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2 years ago

The more active you are, the fewer calories you should eat. A. true B. false

Physics

2 answers:

scoundrel [369]2 years ago

8 0

Answer:False

Explanation:The more active you are, the more calories you burn. Calories are a form of energy and you need to regain that energy by eating more calories

givi [52]2 years ago

3 0

Answer: false

Explanation:

the more active you are, the more calories you are burning. consuming less calories is not ideal, you need to eat more since you are burning more than normal.

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When a loose brick is resting on a wall, it has energy. When the brick is pushed off the wall and is falling down, the amount of

OLga [1]

Potential
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3 years ago

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A marble column of cross-sectional area 1.6 m^2 supports a mass of 26600 kg. The elastic modulus for marble is 5.0 times 10^10 N

vodka [1.7K]

Answer:

Δ L = 2.57 x 10⁻⁵ m

Explanation:

given,

cross sectional area = 1.6 m²

Mass of column = 26600 Kg

Elastic modulus, E = 5 x 10¹⁰ N/m²

height = 7.9 m

Weight of the column = 26600 x 9.8

= 260680 N

we know,

Young's modulus= $\dfrac{stress}{strain}$

stress = $\dfrac{P}{A}$

= $\dfrac{260680}{1.6}$

= 162925

strain = $\dfrac{\Delta L}{L}$

now,

$Y = \dfrac{stress}{strain}$

$\Delta L = \dfrac{162925}{Y}\times L$

$\Delta L = \dfrac{162925}{5 \times 10^10}\times 7.9$

Δ L = 2.57 x 10⁻⁵ m

The column is shortened by Δ L = 2.57 x 10⁻⁵ m

3 0

2 years ago

To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp

motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

$\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0$

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that $f_{max}=\mu N$

Now, we find the acceleration:

$\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}$

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

$x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s$

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3 years ago

Two identical bullets are used. Both are released at the same height - one fired out of a gun, the other is dropped. Ignoring ai

irina [24]

Answer:

Both bullets will hit the ground at the same time.

Explanation:

Let's only analyze the vertical problem.

Any object that is not in the floor or resting in some site is being affected by the gravitational force (remember that we are ignoring air resistance)

Then the acceleration of this object will be equal to the gravitational acceleration:

a = -9.8m/s^2

Where the minus sign is because this acceleration goes down.

To get the velocity equation we need to integrate over time, we will get:

v(t) = ( -9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity.

To get the position equation we need to integrate over time again, we will get:

p(t) = (1/2)*( -9.8m/s^2)*t^2 + v0*t + H

Where H is the initial height.

p(t) = (-4.9 m/s^2)*t^2 + v0*t + H

The object will hit the ground when p(t) = 0

Then we need to solve for t the next equation:

(-4.9 m/s^2)*t^2 + v0*t + H = 0

Notice that the only things we need to know are:

H = initial height (we know that is the same for both bullets)

v0 = initial vertical velocity (also is the same for both bullets)

Notice that the horizontal velocity does not affect this equation, then we will get the same value of t for the dropped bullet and for the fired bullet.

This means that both bullets will hit the ground at the same time.

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2 years ago

Watching television, with theremote control in your hand, which of the following exerts the greatest gravitational force on you?

natulia [17]

A. The couch you are sitting on

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3 years ago

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