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otez555 [7]
3 years ago
7

I will give brainliest to the best answer! I don’t understand vectors. This whole sheet is confusing me and i’m very behind, ple

ase help!!

Physics
1 answer:
Rudik [331]3 years ago
4 0

Explanation:

1. To graphically add vectors, use the tail-to-tip method.  Draw the first vector (it doesn't matter which), then draw the second vector where the first vector ends.  The resultant vector is from the tail of the first vector to the tip of the second vector.

This graph shows two ways to get the resultant: A + B or B + A.

desmos.com/calculator/bqhcclhhqc

2. To algebraically add vectors, split each vector into x and y components.

Aₓ = 5.0 cos 45 = 3.5

Aᵧ = 5.0 sin 45 = 3.5

Bₓ = 2.0 cos 180 = -2.0

Bᵧ = 5.0 sin 180 = 0

The components of the resultant vector are the sums of the components of A and B.

Cₓ = 3.5 + -2.0 = 1.5

Cᵧ = 3.5 + 0 = 3.5

The magnitude of the resultant vector is found with Pythagorean theorem, and the direction is found with tangent.

C = √(Cₓ² + Cᵧ²) ≈ 3.9 m/s

θ = atan(Cᵧ / Cₓ) ≈ 67°

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A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
I= \frac{Q}{\Delta t}
First we need to find the net charge flowing at a certain point of the wire in one second, \Delta t=1.0 s. Using I=0.92 A and re-arranging the previous equation, we find
Q=I \Delta t= (0.92 A)(1.0 s)=0.92 C

Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
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3 years ago
HELP!!!
aniked [119]

Answer:

p = m .v momentum = mass • velocity. [kg • m/s] [kg] [m/s]. Kinetic Energy. KE = 12 • m • v ... 1. A 1500 kg car traveling at 15 m/s to the south collides with a 4500 kg truck that is ... What is the final velocity of the two-vehicle mass? ... m/s. What is the velocity of the joined cars after the collision? ... 5) = (1.5x104+1.5x604) VELVE.

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3 years ago
Which example provides the most complete description of an object's motion?
cupoosta [38]

Answer:

The hiker followed a road heading north for 2 miles in 30 minutes.

Explanation:

In order to describe the motion of an object, distance covered and time taken must be required. The total path covered by an object is called the distance travelled.

The hiker followed a road heading north for 2 miles in 30 minutes. This describes the motion of hiker. The motion shows how fast the hiker is moving.  

Distance, d = 2 miles = 3218.6 m

times, t = 30 minutes = 1800 seconds

So, we can say that the hiker is moving with a speed of 1.78 m/s in north direction.

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3 years ago
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Find the frequency of a wave with the wavelength 3.5 m and the speed is 50 m/s. <br>​
Andru [333]

Answer:

8.57 Hz

Explanation:

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

Wavelength (λ) = 3.5 m

Velocity (v) = 30 m/s

Frequency (f) =?

The velocity, wavelength and frequency of a wave are related according to the equation:

Velocity = wavelength × frequency

v = λ × f

With the above formula, we can simply obtain the frequency of the wave as follow:

Wavelength (λ) = 3.5 m

Velocity (v) = 30 m/s

Frequency (f) =?

v = λ × f

30 = 3.5 × f

Divide both side by 3.5

f = 30 / 3.5

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Thus, the frequency of the wave is 8.57 Hz

7 0
3 years ago
A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

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