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tankabanditka [31]

3 years ago

5

750 kg car zooms away from a red light with an acceleration of 7.8 m/s squared . What is the average net force in Newtons that t

he car experiences?

1 answer:

natta225 [31]3 years ago

5 0

Answer:

<h2>5850 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 750 × 7.8

We have the final answer as

<h3>5850 N</h3>

Hope this helps you

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Encontrar la distancia y desplazamiento de las dos trayectorias si se mueve el móvil Desde A hasta B

Lerok [7]

Answer:

Build Your Confidence

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Explanation:

6 0

3 years ago

A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40

Crank

Answer:

To achieve the velocity of 40 m/sec height will become 4 times

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is $\frac{1}{2}mv^2$

So total energy at the top of the hill $=mgh+0=mgh$

At the bottom of the hill kinetic energy is equal to $\frac{1}{2}mv^2$ and potential energy will be 0

So total energy at the bottom of the hill is equal to $0+\frac{1}{2}mv^2$

Form energy conservation $mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$ , for v = 20 m/sec

$20=\sqrt{2\times 9.8\times h}$

Squaring both side

$19.6h=400$

h = 20.408 m

Now if velocity is 0 m/sec

$40=\sqrt{2gh}$

$19.6h=1600$

h = 81.63 m

So we can see that to achieve the velocity of 40 m/sec height will become 4 times

5 0

3 years ago

You have given a power supply, copper wire, and an iron nail. What should you do to decrease the strength of the electro magnet?

Viefleur [7K]

Explanation:

If you wrap some of the wire around the nail in one direction and some of the wire in the other direction, the magnetic fields from the different sections fight each other and cancel out, reducing the strength of your magnet

5 0

3 years ago

A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk

Murrr4er [49]

Answer:

$\omega_2=5.1rad/s$

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is $L=I\omega$ . If we call 1 the situation when the student is at the rim and 2 the situation when the student is at $r_2=1.39m$ from the center, then we have:

$L_1=L_2$

Or:

$I_1\omega_1=I_2\omega_2$

And we want to calculate:

$\omega_2=\frac{I_1\omega_1}{I_2}$

The total moment of inertia will be the sum of the moment of intertia of the disk of mass $m_D=119.1 kg$ and radius $r_D=3.23m$ , which is $I_D=\frac{m_Dr_D^2}{2}$ , and the moment of intertia of the student of mass $m_S=54.3kg$ at position $r$ (which will be $r_1=r=3.23m$ or $r_2=1.39m$ ) will be $I_{S}=m_Sr_S^2$ , so we will have:

$\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}$

or:

$\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}$

which for our values is:

$\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s$

6 0

3 years ago

A 7.77 kg mass is moving due west at 7.77 m/s. A second mass of 8.88 kg is moving due south at 8.88 m/s. What is the magnitude o

masya89 [10]

Answer:

8.362m/s

Explanation:

Given data

Mass m1= 7.77kg

Velocity v1= 7.77m/s

Mass m2= 8.88kg

Velocity v2= 8.88m/s

Apply the law of conservation of momentum for inelastic collision we have

m1v1+m2v2= (m+m2)V

7.77*7.77+ 8.88*8.88= (7.77+8.88)V

60.3729+78.8544= 16.65V

139.2273= 16.65V

Divide both sides by 16.65

V= 139.2273/16.65

V= 8.362m/s

Hence the final velocity is 8.362m/s

7 0

3 years ago