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Tanzania [10]
3 years ago
11

Which list contains only vector quantities?

Physics
1 answer:
RUDIKE [14]3 years ago
5 0
The answer is D
These are all the vector quantities

Displacement, velocity, acceleration, force, and torque are vectors
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The answer would be B!
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Answer:

acceleration

Explanation:

Even though the initial and final speeds are the same, there has been a change in direction for the object

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What components make up fire​
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Oxygen, heat, and fuel

Explanation:

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A bulldozer attempts to drag a log weighing 500 N along the rough horizontal ground. The cable attached to the log makes an angl
Gemiola [76]

Answer:

T= 224.01 N

Explanation:

in imminent motion we have to :

  • The frictional force reaches its maximum value
  • The system is in balance of forces

Data

W=  500 N :  weight of the log

μs = 0.5

μk = 0.35

α = 30°above the ground :  angle of the cable attached to the log

Newton's first law to the log:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

Forces acting on the log

T: cable tension for impending movement

N: normal force

W : weight

f: frictional force , f= μsN

We apply the formula (1)

∑Fx=0

Tx-f = 0

Tcosα-μsN=0

Tcos30°-0.5N=0 Equation (1)

∑Fy=0

N+Ty-W=0

N+Tsin30°-500=0

N= 500-Tsin30°  Equation (2)

We replace the value of N of the Equation  (2) in the equation (1)

Tcos30°-0.5(500-Tsin30°) = 0

Tcos30°+0.5Tsin30° = 0.5*500

T( cos30°+0.5*sin30°) = 250

(1.116) T = 250

T= 250/1.116

T= 224.01 N

6 0
3 years ago
The strings in a compound bow behave approximately like a
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The speed at which the arrow would be launched is 133.42 m/s

The work-energy theorem asserts that the net work done applied by the forces on a particular object is equivalent to the change in its kinetic energy.

The equation for the work-energy theorem can be computed as:

\mathbf{W =\Delta K.E}

\mathbf{F\Delta x =\dfrac{1}{2} mv^2}

where;

  • Force (F) = 267 N
  • distance Δx = 0.60 m
  • mass (m) = 18 g
  • speed (v) = ???

From the above equation, let make speed(v) the subject of the formula:

∴

\mathbf{v = \sqrt{\dfrac{2(F \Delta x)}{m}} }

\mathbf{v = \sqrt{\dfrac{2(267 \times 0.60)}{0.018}} }

v = 133.42 m/s

Learn more about the work-energy theorem here:

brainly.com/question/17081653

7 0
2 years ago
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