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 )


Answer:
16.6 °C
Explanation:
From the question given above, the following data were obtained:
Temperature at upper fixed point (Tᵤ) = 100 °C
Resistance at upper fixed point (Rᵤ) = 75 Ω
Temperature at lower fixed point (Tₗ) = 0 °C
Resistance at lower fixed point (Rₗ) = 63.00Ω
Resistance at room temperature (R) = 64.992 Ω
Room temperature (T) =?
T – Tₗ / Tᵤ – Tₗ = R – Rₗ / Rᵤ – Rₗ
T – 0 / 100 – 0 = 64.992 – 63 / 75 – 63
T / 100 = 1.992 / 12
Cross multiply
T × 12 = 100 × 1.992
T × 12 = 199.2
Divide both side by 12
T = 199.2 / 12
T = 16.6 °C
Thus, the room temperature is 16.6 °C
Answer:
a force that is able to act at a distance
Explanation:
:)
C Occupational Safety and Health Administration
<h2>Question:</h2>
An automobile is driving uphill. Which form of energy is not involved in this process?
<h2>Choosing:</h2>
electromagnetic
potential
kinetic
chemical
<h2>Answer:</h2>
<u>Electromagnetic</u><u> </u>
<h3>
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