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3 years ago

A 1400kg car is moving at a speed of 25m/s. How much KE does the car have?

Physics

1 answer:

Nady [450]3 years ago

6 0

Answer:

437500Joules

Explanation:

Kinetic energy=1/2mvsquare

1/2 x 1400 x 25 x25

kinetic energy= 437500Joules

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Why is weathering slow in cold, dry places?

Alexxandr [17]

Rate of weathering depends on temperature and moisture. Cold dry places have less water to weather things

Hope this helped!

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6 0

3 years ago

4. Who was the proponent of the Neo-classicism? a) Claude Debussy b) Joseph Maurice Ravel c) Igor Stravinsky d) Arnold Schoenber

nirvana33 [79]

Who was the proponent of the Neo-classicism?

a) Claude Debussy

b) Joseph Maurice Ravel

c) Igor Stravinsky

d) Arnold Schoenberg

4 0

3 years ago

The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm

Nesterboy [21]

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

8 0

3 years ago

An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu

Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where we have

$\alpha = -80.0 rad/s^2$ is the angular acceleration (negative since the rotor is slowing down)

$\omega_f$ is the final angular speed

$\omega_i = 2000 rad/s$ is the initial angular speed

t = 10.0 s is the time interval

Solving for $\omega_f$ , we find the final angular speed after 10.0 s:

$\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s$

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

$\alpha = \frac{\omega_f - \omega_i}{t}$

If we re-arrange it for t, we get:

$t = \frac{\omega_f - \omega_i}{\alpha}$

where here we have

$\omega_i = 2000 rad/s$ is the initial angular speed

$\omega_f=0$ is the final angular speed

$\alpha = -80.0 rad/s^2$ is the angular acceleration

Solving the equation,

$t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s$

6 0

3 years ago

The moon has a diameter of 3.48 x 106 m and is a distance of 3.85 x 108 m from the earth. The sun has a diameter of 1.39 x 109 m

Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

r = Distance between the Earth and the object

Angle subtended is given by

$\theta=\frac{s}{r}$

For the Moon

$\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad$

The angle subtended by the Moon is 0.00903 rad

For the Sun

$\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad$

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

$\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}$

The area ratio is $6.268\times 10^{-6}$

3 0

3 years ago