Answer:
θ = 13.16 °
Explanation:
Lets take mass of child = m
Initial velocity ,u= 1.1 m/s
Final velocity ,v=3.7 m/s
d= 22.5 m
The force due to gravity along the incline plane = m g sinθ
The friction force = (m g)/5
Now from work power energy
We know that
work done by all forces = change in kinetic energy
( m g sinθ - (m g)/5 ) d = 1/2 m v² - 1/2 m u²
(2 g sinθ - ( 2 g)/5 ) d = v² - u²
take g = 10 m/s²
(20 sinθ - ( 20)/5 ) 22.5 = 3.7² - 1.1²
20 sinθ - 4 =12.48/22.5
θ = 13.16 °
Burning of gases is one the example of chemical change
Mary walks:
d 1 = 80 m, d 2 = 125 m, d 3 = 45 m
t = 10 minutes = 600 seconds;
Average speed:
v = ( d 1 + d 2 + d 3 ) / t
v = ( 80 m + 125 m + 45 m ) / 600 s
v = 250 m / 600 s
v = 0.4167 m/s ≈ 0.42 m/s
Answer:
E ) 0.42 meters/second
Pressure = total force/total area
Total force = 660 Newton's
Total area:
Each leg contacts the floor with an area of πr^2=π(0.025m)^2=0.002m^2.
Total contact area for all 3 legs = 0.006 m^2.
Pressure = (660N) / (0.006 m^2)
= 110,000 N/m^2 = 110,000 Pascal's.
Answer:
3.6 × 10⁵ N/C = 360 kN/C
Explanation:
Let R = 2.0 cm be the radius of the sphere and q = -8.0 nC be the charge in it. Let q₁ be the charge at radius r = 1.0 cm. Since the charge is uniformly distributed, the volume charge density is constant. So, q/4πR³ = q₁/4πr³
q₁ = q(r/R)³. The electric field due to q₁ at r is E₁ = kq₁/r² = kq(r/R)³/r² = kqr/R³
The electric field due to the point charge q₂ = 5.0 nC is E₂ = kq₂/r².
So, the magnitude of the total electric field at r = 1.0 cm is
E = E₁ + E₂ = kqr/R³ + kq₂/r² = k(qr/R³ + q₂/r²)
E = 9 × 10⁹(-8 × 10⁻⁹ C × 1 × 10⁻² m/(2 × 10⁻² m)³ + 5 × 10⁻⁹ C/(1 × 10⁻² m)²)
E = 9 × 10⁹(-1 × 10⁻⁵ + 5 × 10⁻⁵)
E = 9 × 10⁹(4 × 10⁻⁵)
E = 36 × 10⁴ N/C = 3.6 × 10⁵ N/C = 360 kN/C
