Answer:
A) 12.08 m/s
B) 19.39 m/s
Explanation:
A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:
mg(sinθ) – F_k = ma
Where; F_k is frictional force due to kinetic friction given by the formula;
F_k = (μ_k) × F_n
F_n is normal force given by mgcosθ
Thus;
F_k = μ_k(mg cosθ)
We now have;
mg(sinθ) – μ_k(mg cosθ) = ma
Dividing through by m to get;
g(sinθ) – μ_k(g cosθ) = a
a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)
a = -3.71 m/s²
We are told that distance d = 24.0 m and v_o = 18 m/s
Using newton's 3rd equation of motion, we have;
v = √(v_o² + 2ad)
v = √(18² + (2 × -3.71 × 24))
v = 12.08 m/s
B) Now, μ_k = 0.10
Thus;
a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)
a = 1.08 m/s²
Using newton's 3rd equation of motion, we have;
v = √(v_o + 2ad)
v = √(18² + (2 × 1.08 × 24))
v = 19.39 m/s
P=25x10^6 andF=750.So plug in everything to solve for A. which is 3x10^-5m^2 OR 0.3mm^2
Answer:
Ionizing radiation is radiation with enough energy so that during an interaction with an atom, it can remove tightly bound electrons from the orbit of an atom, causing the atom to become charged or ionized. ... Forms of electromagnetic radiation.
(from google)
thank you :)
Answer:
(a) F= 6.68*10¹¹⁴ N (-k)
(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Explanation
To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:
F=q* v X B Formula (1 )
q: charge (C)
v: velocity (m/s)
B: magnetic field (T)
vXB : cross product between the velocity vector and the magnetic field vector
Data
q= -1.24 * 10¹¹⁰ C
v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j
B =(1.40 T)i
B =(1.40 T)k
Problem development
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =
= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)
F= 6.68*10¹¹⁴ N (-k)
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =
=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s
F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
The turning moment on the cover of the book is 0.05 Nm.
Explanation:
Given:
Force applied (F) = 0.5 N
Distance covered (d) = 10 cm
Converting Distance covered from cm to meter we get (d)= 0.1 m
To find:
Turning Moment (M) on the cover of the book = ?
Formula to be used:
Turning Moment (M) = F × d
= 0.5 × 0.1
= 0.05 Nm
Thus the turning moment on the cover of the book is found to be 0.05 Nm
