The Celsius temperature scale is based on which of the following

A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 282 N at a speed of v = 0.850 m/s across the warehouse

Elanso [62]

Answer:

$v_{f} = 0.51 \frac{m}{s}$

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 )

$v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}$

$v_{f} = 0.51 \frac{m}{s}$

8 0

3 years ago

If a positively charged particle moves into a magnetic field traveling in a straight line, how would you expect it's motion to c

MrMuchimi

It would begin to curve toward the magnetic pull

6 0

3 years ago

Read 2 more answers

. A car going 50 m/s accelerates to pass a truck. Five seconds later the car is going 80m/s. Calculate the acceleration of the c

sineoko [7]

6 mph/s

Calculating acceleration involves dividing velocity by time — or in terms of SI units, dividing the meter per second [m/s] by the second [s]. Dividing distance by time twice is the same as dividing distance by the square of time. Thus the SI unit of acceleration is the meter per second squared .

8 0

3 years ago

A student was trying to find the relationship between mass and force. He placed four different masses on a table and pulled them

Gwar [14]

Answer:

B. There is a direct proportion between the mass and force listed in the table.

Explanation:

From the table, the values of force increases with increase in the value of mass.

if 5kg=25 N

Finding the contant of proportionality k;

k=25/5=5

thus M=k(F)...........where M is mass in kg and F is force in newton, then

M=5F

This show that for every value of mass, we get the value of Force if we multiply by a contant k=5

This means there is a direct proportionality relation between mass and force in the table.

5 0

3 years ago

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Free Fall: A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses

Slav-nsk [51]

Answer:

v₀₁= 5.525 m / s

Explanation

Freefall Formulas :

The sign of acceleration due to gravity (g) is positive if the object is going down and negative if the object is going up.

vf= v₀+gt

vf²=v₀²+2*g*h

h= v₀t+ (1/2)*g*t²

Where:

h: hight in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

g: acceleration due to gravity in m/s²

Kinematics of the rock from the starting point with vo until it reaches its maximum height:

vf₁= v₀₁-gt₁ :vf₁ =0 to maximum height

0= v₀₁-gt₁

v₀₁ = g*t₁

t₁ =v₀₁ / g Equation (1)

vf₁²= v₀₁²-2*g*h₁ : vf₁ =0 to maximum height

0 = v₀₁²-2*g*h₁

2*g*h₁ = v₀₁²

h₁ = (v₀₁²)/(2g) Equation (2)

Kinematics of the rock when it falls from the maximum height until it touches the floor

h₂= v₀₂t+ (1/2)*g*t₂² v₀₂=vf₁ =0

h₂= 0+ (1/2)*g*t₂²

h₂= (1/2)*g*t₂² Equation (3)

Equation that relates h₁ to h₂

h₂= h₁ + 56.3 , h₁ = (v₀₁²)/(2g)

h₂= (v₀₁²)/(2g) + 56.3 Equation (4)

Equation that relates t₁ to t₂

t₁ + t₂ =4 s

t₂ =4 -t₁

t₂ =4 -(v₀₁/g )

Calculation of v₀₁

We replace equation 4 and equation 5 in equation 3

(v₀₁²)/(2g) + 56.3 = (1/2)*g*(4 -(v₀₁/g ) )²

(v₀₁²)/(2g) + 56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g )+((v₀₁/g )²)

we eliminate (v₀₁²)/(2g) on both sides of the equation

56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g ))

56.3 = 78.4 - 4*v₀₁

4*v₀₁ =78.4-56.3

v₀₁= (78.4-56.3) / ( 4)

v₀₁= 5.525 m / s

7 0

3 years ago