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Ilia_Sergeevich [38]
3 years ago
5

Define the strip attached to the magnet?​

Physics
1 answer:
Aleonysh [2.5K]3 years ago
5 0

Answer:

A stripe of magnetic information that is affixed to the back of a plastic credit or debit card.

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Can Vehicles us bike lanes when they are empty?
Allushta [10]
No it is unsafe. Even if they are empty they cannot
7 0
2 years ago
Three sticks are arranged to form a right triangle. If the lengths of the three sticks are 0.47 m, 0.62 m and 0.78 m, what are t
Svetradugi [14.3K]

The angles in the triangle are 91 degrees, 53 degrees and 36 degrees respectively.

<h3>What is the cosine rule?</h3>

From the cosine rule we know that;

c^2 = a^2 + b^2 - 2abcosC

Since;

a = 0.47 m

b =  0.62 m

c =  0.78 m

Then;

(0.78)^2 = (0.47)^2 + (0.62)^2 - 2(0.47 * 0.62)cosC

0.61 = 0.22 + 0.38 - 0.58 cosC

0.61 - ( 0.22 + 0.38) = - 0.58 cosC

0.01 = - 0.58 cosC

C = cos-1(0.01/-0.58)

C = 91 degrees

Using the sine rule;

b/Sin B = c/Sin C

0.62/sinB = 0.78/sin 91

0.62/Sin B = 0.78

B = sin-1 (0.62//0.78)

B = 53 degrees

Angle A is obtained from the sum of angles in a triangle;

180 - (91 + 53)

A = 36 degrees

Learn more about triangle:brainly.com/question/2773823

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6 0
1 year ago
6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres
sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

2ah=v_f^2-v_0^2

Where:

\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}

If we assume the initial velocity to be 0 we get:

2ah=v_f^2

The acceleration is the acceleration due to gravity:

2gh=v_f^2

Now, we take the square root to both sides:

\sqrt{2gh}=v_f

Now, we substitute the values:

\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f

solving the operations:

280\frac{m}{s}=v

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

F_d=\frac{1}{2}C\rho_{air}Av^2

Where:

\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

F_d=mg

Now, we determine the mass of the raindrop using the following formula:

m=\rho_{water}V

Where:

\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}

The volume is the volume of a sphere, therefore:

m=\rho_{water}(\frac{4}{3}\pi r^3)

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)

Solving the operations:

m=1.39\times10^{-5}kg

Now, we substitute the values in the formula for the drag force:

F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})

Solving the operations:

F_d=1.36\times10^{-4}N

Now, we substitute in the formula:

1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2

Now, we solve for the velocity:

\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2

Now, we substitute the values. We will use the area of a circle:

\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2

Substituting the radius:

\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2

Solving the operations:

70.67\frac{m^2}{s^2}=v^2

Now, we take the square root to both sides:

\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\  \\ 8.4\frac{m}{s}=v \\  \end{gathered}

Therefore, the velocity is 8.4 m/s

7 0
9 months ago
Hàng ngày ta thấy Mặt trời , Mặt trăng quay quanh Trái Đất, khi đó ta đã lấy vật mốc là
marshall27 [118]
I’m sorry I don’t understand this language
5 0
2 years ago
A car traveling on a straight road at 15.0 meters per second accelerates uniformly to a speed of 21.0 meters per second in 12.0
oksano4ka [1.4K]
<h2>Option 3,  216 m is the correct answer.</h2>

Explanation:

We have initial velocity, u = 15 m/s

Time, t = 12 seconds

Final velocity, v = 21 m/s

We have equation of motion v = u + at

Substituting

                     21 = 15 + a x 12

                       a = 0.5 m/s²

Now we have equation of motion v² = u² + 2as

                           21² = 15² + 2 x 0.5 x s

                            s = 216 m

       Displacement = 216 m

Option 3,  216 m is the correct answer.

8 0
3 years ago
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