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yuradex [85]
3 years ago
8

All of the following objects are moving at the same speed. which one bas the greatest kinetic energy? (remember kinetic energy d

epends on mass and speed)
a. car


b. semi tractor-trailer truck


c. motorcycle


d. bicycle
Physics
1 answer:
melomori [17]3 years ago
6 0

B, the semi-tractor trailer truck, since it has the most mass.

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Please answer me!!! asappp​
Anastasy [175]

Answer:

omg this is hard

Explanation:

Yes it is lol

5 0
2 years ago
Help me yall it due in a few minutes :((()
vaieri [72.5K]

Answer:

B. blocks 2 & 3.

Explanation:

Block 1 has equal & opposite forces acting on it.

Block 2 has 5N on one side, 3N on the other. It will move in the direction the 5N of force is pushing.

Block 3 has no opposing force.

6 0
3 years ago
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
Refrigerant-134a enters a compressor at 180 kPa as a saturated vapor with a flow rate of 0.35 m3/min and leaves at 900 kPa. The
Alexus [3.1K]

Answer:

52.5°C

Explanation:

The final enthalpy is determined from energy balance where initial enthalpy and specific volume are obtained from A-12 for the given pressure and state

mh1 + W = mh2

h2 = h1 + W/m

h1 + Wα1/V1

242.9 kJ/kg + 2.35.0.11049kJ/ 0.35/60kg

=287.4 kJ/kg

From the final enthalpy and pressure the final temperature is obtained A-13 using interpolation

i.e T2 = T1 + T2 -T1/h2 -h1(h2 - h1)

= 50°C + 60 - 50/295.15 - 284.79

(287.4 - 284.79)°C

= 52.5°C

7 0
3 years ago
What are the differently types of motion in a bicycle?
Ksivusya [100]
The two types of motion exerted in bicycle are:
1. rotary motion
2. linear motion
7 0
3 years ago
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