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Advocard [28]
3 years ago
12

(3.) A tourist travels 320 km in 3.6 h. What is her average speed for the trip?

Physics
1 answer:
solmaris [256]3 years ago
5 0

Answer:

The average speed of the tourist is 88.89 km/h

Explanation:

Given;

distance of travel, d = 320 km

time of travel, t = 3.6 h

Average speed is given as;

average speed = distance / time

average speed = 320 km / 3.6 h

average speed = 88.89 km/ h

Therefore, the average speed of the tourist is 88.89 km/h

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Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acti
Tems11 [23]

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

\dfrac{dF_{net}}{df}=0

We need to calculate the unknown force

Using formula of net force

\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}

Put the value into the formula

\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}

\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}

The magnitude of net force,

F_{net}=\sqrt{F_{x}^2+F_{y}^2}

F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}

F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}

F_{net}=\sqrt{F^2+4899.78+36.232F}

On differentiating w.r.to F

(\dfrac{dF_{net}}{dF})^2=2F+36.232

0=2F+36.232

F=-\dfrac{36.232}{2}

F=-18.116\ lb

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.

6 0
3 years ago
The Large Hadron collider (LHC) is a huge piece of equipment designed and built in order to make new scientific discoveries. The
Ket [755]

Answer:

D

Explanation:

"The Large Hadron collider (LHC) is a <em>huge piece of equipment designed and built in order to make new scientific discoveries.</em>"

3 0
3 years ago
Two long, parallel transmission lines, 40.0cm apart, carry 25.0-A and 73.0-A currents.A). Find all locations where the net magne
In-s [12.5K]

Answer:

a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm

That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.

b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm

That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.

Explanation:

The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m

The magnetic field in a current carrying wire at a distance r from the wire is given by

B = (μ₀I/2πr)

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.

According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(0.4-x)]

(73/x) = [25/(0.4-x)]

73(0.4-x) = 25x

29.2 - 73x = 25x

73x + 25x = 29.2

98x = 29.2

x = (29.2/98) = 0.298 m

b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?

According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(x-0.4)]

(73/x) = [25/(x-0.4)]

73(x-0.4) = 25x

73x - 29.2 = 25x

73x - 25x = 29.2

48x = 29.2

x = (29.2/48) = 0.608 m

Hope this Helps!!!

5 0
4 years ago
I WILL MARK YOU THE BRAINLIEST
mr Goodwill [35]

Explanation:

700N right

to get the net force

you gotta let one direction be the negative ( the smaller force)

so the total force towards the left is 100N ( 60 + 40= 100)

which is smaller than the right force which is 800 N so you let 100 N be negative

so without even calculating , you can know that it will be moving towards the right because right force > left force

your add both forces ( remember 100 N is negative)

so 800N + ( - 100N)

= 700N

towards the right

hope this helps

this is just one method that helped me understand

please mark it brainliest

3 0
3 years ago
Read 2 more answers
The last is 80 meter per second,please in am in exam​
inna [77]

Answer:

  1. I have fond the answer

Explanation:

but my camera doesn't work

7 0
3 years ago
Read 2 more answers
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