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1 month ago

Sketch a position-time graph for a bear starting

Physics

1 answer:

Sav [3.1K]1 month ago

3 0

Hopefully this clarifies things. The position remains unchanged over the span of 5 seconds, which indicates that it is at rest while time continues to pass. When the slope turns negative, it illustrates that the bear is moving backward, thus changing its direction.

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Richard needs to fly from san diego to halifax, nova scotia and back in order to give an important talk about mathematics. on th

Ostrovityanka [3204]

As the plane heads toward Halifax, the wind speed supports the flight path

resulting in an overall improved speed

Conversely, during the return trip, the wind will resist the plane's motion, decreasing the net speed

The total journey lasts 13 hours

of which 2 hours was dedicated to the mathematics discussion

Consequently, the total flight time is 13 - 2 = 11 hours

Now we apply the formula to calculate the time for traveling to Halifax

$t_1 = \frac{d}{v + 50}$

Time needed to return

$t_2 = \frac{d}{v - 50}$

Let’s look at the total time

$T = t_1 + t_2$

$11 = \frac{d}{v - 50} + \frac{d}{v + 50}$

Here d = 3000 miles

$11 = \frac{3000}{v - 50} + \frac{3000}{v + 50}$

$3.67 * 10^{-3} = \frac{2v}{v^2 - 2500}$

$v^2 - 2500 = 545.45v$

By solving the derived quadratic equation

$v = 550 mph$

the plane's speed calculates to 550 mph

3 0

2 months ago

Read 2 more answers

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

Softa [3030]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change $(\Delta h)=-50 m$

initial speed $(v_1)=15 m/s$

Final Speed $(v_2)=30 m/s$

Specific internal Energy $(\Delta U)=-5 kJ/kg$

according to the first Law of thermodynamics

$Q-W=\Delta H$

$\Delta H=\Delta KE+ \Delta PE +\Delta U$

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

$\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U$

$Q=W+\Delta KE+ \Delta PE +\Delta U$

$Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10$

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

5 0

3 months ago

A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

kicyunya [3294]

Answer:

a) Q = C*emf

b) Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let $V_{R}$ represent the Voltage across the Resistor

$V_{c}$ and

represent the Voltage across the capacitor

Implementing KVL;

$emf - V_{R} - V_{c} = 0\\$

.........................(1)

Since this is a series connection, the same current traverses through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Integrating $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$

Substituting the values of t and I₀ into the aforementioned formula for I

$I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}$

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

$Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf$

4 0

2 months ago

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

serg [3582]

Response:

(a) $W=-19.25J$

(b) $W=-52.8J$

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$