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Mashcka [7]
3 years ago
7

Sam is moving house and is carrying a 300N box of books up a flight of steps 5m high, it takes her 30 seconds. Gary follows her

carrying a bag of clothes doing 1000 J of work; it only takes him 25 seconds. Who provides the biggest power? Show your working.
Physics
1 answer:
ioda3 years ago
7 0

Answer: Sam provides the biggest power

Explanation: <u>Power</u> is defined, in Physics, as the rate of work done. <u>Work</u> is energy transferred to an object due to the force causing the displacement of the object.

So, to determine Power:

For Sam:

P = \frac{work}{time}

P=\frac{F.d}{t}

P=\frac{300.5}{30}

P = 50 Watts

The unit for Power is [P] = J/s = Watt

For Gary:

P = \frac{work}{time}

P=\frac{1000}{25}

P = 40 Watts

Comparing power of Sam and Gary, we can conclude that Sam provides the biggest power of 50 Watts.

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Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 30
dybincka [34]

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

<em><u>Support Cy:</u></em>

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

<em><u>Support Ay:</u></em>

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

4 0
3 years ago
Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at
Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution  

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0
3 years ago
Kim’s experiment showed that chicken eggshells were stronger when she gave the hen feed, to which extra calcium had been added.
astraxan [27]

The step of this scientific method is "draw conclusion".

<h3>What is scientific method?</h3>

The scientific method is the process of objectively establishing facts through testing and experimentation.

The basic scientific process or scientific method involves the following;

  • making an observation
  • forming a hypothesis
  • making a prediction
  • conducting an experiment
  • analyzing the results and
  • drawing conclusion.

Thus, if Kim’s experiment showed that chicken eggshells were stronger when she gave the hen feed, to which extra calcium had been added. The step of this scientific method is "draw conclusion".

Learn more about scientific method here: brainly.com/question/17216882

#SPJ1

6 0
2 years ago
It was a children versus grown-ups competition at school. One event required the adult to throw a basketball as far as he could.
Cerrena [4.2K]
D Because If Your Going To Have A Contest Its Ganna Have To Be The Same Objectives For Both Contenders 
3 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
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