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3 years ago

Which word in the sentence is a gerund?

2 answers:

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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Will give brainliest to correct answer!

ad-work [718]

The international system of units is the designated system of units used by scientist in every part of the world to keep data in the same form and measurements, this is to avoid confusion and the need to convert data when being shared. typically described in meters or kilometer over a time form usually seconds or hours.

6 0

3 years ago

Read 2 more answers

The current in a lamp is 0.5 ampere when it is plugged into a standard wall outlet. What is the resistance of the lamp when it i

lidiya [134]

The resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms)

Explanation:

In the United States Of America the standard voltage is 120 v and their frequency is 60 Hz

Standard wall outlet voltage is 120 V

The current in the lamp is 0.5 ampere

Resistance (R) = V/ I

= 120/0.5

= 240Ω (ohms)

Thus the resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms).

8 0

3 years ago

A sealed container filled with gas is heated. What happens?

Luda [366]

The internal pressure increases as the gas is heated

5 0

3 years ago

Read 2 more answers

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$