Answer:
The initial velocity of the bullet is 1066.63 m/s
Explanation:
The problem involves two different events: A collision between two objects and the stretching of a spring
Since we know the final condition of the system, we start from the end and back to the initial condition
The system of the bullet+block with mass= 4Kg+8gr = 4.008 Kg stretches the spring a distance of x=0.087 m
The work done when stretching a spring of constant k is
This work is equivalent to the change of potential energy of the spring, which is the result of the transformation of the kinetic energy of the total mass of the system
Solving for v' (the velocity of the system after the collision took place)
This is the resulting velocity after the bullet lodged into the block. That event complies with the law of conservation of linear momentum
Since the block is initially at rest ()
Solving for :
The initial velocity of the bullet is 1066.63 m/s
The answer you are looking for is either
A)<span>The ball that went out of the park shows more work because the force applied was greater or
</span><span>C)The ball that went out of the park shows more work because the distance was greater</span>
The weight on earth is equal to m*g, where m is the mass and g is the gravitational force. 6050 (m) * 9.8 (g) = 59290 Newtons.
The duration and intensity of solar radiation or insolation. Both of these factors are in turn governed by the annual change in the position of the Earth's axis relative to the Sun
Yearly changes in the position of the Earth's axis cause the location of the Sun to wander 47° across our skies. Changes in the location of the Sun have a direct effect on the intensity of solar radiation. The intensity of solar radiation is largely a function of the angle of incidence, the angle at which the Sun's rays strike the Earth's surface. If the Sun is positioned directly overhead or 90° from the horizon, the incoming insolation strikes the surface of the Earth at right angles and is most intense. If the Sun is 45° above the horizon, the incoming insolation strikes the Earth's surface at an angle. This causes the rays to be spread out over a larger surface area reducing the intensity of the radiation. Figure 6i-1 models the effect of changing the angle of incidence from 90 to 45°. As illustrated, the lower Sun angle (45°) causes the radiation to be received over a much larger surface area. This surface area is approximately 40% greater than the area covered by an angle of 90°. The lower angle also reduces the intensity of the incoming rays by 30%.
Figure 6i-1: Effect of angle on the area that intercepts an incoming beam of radiation.
We can also model the effect the angle of incidence has on insolation intensity with the following simple equation:
Intensity = SIN (A)
where, A is the angle of incidence and SIN is the sine function found on most calculators. Using this equation we can determine that an angle of 90° gives us a value of 1.00 or 100% (1.00 x 100). Let us compare this maximum value with values determined for other angles of incidence. Note the answers are expressed as a percentage of the potential maximum value.
SIN 80 = 0.98 or 98%
SIN 70 = 0.94 or 94%
SIN 60 = 0.87 or 87%
SIN 50 = 0.77 or 77%
SIN 40 = 0.64 or 64%
SIN 30 = 0.50 or 50%
SIN 20 = 0.34 or 34%
SIN 10 = 0.17 or 17%
SIN 0 = 0.00 or 0%
The yearly changes in the position of the Earth's axis relative to the plane of the ecliptic also causes seasonal variations in day length to all locations outside of the equator. Longest days occur during the June solstice for locations north of the equator and on the December solstice for locations in the Southern Hemisphere. The equator experiences equal day and night on every day of the year. Day and night is also of equal length for all Earth locations on the September and March equinoxes. Figure 6i-2 describes the change in the length of day for locations at the equator, 10, 30, 50, 60, and 70 degrees North over a one-year period. The illustration suggests that days are longer than nights in the Northern Hemisphere from the March equinox to the September equinox. Between the September to March equinox days are shorter than nights in the Northern Hemisphere. The opposite is true in the Southern Hemisphere. The graph also shows that the seasonal (winter to summer) variation in day length increases with increasing latitude.
Figure 6i-2: Annual variations in day length for locations at the equator, 30, 50, 60, and 70° North latitude.
Figure 6i-3 below describes the potential insolation available for the equator and several locations in the Northern Hemisphere over a one-year period. The values plotted on this graph take into account the combined effects of angle of incidence and day length duration (see Table 6h-2). Locations at the equator show the least amount of variation in insolation over a one-year period. These slight changes in insolation result only from the annual changes in the altitude of the Sun above the horizon, as the duration of daylight at the equator is always 12 hours. The peaks in insolation intensity correspond to the two equinoxes when the Sun is directly overhead. The two annual minimums of insolation occur on the solstices when the maximum height of the Sun above the horizon reaches an angle of 66.5°.
The most extreme variations in insolation received in the Northern Hemisphere occur at 90 degrees North. During the June solstice this location receives more potential incoming solar radiation than any other location graphed. At this time the Sun never sets. In fact, it remains at an altitude of 23.5 degrees above the horizon for the whole day. From September 22 (September equinox) to March 21, (March equinox) no insolation is received at 90 degrees North. During this period the Sun slips below the horizon as the northern axis of the Earth has an orientation that is tilted away from the Sun.
Answer:
19.81m/s
Explanation:
To find the speed at which the car leaves the cliff, first we need to know how long it took to reach the ground. We do this with the equation:
Where is the vertical position at a time , since we are looking for the moment when the car reaches the gound: . is the initial vertical position, in this case the height of the cliff: . is gravitational acceleration: .
So replacing the known values:
and since we need the time, we clear for it:
At time the car reaches the gound, according to the problem at a horizontal distance from the base of the cliff of 100m.
To find the velocity we use:
where is the horizontal distance at time (in this case and ), is the initial horizontal distance, we define the cliff as zero in horizontal distance so , and is the velocity of the car when it rolled of the cliff.
replacing the known values:
and we clear for :
The initial velocity (when it rolled of the cliff) is 19.81m/s