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2 years ago

6

A brick is thrown vertically upward with an initial speed of 3.00 m/s from the roof of a building. If the building is 78.4 m tal

l, how much time passes before the brick lands on the ground?

2 answers:

masya89 [10]2 years ago

6 0

Answer:

0.918 sec

Explanation:

time of flight:

t=u²/g

=3²/9.8=0.918sec

natali 33 [55]2 years ago

3 0

Answer:

4.32 seconds

Explanation:

Given:

y = 0 m

y₀ = 78.4 m

v₀ = 3.00 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 = 78.4 + 3.00 t − 4.9 t²

4.9t² − 3t − 78.4 = 0

Solve with quadratic formula:

t = [ 3 ± √(9 − 4(4.9)(-78.4)) ] / 9.8

t ≈ -3.71, 4.32

Since t must be positive, t = 4.32. So it takes 4.32 seconds for the brick to land.

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10. Almost three-quarters of household energy use is due to the following:

Arturiano [62]

Answer: Energy consumption and sustainability is important so that it remain available for future generation.

Explanation:

1. The home furnaces are likely to require fuel like coal, which will directly emit carbon dioxide and carbon monoxide gases. These should be replaced with the electrical furnaces. The old or more power consuming air conditioners should be replaced with new ones.

2. The water heaters should be tankless so their capacity to heat more water could be possible. The water heaters should be electricity saving.

3. Washer and dryers should be water savy and electricity savy. A front-loading washing machine is useful energy saver.

4. The LED lights are more electricity saving than conventional bulbs. Halogen lights are also electricity saving.

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3 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

8 0

2 years ago

Which of the following statements about electromagnetic radiation are TRUE? a. Wavelength and frequency are inversely proportion

Kobotan [32]

Answer: A) Wavelength and frequency are inversely proportional.

Explanation:

From the wave equation;

Velocity= frequency × wavelength

If the above equation is rearranged making the frequency the subject of formula, it would give;

Frequency= velocity/ wavelength.

From the above equation we see that frequency is inversely proportional to the wavelength. This means that for every increase in wavelength there would be a decrease in frequency, and for every increase in frequency there is a reduction in wavelength.

7 0

3 years ago

Which force attracts all matter to each other

likoan [24]

Gravity is the force that attracts all matter to each other.

Explanation:

Sir Isaac Newton discovered Gravity when he saw a falling apple while thinking about the forces of nature.

Gravity is a fundamental force that causes objects to have weight. Gravity acts on all matter and is a function of both mass and distance. Each object attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force of attraction is, however, negligible between most objects because of their small size.

Gravitational force is given as:

$F = \frac{Gm1m2}{r^{2} }$

Where G is gravitational constant and is equal to 6.674×10−11 m³⋅kg⁻¹⋅s⁻²

m₁ and m₂ are the masses of the two objects.

r is the distance between the two objects.

The gravity is what makes an apple fall on the ground and gravity is the force that keeps us on the ground.

Keywords: gravity, Newton, Force, weight

Learn more about gravitational force from brainly.com/question/14321566

#learnwithBrainly

8 0

3 years ago

A certain thin lens is made of glass with refraction index ????lens=1.500. In air, where the index of refraction is 1.000, the l

son4ous [18]

Answer:

The focal length of the lens in ethyl alcohol is 41.07 cm.

Explanation:

Given that,

Refractive index of glass= 1.500

Refractive index of air= 1.000

Refractive index of ethyl alcohol = 1.360

Focal length = 11.5 cm

We need to calculate the focal length of the lens in ethyl alcohol

Using formula of focal length for glass air system

$\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Using formula of focal length for glass ethyl alcohol system

$\dfrac{1}{f'}=(n_{g}-n_{ethyl})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Divided equation (II) by (I)

$\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}$

Where, $n_{g}$ = refractive index of glass

$n_{a}$ = refractive index of air

$n_{ethyl}$ = refractive index of ethyl

Put the value into the formula

$\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}$

$\dfrac{f'}{11.5}=\dfrac{25}{7}$

$f'=\dfrac{25}{7}\times11.5$

$f'=41.07\ cm$

Hence, The focal length of the lens in ethyl alcohol is 41.07 cm.

7 0

3 years ago