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stiv31 [10]
3 years ago
6

Which illustration represents low accuracy but high precision?​

Physics
2 answers:
Misha Larkins [42]3 years ago
7 0

Answer:

D i think

Explanation:

podryga [215]3 years ago
4 0

Answer:

C

Explanation:

a p e x !

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Damage to the ___________ will most likely cause a person to lose the ability to comprehend language. (2 points)
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Answer: Wernicke's Area

Explanation: Wernicke's area is the region of the brain responsible for language <em>interpretation</em>. Broca's area is associated with language <em>production.</em>

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If temperature increases, then pressure __________.
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If the temperature increases, then pressure increases too. (T<span>he molecules in the gas move faster, exerting a greater force. This </span>increases t<span>he </span>pressure<span>.)</span>
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A woman and her dog are out for a morning run to the river, which is located 4.0 KM away. The woman runs at 2.5 M/S in a straigh
VLD [36.1K]

River shore is located at distance

d = 4 km

speed of the woman is given as

v_1 = 2.5 m/s

now the time taken by the woman to cover the distance is

t = \frac{d}{v}

t = \frac{4000}{2.5} = 1600 s

for the same time interval the dog will run to and fro with speed 4.5 m/s

so the total distance moved by the dog is given by

d = v* t

d = 4.5 * 1600

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3 years ago
Define wheel and axle 4 example of wheel and axle​
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3 years ago
Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65
guajiro [1.7K]

Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, \theta =65°

Range of the projectile is, R=520 m.

Acceleration due to gravity, g=9.8\ m/s^2

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

R=\frac{v_{0}^2\sin2\theta}{g}

Plug in all the given values and solve for minimum speed, v_0.

520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

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