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PtichkaEL [24]
1 year ago
7

How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer:

___________ J (round to the nearest whole number)
Physics
1 answer:
zimovet [89]1 year ago
5 0

According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:

W=\Delta K

Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:

K=\frac{1}{2}mv^2

Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:

\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\  \\ \therefore K\approx532,346J \end{gathered}

Therefore, the answer is: 532,346 J.

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Which theory of light is the photon more consistent with
Rudik [331]

Answer: particle theory

Explanation:

4 0
3 years ago
A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swin
Goshia [24]

Answer:

(A) 0.63 J  

(B) 0.15 m

Explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + mh^{2}  

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I = (\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2})^{2}[/tex]

I = (\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2})^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5Iω^{2}

  = 0.5 x 0.07875 x 4^{2} = 0.63 J   0.15 m

(B) from the conservation of energy

   initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

   Ki + Ui = Kf + Uf

   at the maximum height velocity = 0 therefore final kinetic energy = 0

   Ki + Ui = Uf

   Ki = Uf - Ui

 Ki =  mg(H-h)

where (H-h) = rise in the center of mass

     0.63 = 0.42 x 9.8 x (H-h)

   (H-h) = 0.15 m

6 0
3 years ago
A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with
Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\  g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s

\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM

Therefore the speed in RPM will be 62.64 RPM.

5 0
3 years ago
For a series circuit what is the terminal voltage of a battery or power supply equal to in terms of the potential difference or
trapecia [35]

Answer: V=IR

Explanation: for a series circuit connected to a battery supply, the total emf across the circuit is given as

E = I(R + r) and by expanding, we have that E =IR + It

Where r is the internal resistance of the battery

I is the total current flowing in the circuit

R total load resistance in the circuit.

E is the total emf of the circuit.

The total emf is the sum of 2 separate voltages.

"IR" which is the terminal voltage and "Ir" which is the loss voltage.

The teenila voltage is the voltage flowing in the circuit based on the equivalent resistance of the circuit while the loss voltage is the wasted voltage based on the internal resistance of the battery source.

7 0
3 years ago
Where is potential energy the greatest on a rollercoaster?
lidiya [134]
On a roller coaster, the greatest potential energy is at the highest point of the roller coaster
4 0
3 years ago
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