d=vi*t+(1/2)gt²
d=11 m
g=9.8 m/s²
vi=0 m/s
11 m=0 m/s*t+(1/2)9.8 m/s²t²
11 m=4.9 m/s²t²
t²=11 m / 4.9 m/s²
t=√(11 m / 4.9 m/s²)=1.489... s≈1.5 s
Answer: the time the sone is in flight is 1.5 s
Answer:
distance when the weight is 8 kg is 26.66 cm
Explanation:
given data
distance d2 = 10 cm
weight w2 = 3 kg
weight w1 = 8 kg
to find out
distance when the weight is 8 kg
solution
we consider here distance d1 when weight is 8 kg
so equation will be
d1/d2 = w1/w2
d/10 = 8/ 3
so d = 8/3 × 10
so d = 26.66
distance when the weight is 8 kg is 26.66 cm
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
Answer:
KE = 2.03 J
Explanation:
After impact, the kinetic energy of the bullet+block will convert to potential energy
½mv² = mgh
v = √(2gh) = √(2(9.81)(0.00500) = 0.0981 m/s
conservation of momentum during the collision
0.015u + 2.50(0) = (2.50 + 0.015)(0.0981)
u = 16.4481 m/s
KE = ½mv² = ½(0.015)16.4481² = 2.0290499...
KE = 2.03 J
The Soviet Union's Sputnik 1, it was launched October 4th 1957.