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juin [17]
3 years ago
13

A rectangular metal bar has length of 20cm, width of 5cm. Mass of the block is 2kg and its density is 5000kg/m3. Find the height

of the metal bar.
Physics
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer: 0.04m

Explanation:

Mass of the block (m) = 2kg

density (p) = 5000kg/m3

Volume (v) = ?

Recall that density is obtained by dividing mass by volume. i.e density = mass / volume

p = m / v

Make v the subject formula

v = m / p

So, v = 2kg / 5000kg/m3

v = 0.0004 m3

Now given that the volume of the rectangular metal bar is 0.0004m3, find its height:

length of metal bar (l) = 20cm

convert length in centimeters to meters

(If 100cm = 1m

20cm = 20/100 = 0.2m)

width of bar (w) = 5cm

convert width in centimeters to meters

(If 100cm = 1m

5cm = 5/100 = 0.05m)

height (h) = ?

Volume of metal bar = l x w x h

0.0004 m3 = 0.2m x 0.05m x h

0.0004 m3 = 0.2m x 0.05m x h

0.0004 m3 = 0.01m² x h

h = 0.0004 m3 / 0.01m²

h = 0.04m

Thus, the height of the rectangular metal bar is 0.04 meters

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A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk
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Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration\frac{m}{s^{2} }

at: truck acceleration\frac{m}{s^{2} })

ac = at= \frac{vf-vi}{t-ti}  equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 \frac{m}{s}, t = final time =5 s

We replaced the known information in the equation(1):

ac = at = \frac{13-0}{15-0}

ac=ac=\frac{13}{15}  \frac{m}{s}

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

Wc=mc*g

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 \frac{m}{s^{2} }

Wc= 2230*9.8

Wc=21854 N

Normal force calculation:

Newton's first law

sum Fy= 0

N-W=0

N=W

N=21854 N

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N  Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

Ff=0.373*21854

Ff=8151.54 N

Calculation of the tension force in the rope (T):

Newton's Second law

sum Fx= mc*ac

T-Ff=mc*ac

T=2230(\frac{13}{15}) + 8151.54

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0
3 years ago
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