All categories

4 years ago

Question C) needs to be answered, please help (physics)

Physics

1 answer:

Zarrin [17]4 years ago

3 0

(a) Differentiate the position vector to get the velocity vector:

r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k

v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j

(b) The velocity at t = 2.00 s is

v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j

(c) Compute the electron's position at t = 2.00 s:

r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k

The electron's distance from the origin at t = 2.00 is the magnitude of this vector:

||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that

tan(θ) = (-16.0 m/s) / (3.00 m/s) ==> θ ≈ -79.4º

or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.

You might be interested in

A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor

Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

Since the speed of sound is not given in the question, we would assume it to be 340m/s

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

So, the length of the second pipe is L – L’

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

The beat frequency due to the fundamental frequencies of the first and second pipe is

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm

Hence, 1.16 cm were cut from the end of the second pipe

6 0

3 years ago

Blue bumper car, traveling east with a force of 100 N, runs into a yellow bumper car traveling west with a force of 300 N. Becau

Pani-rosa [81]

Answer:

the blue bumper car

Explanation:

because the yellow bumper car is coming in with much more force

and according to Newton, actions and reactions are equal and opposite, so the yellow bumper car will five out 300N while the blue car will give out only 100N

then it almost because a fight between 300N vs 100N

and obviously, the 300N will win and the 100N will be pushed back

4 0

3 years ago

A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise.

Lera25 [3.4K]

The true statement about the CD is:

<h3>b. No net torque acts on it at all.</h3>

$\texttt{ }$

<h3>Further explanation</h3>

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

$\texttt{ }$

Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

$\texttt{ }$

Complete Question:

A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise. Which of the following statements about the CD is true?

a. A net torque acts on it clockwise to keep it moving

b. No net torque acts on it at all.

c. A net torque acts on it counterclockwise to keep it moving

Given:

angular velocity = ω = 5.0 revolutions per second

Asked:

net torque = Στ = ?

Solution:

Constant angular velocity → angular acceleration = α = 0 rad/s²

$\Sigma \tau = I \alpha$

$\Sigma \tau = I (0)$

$\Sigma \tau = 0 \texttt{ Nm}$

$\texttt{ }$

<h3>Conclusion:</h3>

The true statement about the CD is:

b. No net torque acts on it at all.

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

$\texttt{ }$

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

#LearnWithBrainly

8 0

4 years ago

When a phase change of a substance occurs what happens to the temperature of the substance?

Maurinko [17]

C) It does not change. At the point where the phase of a substance changes, there is no temperature change but only the phase change. After this, the temperature then goes up or down depending on the type of phase change occurring.

6 0

3 years ago

A bicyclist moves along a straight line with an initial velocity vo and slows downs. Which of the following the best describes t

xeze [42]

????????????????????

3 0

3 years ago