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2 years ago

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60km/h

Physics

1 answer:

natima [27]2 years ago

4 0

The work done to stop the car is -208.33 kJ

From work-kinetic energy principles, the change in kinetic energy of the car ,ΔK equals the work done to stop the car, W.

W = ΔK = 1/2m(v'² - v²) where

m = mass of car = 1500 kg,
v = initial velocity of car = 60 km/h = 60 × 1000 m/3600 s = 16.67 m/s and
v' = final velocity of car = 0 m/s (since the car stops).

<h3 /><h3>Calculating the work done</h3>

Substituting the values of the variables into the equation, we have

W = 1/2m(v'² - v²)

W = 1/2 × 1500 kg(v(0 m/s)² - (16.67 m/s)²)

W = 750 kg(0 (m/s)² - 277.78 (m/s)²)

W = 750 kg(- 277.78 (m/s)²)

W = -208333.33 J

W = -208.33333 kJ

W ≅ -208.33 kJ

So, the work done to stop the car is -208.33 kJ

Learn more about work done here:

brainly.com/question/9821607

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Which set of atoms will form an ionic compound?

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Answer:

The answer is A)

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3 years ago

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with

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Answer:

Part a)

$I = 1879.7 kg m^2$

Part b)

$\alpha = 0.70 rad/s^2$

Part c)

$I = 153.8 kg m^2$

Part 4)

angular acceleration will be ZERO

Part 5)

$I = 345.6 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about left end of the rod is given as

$I = \frac{m_r L^2}{3} + (\frac{2}{5} m_s R^2 + m_s(R + L)^2)$

So we have

$I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)$

$I = \frac{16}{3}m_r R^2 + (2m_r R^2 + 125 m_rR^2)$

$I = (\frac{16}{3} + 127)m_r R^2$

$I = (\frac{16}{3} + 127)(6.85)(1.44)^2$

$I = 1879.7 kg m^2$

Part b)

If force is applied to the mid point of the rod

so the torque on the rod is given as

$\tau = F\frac{L}{2}$

$\tau = 460(2R)$

$\tau = 460 \times 2 \times 1.44$

$\tau = 1324.8 Nm$

now angular acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{1324.8}{1879.7}$

$\alpha = 0.70 rad/s^2$

Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

$x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)$

$x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}$

so moment of inertia about this position is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})$

$I = m_r R^2(\frac{16}{12} + \frac{25}{4} + 2 + \frac{5}{4})$

$I = 6.85(1.44)^2\times 10.83$

$I = 153.8 kg m^2$

Part 4)

If force is applied parallel to the length of rod

then we have

$\tau = \vec r \times \vec F$

$\tau = 0$

so angular acceleration will be ZERO

Part 5)

moment of inertia about right edge of the sphere is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + 2R)^2 + (\frac{2}{5} m_s R^2 + m_s(R)^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(4R)^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(R^2)$

$I = m_r R^2(\frac{16}{12} + 16 + 2 + 5)$

$I = 6.85(1.44)^2\times 24.33$

$I = 345.6 kg m^2$

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An advanced computer sends information to its various parts via infrared light pulses traveling through silicon fibers (n = 3.50

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Answer:

d = 6.43 cm

Explanation:

Given:

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Solution:

- The amount of time taken for information pulse to travel to memory unit:

t_m = T - t_p

t_m = 2.0 - 0.5 = 1.5 ns

- We will use a basic relationship for distance traveled with respect to speed of light and time:

d = V*t_m

- Where speed of light in silicon medium is given by:

V = c / n

- Hence, d = c*t_m / n

-Evaluate: d = 3*10^8*1.5*10^-9 / 3.50

d = 0.129 m 12.9 cm

- The above is the distance for pulse going to and fro the memory and central unit. So the distance between the two is actually d / 2 = 6.43 cm

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sweet-ann [11.9K]

Answer:

Explanation:

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4 years ago

Read 2 more answers

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60km/h​

Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60km/h