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3 years ago

A 100.0-kg bakery sign hangs from two thin cables as shown.

Physics

1 answer:

harina [27]3 years ago

5 0

Answer:

1. T₁ = 500 N

2. T₂ = 866 N

Explanation:

Please see attached photo for the diagram.

Thus, we can obtain obtained the value of T₁ and T₂ as follow:

1. Determination of T₁

Angle θ = 30

Hypothenus = 100 kg

Opposite = T₁ =?

Sine θ = Opposite /Hypothenus

Sine 30 = T₁ / 100

Cross multiply

T₁ = 100 × Sine 30

T₁ = 100 × 0.5

T₁ = 50 Kg

Multiply by 10 to express in Newton

T₁ = 50 × 10

T₁ = 500 N

2. Determination of T₂

Angle θ = 60

Hypothenus = 100 kg

Opposite = T₂ = ?

Sine θ = Opposite /Hypothenus

Sine 60 = T₂ / 100

Cross multiply

T₂ = 100 × Sine 60

T₂ = 100 × 0.8660

T₂ = 86.6 Kg

Multiply by 10 to express in Newton

T₂ = 86.6 × 10

T₂ = 866 N

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This illustration shows two opposing forces pulling on a wagon. Which description best describes how the wagon will move?

Talja [164]

Answer:

The wagon will move to the right.

Explanation:

From the question given above, the following data were obtained:

Force applied to the left (Fₗ) = 10 N

Force applied to the right (Fᵣ) = 30 N

Direction of the wagon =.?

To determine the direction in which the wagon will move, we shall determine the net force acting on the wagon. This can be obtained as follow:

Force applied to the left (Fₗ) = 10 N

Force applied to the right (Fᵣ) = 30 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 30 – 10

Fₙ = 20 N to the right

From the calculations made above, the net force acting on the wagon is 20 N to the right. Hence the wagon will move to the right.

8 0

2 years ago

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr

slava [35]

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

W = $F_{total} .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} .d =\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}$

$F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_{sprinter} = F_{total} + F_{wind} = 39.7 + 30 = 69.68 N$

7 0

3 years ago

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An electromagnetic wave is propagating in the positive x direction. at a given moment in time, the magnetic field at the origin

antiseptic1488 [7]

Since in an electromagnetic wave the electric and magnetic fields are perpendicular to each other and perpendicular to the direction of motion, the electric field has to point in the z direction.

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3 years ago

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A spherical drop of water carrying a charge of 42 pC has a potential of 620 V at its surface (with V = 0 at infinity). (a) What

iren [92.7K]

Answer:

0.0006091222 m

Explanation:

q = Charge = 42 pC

V = Voltage = 620 V

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Electric potential is given by (at r = R)

$V=\dfrac{q}{4\pi\epsilon R}\\\Rightarrow R=\dfrac{q}{4\pi\epsilon V}\\\Rightarrow R=\dfrac{42\times 10^{-12}}{4\pi\times 8.85\times 10^{-12}\times 620}\\\Rightarrow R=0.0006091222\ m$

The radius of the drop is 0.0006091222 m

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3 years ago

What increases your ability to see at night

PtichkaEL [24]

The answer would be increased number <span />of rod cells.Hope this helps :)

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3 years ago