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2 years ago

How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?

Physics

1 answer:

podryga [215]2 years ago

4 0

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

$\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J$

Final velocity = 40 m/s

$\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J$

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

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the millersburg ferry (m=13000.0 kg loaded) puts its engines in full reverse and stops in 65 seconds. if the speed before brakin

kenny6666 [7]

The braking force is -400 N

Explanation:

We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:

$F \Delta t = m(v-u)$

where in this problem, we have:

F is the force applied by the brakes

$\Delta t = 65 s$ is the time interval

m = 13,000 kg is the mass of the ferry

u = 2.0 m/s is the initial velocity

v = 0 is the final velocity

And solving for F, we find the force applied by the brakes:

$F=\frac{m(v-u)}{\Delta t}=\frac{(13000)(0-2.0)}{65}=-400 N$

where the negative sign indicates that the direction is backward.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

4 0

2 years ago

A metal bar moves through a magnetic field. the induced charges on the bar are

Dmitry [639]

I would say its a positive cgarge

7 0

2 years ago

one horsepower is a unit of power equal to 746w. how much energy can a 150-horsepower engine transform in 10.0s?

Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

(150 horsepower) x (746 watt/HP) x (1 joule/sec / watt) x (10 sec)

= (150 x 746 x 1 x 10) joule = 1,119,000 joules .
if correct plz mark brainly

8 0

1 year ago

A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and

scoray [572]

Answer:

$F_{allow} = 208.15kN$

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

$K_{allow} =\frac{K_c}{N}$

We know that $K_c$ is 33Mpa*m^{0.5} and our Safety factor is 2,

$K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}$

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

$\alpha = 0.5*L_c = 0.5*4mm = 2mm$

<em>For the panel</em>

$\gamma = 0.5*W = 0.5*250mm = 125mm$

To find now the goemetry factor we need to use this equation

$\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1$

That allow us to determine the allowable nominal stress,

$\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}$

$\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}$

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

$F_{allow}=\sigma_{allow}*L_c*W$

$F_{allow} = 208.15*250*4$

$F_{allow} = 208.15kN$

That is the maximum tensile load before a catastrophic failure.

4 0

2 years ago

At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti

dalvyx [7]

Answer:

Part(a): The angular acceleration is $5.63~rad~s^{-2}$ .

Part(b): The angular displacement is $2629~rad$ .

Explanation:

Part(a):

If $\omega_{1},~\omega_{2}~and~\alpha$ be the initial angular speed, final angular speed and angular acceleration of the centrifuge respectively, then from rotational kinematic equation, we can write

$\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)$

where ' $t$ ' is the time taken by the centrifuge to increase its angular speed.

Given, $\omega_{i} = 250~rad~s^{-1}$ , $\omega_{f} = 750~rad~s^{-1}$ and $t = 9.5~s$ . From equation ( $I$ ), the angular acceleration is given by

$\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}$

Part(b):

Also the angular displacement ( $\Delta \theta$ ) can be written as

$&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad$

8 0

2 years ago