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3 years ago

7. A 2.0 kg block, starting from rest, is pushed by a

Physics

1 answer:

11Alexandr11 [23.1K]3 years ago

7 0

Answer:

(A) 0.8 kgm/s

Explanation:

because of the even ground it would only slow down

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A ball is rolling horizontally at 3.00 meters per second as it leaves the edge of a tabletop 0.750 meter above the floor. The ba

NikAS [45]

Answer: 9.81m/s^2

Explanation: since it’s free fall and friction is neglected, the magnitude of the ball’s acceleration will be equal to the general acceleration due to gravity which is 9.81m/s^2

8 0

3 years ago

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The formula for distance divided by time

azamat

It would be: Speed = Distance / Time

4 0

3 years ago

In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope

zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude $g$ .

If we have the initial velocity $v_o$ and its angle $\theta$ , we can obtain the vertical component of the velocity $v_{oy}$ using trigonometry:

$v_{oy}=v_osin\theta$

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

$v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }$

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

$g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

Finally, from the equation of horizontal motion with constant speed, we have that:

$x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}$

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

$v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m$

In words, the projectile travels 7.49m horizontally before it lands.

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4 years ago

Healthy pregnant women should get at least 150 minutes of moderate-intensity aerobic activity, spread over a 7-day period, each

kirill115 [55]

Healthy pregnant women should get at least 150 minutes of moderate-intensity aerobic activity, spread over a 7-day period, each week - True

<u>Explanation:</u>

Women who are healthy but moderately active must be engaged with aerobic activities moderately. they should be engaged with this activities at least 2 and a half hours in all 7 days of a week. They must be engaged with this activity during the pregnancy time and also in the period after their delivery.

During the time of pregnancy all women must be healthy and physically active. The ways for maintaining a very good health in the pregnancy time will be guided by the health care providers. Women should be highly active during and after pregnancy time. This helps in having a baby that is also very healthy.

3 0

3 years ago

Hello....I need help with this question.

musickatia [10]

Answer:

3.2 m/s²

Explanation:

Acceleration can be calculated as:

v = u + at (where v is final velocity, u is initial velocity, a is acceleration and t is time)

25 m/s = 9 m/s + a(5 s) (a is unknown)

16 m/s = a(5 s)

a = 3.2 m/s²

We assume that this is a uniform acceleration (meaning that the velocity increases at an equal rate for those 5 seconds).

6 0

4 years ago

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