Answer:
210
Explanation:
A ball rolls horizontally off the cliff at a speed of 30 m/s. It takes 7 seconds for the ball to hit the ground. What is the height of the cliff and the horizontal distance traveled by the ball?
S = (1/2)*9.8 m/s^2 * 7^2 = 240.1 m if the ball is very dense so air resistance, and therefore terminal velocity, can be ignored.
S = v * t = 30 m/s * 7 s = 210 m for the horizontal distance, again assuming negligible air resistance.
Melting, of course. Just as how an ice cube melts to water.
Answer:
B. 2 m/s
B. Acceleration = 4.05 m/s² and Tension = 297.5 N.
Explanation:
A force is applied on a mass m whose acceleration is 4 m/s
Force = mass × acceleration
a = F/m = 4 m/s
4 m/s = F/m
F = 4 m/s (m)
If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:
2F/4 m = F/ 2m
4m/s (m) / 2m = 2 m/s
a = 2 m/s
2.
Given that
mass
= 30 kg
mass
= 50 kg
= 0.1
From the question; we can arrive at two cases;
That :
----- equation (1)
---- equation (2)
50 a = 50 g - T
30 a = T - 30 g sin 30 - 4 × 30 g cos 30
By summation
80 a =
g
80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)
80 a = 324 m/s ²
a = 324/80
a = 4.05 m/s²
From equation , replace a with 4.05
50 × 4.05 = 50 × 10 - T
T = 500 -202.5
T =297.5 N
Answer:
r = 3.787 10¹¹ m
Explanation:
We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration
F = ma
G m M / r² = m a
The centripetal acceleration is given by
a = v² / r
For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship
v = d / t
The distance traveled Esla orbits, in a circle the distance is
d = 2 π r
Time in time to complete the orbit, called period
v = 2π r / T
Let's replace
G m M / r² = m a
G M / r² = (2π r / T)² / r
G M / r² = 4π² r / T²
G M T² = 4π² r3
r = ∛ (G M T² / 4π²)
Let's reduce the magnitudes to the SI system
T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)
T = 1.03 10⁸ s
Let's calculate
r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]
r = ∛ (21.44 10³⁵ / 39.478)
r = ∛(0.0543087 10 36)
r = 0.3787 10¹² m
r = 3.787 10¹¹ m
Answer:
The oxidation number of a monatomic (composed of one atom) ion is the same as the charge of the ion. For example, the oxidation numbers of K+, Se2−, and Au3+ are +1, -2, and +3, respectively. The oxidation number of oxygen in most compounds is −2.
Explanation: