3D%2012.5%20A%7D)
Therefore, the magnetic field (
) at 'P' due to the top wire
![B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}](https://tex.z-dn.net/?f=B_%7Bt%7D%20%3D%20%5Cdfrac%7B%5Cmu_%7B0%7DI_%7Bt%7D%7D%7B2%20%5Cpi%20d%7D)
and the magnetic field (
) at 'P' due to the bottom wire
![B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}](https://tex.z-dn.net/?f=B_%7Bb%7D%20%3D%20%5Cdfrac%7B%5Cmu_%7B0%7DI_%7Bb%7D%7D%7B2%20%5Cpi%20d%7D)
Therefore taking the value of
the net magnetic field (
) at the midway between the wires will be
![\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T](https://tex.z-dn.net/?f=%5Clarge%7BB_%7BM%7D%20%3D%20%5Cdfrac%7B4%20%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%7D%7B2%20%5Cpi%20d%7D%20%28I_%7Bt%7D%20-%20I_%7Bb%7D%29%20%3D%20%5Cdfrac%7B2%20%5Ctimes%2010%5E%7B-7%7D%7D%7Bd%7D%20%3D%20%5Cdfrac%7B41.4%20%5Ctimes%2010%20%5E%7B-4%7D%7D%7Bd%7D%7D%20T)
Answer:
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.
how much work is done on the monitor by (a) friction, (b) gravity
work(friction) = 453.5J
work(gravity) = -453.5J
Explanation:
Given that,
mass = 14kg
displacement length = 5.50m
displacement angle = 36.9°
velocity = 2.30cm/s
F = ma
work(friction) = mgsinθ .displacement
= (14) (9.81) (5.5sin36.9°)
= 453.5J
work(gravity)
= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)
= 126.9°
work(gravity) = (14) (9.81) (5.5cos126.9°)
= -453.5J