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MariettaO [177]

2 years ago

7

The diagram shows a motion map. A long black arrow pointed right labeled x. Above that are 4 green arrows, each shorter ending i

n one dot. Above the green arrows are 2 yellow arrows pointing left. Farther up is 3 green arrow pointed left, each one longer. Above the green arrows are 3 small yellow arrows pointing left. Which best describes the motion of the object between 1 and 4 seconds? The object has decreasing acceleration and increasing velocity. The object has positive acceleration and eventually stops. The object has decreasing acceleration and decreasing velocity. The object has negative acceleration and eventually stops.

1 answer:

never [62]2 years ago

7 0

Answer:

Is D.the object is negative

Explanation:

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Li is riding her bicycle at 8.0 m/s. She slows down to 4.0 m/s. Her change in velocity is m/s. If Li takes 2 seconds to make thi

forsale [732]

You will have to use this formula:
$v = vo + a \times t$

Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs

Then:

-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2

Ps: It's value is negative because the she was in retrograde motion.

Answer: Her acceleration is -2 m/s^2.

4 0

2 years ago

Read 2 more answers

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

2 years ago

Two resistors ( 3 ohms & 6 ohms) in a series circuit with a power supply = 12 volts. The current through resistor 6 ohms is

Scorpion4ik [409]

In a series circuit . . .

-- The total resistance is the sum of the individual resistors.

-- The current is the same at every point in the circuit.

The total resistance in this circuit is (3Ω + 6Ω ) = 9Ω

The current at every point is (V/R) = (12v / 9Ω ) = <em>1.33 A</em> .

Pick choice<em> (a)</em>.

6 0

1 year ago

One component of a metal sculpture consists of a solid cube with an edge of length 38.9 cm. The alloy used to make the cube has

vovangra [49]

Answer:

The mass of the cube is 420.8 kg.

Explanation:

Given that,

Length of edge = 38.9 cm

Density $\rho= 7.15 \times10^{3}\ kg/m^3$

We need to calculate the volume of cube

Using formula of volume

$V = 38.9^3$

$V=0.058863\ m^3$

We need to calculate the mass of the cube

Using formula of density

$\rho = \dfrac{m}{V}$

$m = V\times\rho$

$m =0.058863\times7.15 \times10^{3}$

$m=420.8\ kg$

Hence, The mass of the cube is 420.8 kg.

7 0

2 years ago

The engine on a fighter airplane can exert a force of 105,840 N (24,000 pounds). The take-off mass of the plane is 16,875 kg. (I

FrozenT [24]

Answer:

The acceleration you can get with that engine in your car is around 70,56 $(\frac{m}{s^{2} })$ or 7,26 $(\frac{ft}{s^{2} } )$ using 1500kg of mass or 3306 pounds

Explanation:

Using the equation of the force that is:

$F=m*a$

So, you notice that you know the force that give the engine, so changing the equation and using a mass of a car in 1500 kg or 3306 pounds

$a=\frac{F}{m} =\frac{105840 N }{1500 (kg) }$

$a=\frac{105840 (\frac{kg*m}{s^{2} } )} {1500 kg }$

<em> Note: N or Newton units are: $\frac{kg * m}{s^{2} }$ </em>

$a= 70,54 \frac{m}{s^{2} }$

Also in pounds you can compared

$a= \frac{2400 lf }{3 306 lf}$

Note: lf in force units are: $\frac{lf*ft}{s^{2} }$

$a=7,26 \frac{ft}{s^{2} }$

8 0

2 years ago