A car of mass 200 kg, moving with a forward acceleration of 3 m/s-ıs acted upon by

A 920-kg compact car moving at 92 m/s has approximately 3,893,440 Joules of kinetic energy. What is the change in kinetic energy

tensa zangetsu [6.8K]

Answer:

Change in kinetic energy = 3297280 J

Explanation:

Given that,

Mass, m = 920 kg

Speed of a car, v = 92 m/s

Kinetic energy, K = 3,893,440 J

If the speed of a car, V = 36 m/s

Net kinetic energy is given by :

$K_n=\dfrac{1}{2}mV^2\\\\=\dfrac{1}{2}\times 920\times (36)^2\\\\K_n=596160\ J$

The change in kinetic energy = 3,893,440 - 596160

= 3297280 J

So, the change in kinetic energy of the car is 3297280 J.

3 0

3 years ago

Hydraulic engineers often use, as a unit of volume of water, the "acre-foot", defined as the volume of water that will cover 1 a

Alex17521 [72]

Answer:

Volume = 1,015 acre-feet (Approx)

Explanation:

Given:

Rain = 1.7 in

Time = 30 min

Area = 29 km²

Find:

Volume in acre-feet

Computation:

1 km = 1,000 m

1 m = 3.28 feet

1 km² = 247.105 acre

d = 1.7 in = 1.7 / 12 = 0.14167 ft

Area = 29 × 247.105 = 7,166.045 acre

Volume = 7,166.045 acre × 0.14167 ft

Volume = 1,015 acre-feet (Approx)

7 0

3 years ago

What is the approximate efficiency of the engine?

Elodia [21]

Answer:91% I think

Explanation:

8 0

3 years ago

Read 2 more answers

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

4 years ago

Which of the following describes the effect of a convex mirror on light rays?

natulia [17]

B. It reflects light rays outward.

A convex mirror looks like this ⊂. So when light rays hit the outward curves, it bends the light rays outwards, and backwards, and it would appear to be coming from its center called the focus.

3 0

3 years ago