Answer:
no I dont sorry lol. ............
Answer:
<h2>9.52 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
f is the force
m is the mass
From the question we have
We have the final answer as
<h3>9.52 m/s²</h3>
Hope this helps you
Answer:
1.35 m
Explanation:
Taking down to be positive, given:
Δx = Δy / tan 30.0º
v₀ₓ = 4.50 m/s
v₀ᵧ = 0 m/s
aₓ = 0 m/s²
aᵧ = 10 m/s²
Find: Δy
First, find the time it takes to land in terms of Δy.
Δy = v₀ t + ½ at²
Δy = (0 m/s) t + ½ (10 m/s) t²
Δy = 5t²
Next, find Δx in terms of t.
Δx = v₀ t + ½ at²
Δx = (4.50 m/s) t + ½ (0 m/s) t²
Δx = 4.50t
Substitute:
Δy = 5 (Δx / 4.50)²
20.25 Δy = 5 (Δx)²
4.05 Δy = (Δx)²
4.05 Δy = (Δy / tan 30.0º)²
4.05 Δy = 3 (Δy)²
1.35 = Δy
The basketball was thrown from an initial height of 1.35 m.
Graph: desmos.com/calculator/ujuzdo9xpr
Answer:
Q = 142.324kJ
Explanation:
Data:
M = 500g = 0.5kg
T1 = 10°C = (10 + 273.15)K = 285.15K
T2 = 80°C = (80 + 273.15)K = 353.15K
Q = ?
C = 4186J/kg.K
Q = mc(T2 - T1)
Q = 0.5 * 4186 * (353.15 - 285.15)
Q = 0.5 * 4186 * 68
Q = 142324J
Q = 142.324kJ.
Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
