Answer:
Explanation:
No, the reading is not expected to be accurate. This is because Relative to Earth, the air on Mars is extremely thin. The Martian atmosphere is primarily carbon dioxide with a much lower surface pressure, and Mars does not have oceans and an Earthlike hydrological cycle so latent heat release is not as important as it is for Earth.
Answer:
Bacteria, Hydra, Copperheads, Blackworms, and Strawberries
Explanation:
Answer:
A) 1.53s
B) 19.8m
C) 2.869m
Explanation:
A) The time of flight for a projectile can be calculated using the formula:
t = 2μsinθ/g
Where; u = velocity
θ = angle
g = acceleration due to gravity (9.8m/s^2)
t = 2 × 15 × sin 30°/9.8
t = 30sin30°/9.8
t = 30 × 0.5/9.8
t = 15/9.8
t = 1.53s
B) The horizontal range (distance) for a projectile can be calculated using the formula:
Range = u²sin2θ/ g
Range = 15² sin 2 × 30 / 9.8
Range = 225 sin 60/9.8
Range = 225 × 0.8660/9.8
Range = 194.855/9.8
Range = 19.8m
C) The maximum height for a projectile can be calculated using the formula:
h = u²sin²θ/2g
h = 15² (sin 30)² / 2 × 9.8
h = 225 × 0.25 / 19.6
h = 56.25/19.6
h = 2.869m
Answer:
(a) W = 8.66 J
(b) Velocity = 2.40 m/s
Explanation:
(a) Work done is given as the product of force and displacement. That is:
W = F * d * cos(A)
Where F = force applied
d = distance moved
A = angle of ramp
Therefore, work done is:
W = 20 * 0.5 * cos30
W = 8.66 J
(b) Work done is equal to change in Kinetic energy. Since the initial kinetic energy is zero:
W = KE(final)
W = ½ * m * v²
Where v = final velocity
=> 8.66 = ½ * 3 * v²
v² = 5.773
v² = 2.40 m/s
Answer:
<h3>
a)</h3>
<u>=> R= 6 Ohms(Ω)</u>
<h3>b)</h3>
<em>these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,</em>
<em>R and 100 can interchange places</em>
<u>=> R = 576 Ω</u>
<u></u>
By Ohm's Law:
=> 240 = I × 576
=>
=> I = 0.417 A
<h3 /><h3>c)</h3>
I don't know it's resistance,... so sorry
<h3>d)</h3>
The brightness of the bulb in series is <em><u>less than</u></em> when they're placed individually.
For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.
Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.
So, the effective resistance of some bulbs in series <u>is more</u> than the individual resistance.
And
<em>Brightness, i. e., Power</em>
If resistance increases, Power decreases.
Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.
