Answer:
Explanation:
We apply Newton's second law at the crate :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m=90kg : crate mass
F= 282 N
μk =0.351 :coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Crate weight (W)
W= m*g
W= 90kg*9.8 m/s²
W= 882 N
Friction force : Ff
Ff= μk*N Formula (2)
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W = 0
N = W
N = 882 N
We replace the data in the formula (2)
Ff= μk*N = 0.351* 882 N
Ff= 309.58 N
We apply the formula (1) in x direction:
∑Fx = m*ax , ax=0
282 N - 309.58 N = 90*a
a= (282 N - 309.58 N ) / (90)
a= - 0.306 m/s²
Kinematics of the crate
Because the crate moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d Formula (3)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = 0.850 m/s
d = 0.75 m
a= - 0.306 m/s²
We replace the data in the formula (3)
vf²=(0.850)²+(2)( - 0.306 )(0.75 )
Complete Question:
In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.
a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.
b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.
c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.
d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?
Answer:
force, 1.318 ₓ 10⁻⁴
direction, 18.435°
Explanation:
The attached file gives a breakdown step by step solution to the questions
Answer:
Explanation:
Given Data:
Numbers of times Telescope cycled around the earth in 6 years=37,000 times
Total Distance traveled in 6 years by the Hubble Space Telescope=1,280,000,000 Km
Find:
Kilometers in one Orbit=?
Solution:
Kilometers in 37,000 Orbits=1,280,000,000 Km
Kilometers in 1 Orbit=1,280,000,000/37,000
In Scientific Notation:
Kilometers in 1 Orbit=34594.594 Km
Kilometers in 1 Orbit in Scientific notation:
Newton's law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers
Answer: a) 3.85 days
b) 10.54 days
Explanation:-
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = time taken for decomposition = 3 days
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 
First we have to calculate the rate constant, we use the formula :
Now put all the given values in above equation, we get
a) Half-life of radon-222:
Thus half-life of radon-222 is 3.85 days.
b) Time taken for the sample to decay to 15% of its original amount:
where,
k = rate constant = 
t = time taken for decomposition = ?
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process = 
Thus it will take 10.54 days for the sample to decay to 15% of its original amount.
