Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
Answer:
5.3 cm
Explanation:
This question is an illustration of real and apparent distance.
From the question, we have the following given parameters
Real Distance, R = 8.0cm
Refractive Index, μ = 1.5
Required
Determine the apparent distance (A)
The relationship between R, A and μ is:
μ = R/A
i.e.
Refractive Index = Real Distance ÷ Apparent Distance
Substitute values in the above formula
1.5 = 8/A
Multiply both sides by A
1.5 * A = A * 8/A
1.5A = 8
Divide both side by 1.5
1.5A/1.5 = 8/1.5
A = 8/1.5
A = 5.3cm
Hence, the letters would appear at a distance of 5.3cm
Answer:
(a) 11.66 square meters per liter
(b) 11657.8 per meters
(c) 0.00211 gal per square feet
Explanation:
(a) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 1gal/3.7854L = 11.66m^2/L
(b) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 264.17gal/1m^3 = 11657.8/m
(c) Inverse of 475ft^2/gal = 1/475ft^3/gal = 0.00211gal/ft^3
<span>e=ca{\displaystyle e={\frac {c}{a}}}.</span>
Answer:
A.The positive z-direction
Explanation:
We are given that
Linear charge density of long line which is located on the x-axis=
Linear charge density of another long line which is located on the y-axis=
We have to find the direction of electric field at z=a on the positive z-axis if and are positive.
The direction of electric field at z=a on the positive z-axis is positive z-direction .
Because and are positive and the electric field is applied away from the positive charge.
Hence, option A is true.
A.The positive z-direction
