For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
Answer:0.53s
Explanation:
Answer:
5.87 m, 75.5°
Explanation:
Let's say C is the resultant vector.
The x component is the sum of the x components of A and B.
Cx = Ax + Bx
Cx = 10.1 cos 60° + 7.3 cos 275°
Cx = 5.69
The y component is the sum of the y components of A and B.
Cy = Ay + By
Cy = 10.1 sin 60° + 7.3 sin 275°
Cy = 1.47
The magnitude is found with Pythagorean theorem, and the angle is found with trigonometry.
C² = Cx² + Cy²
C² = 5.69² + 1.47²
C = 5.87
θ = atan(Cy / Cx)
θ = atan(5.69 / 1.47)
θ = 75.5°
It is an important part of many cells and processes such as amino acids, proteins and even our DNA. It is also needed to make chlorophyll in plants, which is used in photosynthesis to make their food.
Hope this helps! ;)
Answer: 94.9 metres
Explanation:
Given that,
Force between the two charges (F) = 24x10² N
Negative charge (q1) = 8x10-5C
Positive charge (q2) = 3x10C
Distance between charges (r) = ?
Apply the formula for
F= (kq1q2) / r²
Recall that k is Coulomb's constant with a value of 9 × 10^9 Nm2C−2
Then,
24x10² N = (9 × 10^9 Nm2C−2 x 8x10-5C x 3x10C) / r²
24x10² N = (2.16 x 10^7) / r²
r² = (2.16 x 10^7 / 24x10² N)
r² = 9000
To get the value of r, find the square root
r = √9000
r = 94.868m (rounded to the nearest tenth as 94.9m)
Thus, a distance of 94.9 metres separates the charges.