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3 years ago

Which of the following is equivalent to 2,100 grams?

2 answers:

6 0

D is the correct answer. 10 decigrams are equivalent to 1 gram, so 2,100 grams are equivalent to 2100 x 10. This makes 21,000.

Reptile [31]3 years ago

6 0

21,000 decigrams is was hard for me too. ha ha ha

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Calculate the average speed in metres per second from Glasgow to Edinburgh

mariarad [96]

This is the same question as the one previously but with more details, so I will just use my previous answer.

1800 to 1820 is 20 minutes.1830 to 1838 is 8 minutes.1840 to 1905 is 25 minutes.
The total time travelled is 20+8+25 = 53 minutes = 3180 seconds.
The distance between Glasgow and Edinburgh is 28 + 12 + 34 = 74 km = 74000 m.

So, the average speed is 74000m/3180s = 23.27 m/s (4 s.f.)

5 0

3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

3 years ago

Assuming that each of the following objects is a typical example of its class, rank them by increasing density.

inysia [295]

molecular cloud <interstellar cloud <1 Msun protostar <1 Msun star <intercloud gas

Explanation:

Molecular cloud- They are a variety of interstellar cloud in which molecular hydrogen can sustain themselves. They have a very low temperature ranging from -440 to -370 degrees Fahrenheit or between 10 to 50 Kelvin. Owing to their extremely low temperature, they appear mostly dark when viewed through telescopes.

Interstellar cloud- They are a congregation of a large number of interstellar gases, dust and plasma in any galaxy or universe. They have varying temperature depending on their proximity to a star. E.g. Neutral hydrogen atom clouds have a temperature of around just 100 Kelvin while those in the near vicinity of a star have temperatures as high as 10,000 Kelvin.

1 Msun star- These stars have temperature anywhere between 5300 and 6000 Kelvin. The main source of such high surface temperature is nuclear fusion process where elemental hydrogen molecules are fused to form helium molecules.

1 Msun protostar- protostar is rather a young star which is still in formation phase (i.e. gathering mass from the parent molecular cloud). They have temperature anywhere between 2000-3000 kelvin and are accompanied by dust usually.

Intercloud gas- These are the remainder gases that are spread throughout the interstellar space. This Intercloud gas is divided into warm intercloud medium and extremely hot coronal gas with temperatures comparing to Sun’s corona. Warm intercloud forms the dominant part of intercloud gas with a temperature around 8000 Kelvin.

8 0

3 years ago

A company records authors reading their books aloud. These recordings are shared with libraries around the world for people to l

9966 [12]

Answer:

Explanation:

the loud noise can reduce the quality of the analog signal

5 0

2 years ago

Read 2 more answers

A shopper pushes a grocery cart for a distance of 18m at a constant speed on level ground, against a 37.5N frictional force. he

goldfiish [28.3K]

d = distance to which the grocery cart is pushed = 18 m

f = frictional force = 37.5 N

θ = angle of force below the horizontal = 27.5 deg

W = gravitational force in downward direction

Θ = angle between gravitational force in down direction and displacement in horizontal direction = 90

U = work done on the cart by gravitational force

work done on the cart by gravitational force is given as

U = W d CosΘ

inserting the values

U = W (18) Cos90

U = 0 J

5 0

3 years ago