Answer:
d_2 = 4d_1
Explanation:
The range or horizontal distance covered by a projectile projected with a velocity U at an angel of θ to the horizontal is given by
R = U²sin2θ/g
Let the range or horizontal distance of ball 1 with initial velocity U projected at an angle θ = 55° be
d_1 = U²sin2θ/g
Let the range or horizontal distance of ball 2 with initial velocity V = 2U projected at an angle θ = 55° be
d_2 = V²sin2θ/g
= (2U)²sin2θ/g
= 4U²sin2θ/g
= 4d_1 (since d_1 = U²sin2θ/g)
So, the ball 2 lands a distance d_2 = 4d_1 from the initial point.
Answer: The smallest effort = 300N
Explanation:
Using one of the condition for the attainment of equilibrium:
Clockwise moment = anticlockwise moments
900 × 1 = 3 × M
Where M = the weight of the strong man
3M = 900
M = 900/3 = 300N
Therefore, 300N is the smallest effort that the strongman can use to lift the goat
Answer:
No, it's not there.
Explanation:
For a machine to be 100% efficient, it has to be with an output which is equal to its input. But machines have an out put less than an input, hence efficiency below 100%.
Answer:
very small solid particles called interstellar dust.
Explanation:
In the space between the stars there is gas and dust, which represent at least 20% of the mass of our galaxy. In the Milky Way it is considered that there is a gas density of approximately 0.2 to 0.5 atoms / cm3 in the surroundings of the Sun; with respect to the dust an average of 1 g / cm3 is estimated.
Gas is about atoms and molecules, mainly hydrogen; In order of abundance, helium, carbon, oxygen, nitrogen and iron follow. On the other hand, the dust is tiny particles, generally smaller than 10 microns; the dust does not shine and therefore it is only distinguished when it is projected on bright regions (nebulae or clusters).
Interstellar matter is mainly concentrated towards the plane of the galaxy, in the strip corresponding to the Milky Way; there you can see bright nebulas of diffuse character called nebulas. These nebulae are classified according to three types: (a) bright or emission nebulae, (b) reflection nebulae and (c) planetary nebulae.
Hydrogen appears both ionized and neutral; The bright nebulae are composed of ionized hydrogen and other ionized elements. Non-ionized (neutral) hydrogen is found in the spiral arms of the Milky Way and can be detected through radio waves.
Answer:
The resolution of an analog-to-digital converter is 24.41 mV
Explanation:
Resolution of an analog-to-digital = (analogue signal input range)/2ⁿ
where;
n is the number or length of bit, and in this question it is given as 12
Also, the analogue signal input range is 100V
Resolution of an analog-to-digital = 100V/2¹²
2¹² = 4096
Resolution of an analog-to-digital = 100V/4096
Resolution of an analog-to-digital = 0.02441 V = 24.41 mV
Therefore, the resolution of an analog-to-digital converter is 24.41 mV
