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Elena L [17]
1 year ago
12

Which of these is the most effective way for Leanna to cool down after an intense bike ride

Physics
1 answer:
Sonja [21]1 year ago
6 0
I am pretty sure the answer would be too stretch
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A quantity y is to be determined from the equation y=(px)/q^2
ki77a [65]

Answer:

heya answer option b

Explanation:

please mark me brainliest

4 0
1 year ago
Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

 They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole

            - T_{y} L + W₂ L + W₁ L / 2 = 0

            T_{y} = W₂ + W₁ / 2

            T_{y} = 120 + 150/2

            T_{y} = 195 lb

we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

             T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

4 0
2 years ago
A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo
padilas [110]

Answer:

 a)    λ = 189.43 10⁻⁹ m  b)    λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

      d sin θ= m λ

Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

        tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

      tan θ = sin θ / cos θ = sin θ

We replace

      d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

     d = λ  L / y

     d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

     d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

     λ  = d y / L

    λ  = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

    λ = 189.43 10⁻⁹ m

Part B

The compound wavelength B

      λ  = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

      λ = 269.19 10⁻⁹ m

4 0
2 years ago
Aaron Agin nodded off while driving home from play practice this past Sunday evening. His 1500-kg car hit a series of guardrails
Inessa [10]

Answer: 6.48m/s

Explanation:

First, we know that Impulse = change in momentum

Initial velocity, u = 19.8m/s

Let,

Velocity after first collision = x m/s

Velocity after second collision = y m/s

Also, we know that

Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).

5700 = 1500(19.8 - x)

5700 = 29700 - 1500x

1500x = 29700 - 5700

1500x = 24000

x = 24000/1500

x = 16m/s

Also, at the second guard rail. impulse = ft, so that

Impulse = 79000 * 0.12

Impulse = 9480

This makes us have

Impulse = m(x - y)

9480 = 1500(16 -y)

9480 = 24000 - 1500y

1500y = 24000 - 9480

1500y = 14520

y = 14520 / 1500

y = 9.68

Then, the velocity decreases by 3.2, so that the final velocity of the car is

9.68 - 3.2 = 6.48m/s

5 0
1 year ago
A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the
Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = \frac{\pi d^{2} }{4}

where d is the diameter of the wire

1.613 x 10^-7 = \frac{3.142* d^{2} }{4}

6.448 x 10^-7 = 3.142 x d^{2}

d^{2} =\sqrt{ 2.05*10^-7}

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0
1 year ago
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