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taurus [48]
3 years ago
10

Which word in the sentence is a gerund?

Physics
2 answers:
OleMash [197]3 years ago
8 0
A gerund usually ends wit -ing, so we must find the word that includes that.

Loading includes ing at the end.

Final answer: b. loading
aliina [53]3 years ago
4 0
A <span>gerund is a word ending in ing.
B)Loading is the gerund.
Answer: B
</span>
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An object floats in a beaker as shown. when it was put into the beaker, it displaced an amount of water into the graduated cylin
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answer is  C. 10 g

Explanation:

: When an object floats, it displaces an amount of water that has the same mass as itself. Therefore, the mass of the water in the graduated cylinder is equal to the mass of the object. We can see that there are 10 mL of water in the graduated cylinder. We also know that the density of water is 1 g/mL. Since each mL of water has a mass of 1 g, then 10 mL of water has a mass of 10 g. If the mass of the displaced water is 10 g, then the mass of the floating object is also 10 g.

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3 years ago
Which of the following statements is false?
agasfer [191]

A & B

Observe the path of the light ray as it passes through the lenses as shown in the attached images. Concave lenses diverge light rays while the convex lens converges the light rays.

Explanation:

Real images are formed where the rays converge, a property of images by convex lenses. Convex lenses can be used to magnify objects. If the image occurs before the focal point of the lens then the image will be upright but smaller. The images inverts and gets bigger past the focal point.

Virtual images are property of concave lenses. These images appear closer but smaller than the real object.

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6 0
3 years ago
Read 2 more answers
HELP PLS WHAT IS Y COMPONENT
jeka94

Answer:

11.5

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6 0
3 years ago
2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas
Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

ds=\dfrac{dQ}{dt}

At constant pressure,

dQ=C_{p}dt

\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}

\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}

Put the value into the formula

14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})

\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}

0.693=ln\dfrac{T_{2}}{T_{1}}

ln2=ln\dfrac{T_{2}}{T_{1}}

T_{2}=2T_{1}...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

\Delta Q=C_{p}\Delta T

Put the value into the formula

6236=2.5\times8.3144(T_{2}-T_{1})

6236=20.786(T_{2}-T_{1})

T_{2}-T_{1}=\dfrac{6236}{20.786}

T_{2}-T_{1}=300

Put the value of T₂

2T_{1}-T_{1}=300

T_{1}=300\ K

Put the value of T₁ in equation (I)

T_{2}=2\times300

T_{2}=600\ K

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0
3 years ago
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