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nignag [31]
3 years ago
11

A cart is propelled over an xy plane with acceleration compo- nents ax 4.0 m/s2 and ay 2.0 m/s2. Its initial velocity has com- p

onents v0x 8.0 m/s and v0y 12 m/s. In unit-vector notation, what is the velocity of the cart when it reaches its greatest y coordinate?
Physics
1 answer:
Tasya [4]3 years ago
4 0

Answer:

(-16.0\ \hat i+0\ \hat j)\ \rm m/s.

Explanation:

<u>Given:</u>

  • x-component of acceleration of the cart, \rm a_x = 4.0\ m/s^2.
  • y-component of acceleration of the cart, \rm a_y = 2.0\ m/s^2.
  • x-component of initial velocity of the cart, \rm v_{0x} = 8.0\ m/s.
  • y-component of initial velocity of the cart, \rm v_{0y} = 12\ m/s.

Let the x and y components of the final velocity of the cart when the cart reaches its greatest y coordinate be \rm v_x\ and \ v_y respectively.

When the cart reaches its greatest y-coordinate then the y component of its final velocity should be zero, \rm v_y =0.\\

Let the cart reaches its greatest y-coordinate at time t, then using the following equation:

\rm v_y = v_{0y}+a_yt\\0=12+2\times t\\\Righttarrow t=\dfrac{-12}{2}=-6\ s.

At this time, the x component of the velocity of the particle is given by

\rm v_x = v_{0x}+a_xt=8.0+4.0\times (-6)=8.0-24.0=-16.0\ m/s.

Thus, the velocity of the cart when it reaches its greatest y coordinate, in unit-vector notation, is given by

\vec v = v_x\hat i+v_y\hat j=(-16.0\ \hat i+0\ \hat j)\ \rm m/s.

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When going round a corner your direction changes which means your velocity changes which means there is an acceleration.
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A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
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Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

Distance, d = 269 cm

We need to find the time interval of the marble. Speed is distance per unit time.

v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s

Hence, the time interval of the marble is 6.09 seconds.

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4 0
2 years ago
If a system absorbs 155.0 kJ of heat from its surroundings and the system does 45.0 kJ of work on its surroundings, what is the
Ede4ka [16]

The change in the internal energy of the system is 110 kJ.

<h3>What is internal energy?</h3>

Internal energy is defined as the energy associated with the random, disorder motions of molecules.

calculate the change in internal energy, we apply the formula below.

Formula:

  • ΔU = Q-W.................... Equation 1

Where:

  • ΔU = Change in internal energy
  • Q = Heat absorbed from the surroundings
  • W = work done by the system

From the question,

Given:

  • Q = 115 kJ
  • W = 45. 0 kJ

Substitute these values into equation 1

  • ΔU = 155-45
  • ΔU = 110 kJ

Hence, The change in the internal energy of the system is 110 kJ.

Learn more about  change in internal energy here: brainly.com/question/4654659

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2 years ago
Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-
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Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

E = 6.05×10⁻³ J.

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