Answer:
W ≅ 292.97 J
Explanation:
1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)
Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;
W = (Fcosθ)d
Given that:
Tension of the force = 62 N
angle of incline θ = 34°
distance d =5.7 m.
Then;
W = 62 × cos(34) × 5.7
W = 353.4 cos(34)
W = 353.4 × 0.8290
W = 292.9686 J
W ≅ 292.97 J
Hence, the work done by tension before the block goes up the incline = 292.97 J
The time taken by the pulse to travel from one support to the other is 0.208 s.
<h3>Given:</h3>
The mass of the cord is m = 0.65 kg.
The distance between the supports is, d = 8.0 m.
The tension in the cord is T = 120 N.
The time taken by the pulse to travel from one support to the other is given as,
Here, v is the linear velocity of a pulse. Its value is,
Then,
Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.
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Answer:
procedure in which radio waves and a powerful magnet linked to a computer
Explanation:
Listen to pronunciation. (mag-NEH-tik REH-zuh-nunts IH-muh-jing) A procedure in which radio waves and a powerful magnet linked to a computer are used to create detailed pictures of areas inside the body. These pictures can show the difference between normal and diseased tissue.
Answer:
P = 1.45 hp or 1.94kW
Explanation:
Given:
v = 2.5m/s
uk = 0.06
m = 60kg
Fk = uk*m*g*fsin(15)
Fk = 0.06*85*9.81*60*sin(15)
Fk = 776.15 N
Power=force*speed
P = F*v
P = 776.15N*2.5m/s
P = 1940.36 W
1 Horsepower = 0.7457 Kilowatts
P = 1.45 hp
