PHYSICS COLLEGE

A rocket is attached to a toy car that is confined to move in the x-direction ONLY. At time to = 0 s, the car is not moving but the rocket is lit, so the toy car accelerates in the +x-direction at 5.35 m/s2. At t; = 3.60 s, the rocket's fuel is used up, and the toy car begins to slow down at a rate of 1.95 m/s2 because of friction. A very particular physics professor wants the average velocity for the entire trip of the toy car to be +6.50 m/s. In order to make this happen, the physics professor plans to push the car (immediately after it comes to rest by friction) with a constant velocity for 4.50 sec. What displacement must the physics professor give the car (immediately after it comes to rest by friction) in order for its average velocity to be +6.50 m/s for its entire trip (measured from the time the rocket is lit to the time the physics professor stops pushing the car)?

Answers

Answer 1

Answer:

Answer:

What displacement must the physics professor give the car

= 12.91 METERS

Explanation:

Check the attached file for explanation

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The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensity if the frequency is increased to 2.20 kHz while a constant displacement amplitude is maintained.(b) Calculate the intensity if the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled.

Answers

Final answer:

The intensity of sounds is dependent on the square of the amplitude, not the frequency. Therefore, the intensity of sound remains the same when frequency is altered but the amplitude is constant. When the amplitude is quadrupled, the intensity of the sound becomes sixteen times greater.

Explanation:

In the field of physics, the intensity of a sound wave is defined as the power per unit area carried by the wave. This question involves calculating the change in sound wave intensity when the frequency and displacement amplitude of the source are altered.

(a) When the frequency is increased to 2.20 kHz while keeping the displacement amplitude constant, the intensity does not change, as the intensity in this case is not dependent on the frequency but on the square of the amplitude. Therefore, the intensity remains 0.750 W/m2.

(b) When the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled, the intensity changes. Since the intensity of a sound wave is proportional to the square of the amplitude, by quadrupling the amplitude, the intensity will become 16 times greater (since 4 squared is 16). Hence, the new intensity will be 16 * 0.750 = 12 W/m2.

Learn more about Sound Wave Intensity here:

brainly.com/question/31851162

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a. A nucleus is made up of protons and neutrons. Protons have positive charges and neutrons have no charge. The strong nuclear force holds the nucleus together because it acts against another force inside the nucleus. What force is the strong nuclear force counteracting?

Answers

Answer:

The electromagnetic force tends to push the protons apart. (Like forces repel).

Explanation:

Answer:its the electromagnetic force

Explanation:

Protons are positive so repel themselves

A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?

Answers

Answer:

The no. of electrons is $7.59* 10^(21)$

Solution:

According to the question:

The rate at which the charge is delivered is given by:

$(dQ)/(dt) = - 0.75$

Now,

$\int_(0)^(Q)dQ = - 0.75\int_(0)^(27 min) dt$

$Q = -0.75t|_(0)^(27 min)$

$Q= -0.75* 27* 60 = - 1215 C$

No. of electrons, n can be calculated from the following relation:

Q = ne

where

e = electronic charge = $1.6* 10^(- 19) C$

Thus

$n = (Q)/(e)$

$n= (1215)/(1.6* 10^(- 19))$

$n = 7.59* 10^(21)$

Nichrome wire, often used for heating elements, has resistivity of 1.0 × 10-6 Ω ∙ m at room temperature. What length of No. 30 wire (of diameter 0.250 mm) is needed to wind a resistor that has 50 ohms at room temperature?

Answers

Answer:

Length = 2.453 m

Explanation:

Given:

Resistivity of the wire (ρ) = 1 × 10⁻⁶ Ω-m

Diameter of the wire (d) = 0.250 mm = 0.250 × 10⁻³ m

Resistance of the wire (R) = 50 Ω

Length of the wire (L) = ?

The area of cross section is given as:

$A=(1)/(4)\pi d^2\n\nA=(1)/(4)*\ 3.14* (0.250* 10^(-3))^2\n\nA=0.785* 6.25* 10^(-8)\n\nA=4.906* 10^(-8)\ m^2$

We know that, for a constant temperature, the resistance of a wire is directly proportional to its length and inversely proportional to its area of cross section. The constant of proportionality is called the resistivity of the wire. Therefore,

$R=\rho (L)/(A)$

Expressing the above in terms of length 'L', we get:

$L=(RA)/(\rho)$

Plug in the given values and solve for 'L'. This gives,

$L=(50* 4.906* 10^(-8))/(1* 10^(-6))\ m\n\nL=(2.453)/(1)=2.453\ m$

Therefore, length of No. 30 wire (of diameter 0.250 mm) is 2.453 m.

What is an atomic nucleus

Answers

Answer:

The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.

Explanation:

The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which data set represents constant acceleration?

Answers

Answer:

The first graph is showing the constant acceleration (1 m/s)

Explanation:

The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4

The last graph is showing constant velocity therefore there is no acceleration (a = 0)

Answer:

Explanation:

on edge

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