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julia-pushkina [17]

3 years ago

9

your shopping cart has a mass of 65 kilograms. in order to accelerate the shopping cart down an aisle at 0.3 m/sec squared , wha

t force would you need to use or apply to the cart

1 answer:

frez [133]3 years ago

7 0

Assuming there is no friction between the floor and the shopping cart we can solve for the force by using F=m*a.
<span>With this F=65*(.3) which gives is the force to be 19.5N</span>

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A 763 kg car moving at 26 m/s brakes to a stop. The brakes contain about 15 kg of iron that absorb the energy. What is increase

Neko [114]

Answer:

$\Delta T=38.20^{\circ}$

Explanation:

It is given that,

Mass of the car, m = 763 kg

Speed of the car, v = 26 m/s

Mass of the iron, m' = 15 kg

Specific heat of iron, c = 450 J/kg

When the car is in motion, it will possess kinetic energy. It is given by :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 763\times (26)^2$

K = 257894 J

Since, energy is absorbed by the brakes. The kinetic energy of the car is absorbed by the brakes. So,

$K=mc\Delta T$

$\Delta T$ is the increase in temperature of the brakes

$\Delta T=\dfrac{K}{m'c}$

$\Delta T=\dfrac{257894}{15\times 450}$

$\Delta T=38.20^{\circ}$

So, the increase in temperature of the brakes is 38.20 degrees Celsius. Hence, this is the required solution.

3 0

3 years ago

A squeeze bottle squeezes when pressed. It regains its shape when pressed. It regains its shape when pressure from your hand wit

Vaselesa [24]

Answer:

the bottle would stay in its squashed shape, and it will only regain its shape if u remove the cork

Explanation:

because theres not enough air inside to let the bottle to go back its normal form. Its like when u let the air out of a balloon, theres no air inside to let it stay in its 'big or expanded' form.

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4 years ago

A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used

zloy xaker [14]

Answer:

The compression is $\sqrt{2} \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_{potential} = \frac{1}{2} k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_{kinetic} = \frac{1}{2} m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.

$\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2$

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

$2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2$

But, in the left side we can use the previous equation to obtain:

$2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2$

$D^2 = \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k}$

$D^2 = 2 \ d^2$

$D = \sqrt{2 \ d^2}$

$D = \sqrt{2} \ d$

And this is the compression we are looking for

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3 years ago

Read 2 more answers

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similarly kirchoff's law in right side loop

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also by junction law we know that

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$-15 = 13I_2 - 10.25I_3$

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$I_1 = 0.492 A$

$I_2 = -0.428 A$

$I_3 = 0.920 A$

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3 years ago

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Scrat [10]

It is b. sodium because it is in group 1

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