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Nastasia [14]
3 years ago
12

How is a vector represented in symbol form?

Physics
1 answer:
VikaD [51]3 years ago
6 0
I believe it’s just a “v” with an arrow above it.
You might be interested in
A cylindrical wire has a length of 2.80 m and a radius of 1.03 mm. It carries a current of current of 1.35 A, when a a voltage o
aleksley [76]

Answer:

0.023 Ohms

Explanation:

Given data

Length= 2.8m

radius= 1.03mm

current I= 1.35 A

voltage V= 0.032V

We know that from Ohm's law

V= IR

Now  R= V/I

Substitute

R= 0.032/1.35

R= 0.023 Ohms

Hence the resistance is 0.023 Ohms

5 0
3 years ago
En un m.A.S. La amplitud tiene un valor de 10 centimetros y el periodo es de 2 segundos calcular el valor de la velocidad de 0.8
Ivanshal [37]

Answer:

v1=18.46m/s

v2=29.8cm/s

Explanation:

We know that

A=10cm\\T=2s

the equation of the motion is

x=Acos(\omega t)\\

we can calculate w by using

\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi

Hence, we have that

x=10cm*cos(\pi t)\\

the speed will be

v=-\omega*Asin(\omega t)\\|v(0.8)|=|\pi*10cm*sin(\pi *0.8)|=18.46\frac{cm}{s}\\|v(1.4)|=|\pi*10cm*sin(\pi *1.4)|=29.8\frac{cm}{s}

hope this helps!

6 0
3 years ago
A current of 2.0A flows through a light bulb. What is the amount of charge flowing through the light bulb in 40 seconds?
lidiya [134]

Answer:

Quantity of charge = 80 Coulombs

Explanation:

Given the following data;

Current = 2 A

Time = 40 seconds

To find the amount of charge flowing through the light bulb;

Mathematically, the quantity of charge passing through a conductor is given by the formula;

Quantity of charge = current * time

Substituting into the formula, we have;

Quantity of charge = 2 * 40

Quantity of charge = 80 Coulombs

5 0
2 years ago
2. A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.5 . At 30.0 s after blast
harina [27]

Answer:

a)The highest point reached by the rocket is 1412 m

b)The rocket crashes after 54.7 s

Explanation:

Hi there!

The equations of height and velocity of the rocket are the following:

h = h0 + v0 · t + 1/2 · a · t² (while the engines work).

h = h0 + v0 · t + 1/2 · g · t² (when the rocket is in free fall).

v = v0 + a · t (while the engines work).

v = v0 + g · t (when the rocket is in free fall).

Where:

h = height of the rocket at a time t.

h0 = initial height of the rocket.

v0 = initial velocity.

t = time.

a = acceleration due to the engines.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the rocket at a time t.

First, let's find the velocity and height reached by the rocket until the engines fail:

h = h0 + v0 · t + 1/2 · a · t²

Let's set the origin of the frame of reference at the launching point so that h0 = 0. Since the rocket starts from rest, v0 = 0. So after 30.0 s the height of the rocket will be:

h = 1/2 · a · t²

h = 1/2 · 2.5 m/s² · (30.0 s)²

h = 1125 m

Now let's find the velocity of the rocket at t = 30.0 s:

v = v0 + a · t (v0 = 0)

v = 2.5 m/s² · 30.0 s

v = 75 m/s

After 30.0s the rocket will continue to ascend with a velocity of 75 m/s. This velocity will be gradually reduced due to the acceleration of gravity. When the velocity is zero, the rocket will start to fall. At that time, the rocket is at its maximum height. So, let's find the time at which the velocity of the rocket is zero:

v = v0 + g · t

0 = 75 m/s - 9.8 m/s² · t (v0 = 75 m/s because the rocket begins its free-fall motion with that velocity).

-75 m/s / -9.8 m/s² = t

t = 7.7 s

Now, let's find the height of the rocket 7.7 s after the engines fail:

h = h0 + v0 · t + 1/2 · g · t²

The rocket begins its free fall at a height of 1125 m and with a velocity 75 m/s, then, h0 = 1125 m and v0 = 75 m/s:

h = 1125 m + 75 m/s · 7.7 s - 1/2 · 9.8 m/s² · (7.7 s)²

h = 1412 m

The highest point reached by the rocket is 1412 m

b) Now, let's calculate how much time it takes the rocket to reach a height of zero (i.e. to crash) from a height of 1412 m.

h = h0 + v0 · t + 1/2 · g · t² (v0 = 0 because at the maximum height the velocity is zero)

0 = 1412 m - 1/2 · 9.8 m/s² · t²

-1412 m / -4.9 m/s² = t²

t = 17 s

The rocket goes up for 30.0 s with an acceleration of 2.5 m/s².

Then, it goes up for 7.7 s with an acceleration of -9.8 m/s².

Finally, the rocket falls for 17 s with an acceleration of -9.8 m/s²

The rocket crashes after (30.0 s + 7.7 s + 17 s) 54.7 s

6 0
3 years ago
. Which of the following is a vector describing how fast an object moves over a long time interval?. . average speed. instantane
gulaghasi [49]
"Average velocity" is the vector among the choices given in the question that describes <span>how fast an object moves over a long time interval. The correct option among all the options that are given in the question is the last option or the fourth option. I hope the answer has helped you.</span>
8 0
3 years ago
Read 2 more answers
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