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kumpel [21]
3 years ago
6

The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t

o the work of the resistance force; equal to the mean resistance force?
Physics
1 answer:
maxonik [38]3 years ago
3 0

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s

Work done is equal to the change in kinetic energy. It can be given as follows :

W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J

So, the required work done is 32.99 J.

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In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the
ololo11 [35]

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

7 0
3 years ago
PLEASE HELP THIS IS DUE LIKE HELP ME PLEASE
marysya [2.9K]

Answer:

At the end points of motion (either side) the velocity must be zero because the velocity is changing from - to + (it can't turn around around without passing thru zero,

The velocity will then increase to the midpoint of the motion.

m g h = 1/2 m v^2    where h is the vertical distance thru which the pendulum travels

5 0
3 years ago
A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
4 years ago
What are the names of the 4 types of fronts? How are they created?
jeka57 [31]

Answer:

Stationary Front, warm front, cold front, Occluded Front.

Explanation:

Stationary Front. When the surface position of a front does not change (when two air masses are unable to push against each other; a draw), a stationary front is formed.

cold front is the leading edge of a cooler mass of air at ground level that replaces a warmer mass of air and lies within a pronounced surface trough of low pressure. It often forms behind an extratropical cyclone (to the west in the Northern Hemisphere, to the east in the Southern), at the leading edge of its cold air advection pattern—known as the cyclone's dry "conveyor belt" flow. Temperature differences across the boundary can exceed 30 °C (86 °F) from one side to the other. When enough moisture is present, rain can occur along the boundary. If there is significant instability along the boundary, a narrow line of thunderstorms can form along the frontal zone. If instability is weak, a broad shield of rain can move in behind the front, and evaporative cooling of the rain can increase the temperature difference across the front. Cold fronts are stronger in the fall and spring transition seasons and weakest during the summer.

A warm front is a density discontinuity located at the leading edge of a homogeneous warm air mass, and is typically located on the equator-facing edge of an isotherm gradient. Warm fronts lie within broader troughs of low pressure than cold fronts, and move more slowly than the cold fronts which usually follow because cold air is denser and less easy to remove from the Earth's surface. This also forces temperature differences across warm fronts to be broader in scale. Clouds ahead of the warm front are mostly stratiform, and rainfall gradually increases as the front approaches. Fog can also occur preceding a warm frontal passage. Clearing and warming is usually rapid after frontal passage. If the warm air mass is unstable, thunderstorms may be embedded among the stratiform clouds ahead of the front, and after frontal passage thundershowers may continue. On weather maps, the surface location of a warm front is marked with a red line of semicircles pointing in the direction of travel.

In meteorology, an occluded front is a weather front formed during the process of cyclogenesis. The classical view of an occluded front is that they are formed when a cold front overtakes a warm front, such that the warm air is separated (occluded) from the cyclone center at the surface. The point where the warm front becomes the occluded front is called the triple point; a new area of low-pressure that develops at this point is called a triple-point low. A more modern view of the formation process suggests that occluded fronts form directly during the wrap-up of the baroclinic zone during cyclogenesis, and then lengthen due to flow deformation and rotation around the cyclone.

3 0
3 years ago
Read 2 more answers
A real spring does not oscillate forever. Instead, it eventually comes to a stop. Does this violate the law of conservation of e
ss7ja [257]
It does not violate the law of conservation of energy. The oscillation stops when the energy is lost and the energy is lost because it becomes heat that is created by the air resistance and many other forces found in the surrounding of the oscillating spring.
5 0
3 years ago
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