A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
1 answer:
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<h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>
Explanation:
<h2>Given:</h2>m = 1,250 kg
= 30 km/hr
F = 1,700 N
t = 10 s
<h2>Required:</h2>Final velocity
<h2>Equation:</h2><h3>Force</h3>F = ma
where: F - force
m - mass
a - acceleration
<h3>Acceleration</h3>a =
where: a - acceleration
- initial velocity
- final velocity
t - time elapsed
<h2>Solution:</h2><h3>Solve for acceleration using the formula for force</h3>F = ma
Substitute the value of F and m
(1700 N) = (1250 kg)(a)
a =
a = 1.36 m/s²
<h3>Solve for final velocity using the formula for acceleration</h3>- Convert 30 km/hr to m/s
=
=
- Substitute the value of a,
and t
a =
- Convert to km/hr
=
=
Answer:
The greatest mass, that the weight C can have such that block A does not move is approximately 23.259 kg
Explanation:
The given parameters are;
The mass of the block A= 58 kg
The coefficient of static friction between the block and the surface, = 0.300
The coefficient of static friction between the rope and the fixed peg, B = 0.310
Let T represent the tension in the rope
Therefore, when the rope is static, T = The normal reaction at the peg, B,
The angle of inclination of the rope holding the block A = arctan(3/4) ≈ 36.87°
The length of the rope = √(0.4² + 0.3²) = 0.5
∴ sin(θ) = 3/5 = 0.6
cos(θ) = 4/5 = 0.8
The vertical component of the tension in the rope = T × sin(θ) = 0.6·T
The horizontal component of the tension in the rope = T × cos(θ) = 0.8·T
The friction force = μ×(W - 0.6·T) = 0.300×(58×9.8 - 0.6·T) = 170.52 - 0.18·T
The block will start to move when we have;
The horizontal component of the tension in the rope = The friction force
∴ 0.8·T = 170.52 - 0.18·T
0.8·T + 0.18·T = 170.52
0.98·T = 170.52
T = 170.52/0.98 = 174
Therefore, the tension in the rope = T = 174 N = The normal reaction at the peg, B
The frictional force at the peg, =
×
= 0.310 × 174 N = 53.94 N
The weight of the mass, ,
= The frictional force at the peg,
+ The tension in the rope
∴ The weight of the mass, ,
= 53.94 N + 174 N = 227.94 N
Weight, W = Mass, m × The acceleration due to gravity, g, from which we have;
m = W/g
Where;
g = 9.8 m/s²
∴ =
/g = 227.94 N/(9.8 m/s²) ≈ 23.259 kg.
The greatest mass, that the weight C can have such that block A does not move = ≈ 23.259 kg.