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AlladinOne [14]
2 years ago
14

Can someone help label these?

Physics
1 answer:
seropon [69]2 years ago
7 0
A. reactants
B. subscript
C. coefficient
D. products
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A force of 50 newtons causes a sled to accelerate at a rate of 5 meters per second. What is the mass of the sled.
notka56 [123]
F=ma
50=m(5)
m=10kg
hence,ans is B
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3 years ago
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Formula for percentage error​
GarryVolchara [31]

Answer:

PE = (|accepted value – experimental value| \ accepted value) x 100%

Explanation:

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2 years ago
Which statement is true according to Newton's first law of motion?
katrin2010 [14]
C) In the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.

hope this helps and have a great day :)







8 0
3 years ago
Read 2 more answers
A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A
Y_Kistochka [10]

Answer:

The tensions in T_{BC} is approximately 4,934.2 lb and the tension in T_{BD} is approximately  6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = T_{BC}

The tension in rope segment BD = T_{BD}

The tension in rope segment AB = T_{AB} = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = T_{AB} = 1200 lb

T_{BC} × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

T_{BC} × sin(26.0°) + T_{BD} × sin(21.0°) = 0...........................(2)

Which gives;

T_{BC} × sin(26.0°) = - T_{BD} × sin(21.0°)

T_{BC} = - T_{BD} × sin(21.0°)/(sin(26.0°))  ≈ - T_{BD} × 0.8175

Substituting the value of, T_{BC}, in equation (1), gives;

- T_{BD} × 0.8175 × cos(26.0°) + T_{BD} × cos(21.0°) = 1200 lb

- T_{BD} × 0.7348  + T_{BD} ×0.9336 = 1200 lb

T_{BD} ×0.1988 = 1200 lb

T_{BD} ≈ 1200 lb/0.1988 = 6,035.6938 lb

T_{BD} ≈ 6,035.6938 lb

T_{BC} ≈ - T_{BD} × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

T_{BC} ≈ -4934.1733 lb

From which we have;

The tensions in T_{BC} ≈ -4934.2 lb and  T_{BD} ≈ 6,035.7 lb.

8 0
3 years ago
Perform the following calculations and give your answer with the correct number of significant figures
love history [14]

Answer:

see below

Explanation:

a. 0.1886 x 12 =2.2632

This has 2 sig figures so the answer can only have 2 sig figures

2.3

b. 2.995 - 0.16685 =2.82815

The most accurate in the problem is to thousands place so our answer can only be accurate to the thousands place

2.828

c. 910 x 0.18945=172.3995

The  least number of significant figures is 3 so the answer can only have 3 significant figures

172

3 0
3 years ago
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