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Kryger [21]
3 years ago
7

How?? anyone?!...........

Physics
1 answer:
Mariana [72]3 years ago
8 0
-- find the horizontal and vertical components of F1.

-- find the horizontal and vertical components of F2.

-- find the horizontal and vertical components of F3.

-- add up the 3 horizontal components; their sum is the horizontal component of the resultant.

-- add up the 3 vertical components; their sum is the vertical component of the resultant.

-- the magnitude of the resultant is the square root of (vertical component^2 + horizontal component^2)

-- the direction of the resultant is the angle whose tangent is (vertical component/horizontal component), starting from the positive x-direction.
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A person takes a trip, driving with a constant speed of 94.5 km/h except for a 22.0 min rest stop. If the person's average speed
qwelly [4]
From the average speed you can fix an equation:

Average speed = distance / time

You know the average speed = 65.1 kg / h, then

65.1 = distance / total time,

where total time is the time traveling plus 22.0 minutes

Call t the time treavelling and pass 22 minutes to hours:

65.1 = distance / [t + 22/60] ==> distance = [t + 22/60]*65.1

 
From the constant speed, you can fix a second equation

Constant speed = distance / time traveling

94.5 = distance / t ==> distance = 94.5 * t

The distance is the same in both equations, then you have:

[t +22/60] * 65.1 = 94.5 t

Now you can solve for t.

65.1t + 22*65.1/60 = 94.5t

94.5t - 65.1t = 22*65.1/60

29.4t = 23.87

t = 23.87 / 29.4

t  = 0.812 hours

distance = 94.5 km/h * 0.812 h = 76.7 km

Answers: 1) 0.81 hours, 2) 76.7 km


4 0
4 years ago
Janelle is exploring the relationship between the brightness of a light bulb and the current that powers it. When applying these
Vlada [557]

Answer:B

Explanation:

Y is the brightest bulb

7 0
3 years ago
Kate is riding on a train and notices that the wildflowers by the side of the tracks seem to be moving by much faster than the m
madreJ [45]

Answer: motion parallax

Explanation:

Motion parallax refers to a form of depth perception whereby objects that are closer to an individual appears to move at a faster speed than the objects that are far.

Therefore, Kate is riding on a train and notices that the wildflowers by the side of the tracks seem to be moving by much faster than the mountains in the distance is an example of motion parallax.

6 0
3 years ago
Beth, a construction worker, attempts to pull a stake out of the ground by pulling on a rope that is attached to the stake. The
AnnyKZ [126]

Answer:

Fy=107.2 N

Explanation:

Conceptual analysis

For a right triangle :

sinβ = y/h formula (1)

cosβ = x/h formula (2)

x: side adjacent to the β angle

y:  opposite side of the β angle

h: hypotenuse

Known data

h = T = 153.8 N : rope tension

β= 44.2°with the horizontal (x)

Problem development

We apply the formula (1) to calculate Ty : vertical component of the rope force.

sin44.2° = Ty/153.8 N

Ty = (153.8 N ) *(sen44.2°)= 107.2 N  directed down

for equilibrium system

Fy= Ty=107.2 N

Fy=107.2 N upward component of the force acting on the stake

8 0
3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

5 0
4 years ago
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