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nata0808 [166]
3 years ago
10

Need help with this one question!​

Physics
2 answers:
Lelu [443]3 years ago
8 0

Answer:

Actual Mechanical Advantage

Explanation:

AMA=mg/F(applied)

lisov135 [29]3 years ago
4 0
I think the following ratio is called, “actual mechanical advantage.”
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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.

8 0
4 years ago
How far apart must two protons be if the electrical force acting on either one is equal to its weight on the earth's surface whe
BARSIC [14]

Answer:

G M m / R^2 = F   force of attraction for proton

K e^2 / r^2 = F     force of attraction between 2 protons

G M m / R^2 = K e^2 / r^2

r^2 = K e^2 R^2 / (G M m)

r^2 = 9E9 * (1.6E-19)^2 * (6.37E6)^2 / (6.67E-11 * 5.98E24 * 1.7E-27)

r^2 = 9 * 2.56 * 40.6 / (6.67 * 5.98 * 1.7) * 10^-19

r^2 = 1.38 * 10^-30

r = 1.17E-15 m

7 0
3 years ago
You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 km/h (65 mi/h), and the trip take
valkas [14]

Answer: 2 hr 45 min

Explanation:

Given

The average speed between San Diego and Los Angeles is 105\ kmph

time taken for this trip is  1\ hr\ 50\ min\ or\ \frac{11}{6}\ hr

Distance between San Diego and Los Angeles is

\Rightarrow d=105\times \dfrac{11}{6}\\\\\Rightarrow d=192.5\ km

For Friday afternoon, the average speed is  70\ km/h

time taken to complete the trip is

\Rightarrow t=\dfrac{192.5}{70}\\\\\Rightarrow t=2.75\ hr\ or\ 2\ hr\ 45\ min

3 0
3 years ago
Lord Beckett and members of the EIT Co. spot the Black Pearl in the distance making its way towards land. As a
KATRIN_1 [288]

Answer:

Explanation:

The cannonball goes a horizontal  distance of 275 m . It travels a vertical distance of 100 m

Time taken to cover vertical distance =  t ,

Initial velocity u = 0

distance s = 100 m

acceleration a = 9.8 m /s²

s = ut + 1/2 g t²

100 = .5 x 9.8 x t²

t = 4.51 s

During this time it travels horizontally also uniformly so

horizontal velocity Vx = horizontal displacement / time

= 275 / 4.51 = 60.97 m /s

Vertical velocity Vy

Vy = u + gt

= 0 + 9.8 x 4.51

= 44.2 m /s

Resultant velocity

V = √ ( 44.2² + 60.97² )

= √ ( 1953.64 + 3717.34 )

= 75.3 m /s

Angle with horizontal Ф

TanФ = Vy / Vx

= 44.2 / 60.97

= .725

Ф = 36⁰ .

6 0
3 years ago
a student determines the circumference of a golf ball. a man pulls a sledge of mass 25 kg across level ground with a horizontal
Travka [436]

Explanation:

Fnet = Ft - Ff = 60N - 20N = 40N

a = Fnet / m = 40N / 25kg = 1.6m/s².

4 0
3 years ago
Read 2 more answers
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