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siniylev [52]
3 years ago
6

A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q

on the rim, vertically below P, 0.19 s later.
(a) What is the distance ?
(b) How far away from the dart board is the dart released?
Physics
1 answer:
Kryger [21]3 years ago
3 0
D=s(t) so it would be d=10(.19) d=.19 FOR BITH SNDWERS
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A 1 kg particle moves upward from the origin to (23) m. Wit is work done by the force of gravity which is in - y direction s B.
sveta [45]

Answer:

Explanation:

mass, m = 1 kg

Position (2, 3 ) m

height, h = 2 m

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

Here, no force is acting in horizontal direction, the force of gravity is acting in vertical direction, so the work done by the gravitational force is to be calculated.

Force  mass x acceleration due to gravity

F = 1 x 9.8 = 9.8 N

Work = force x displacement x CosФ

Where, Ф be the angle between force vector and the displacement vector.

Here the value of Ф is 180° as the force acting vertically downward and the displacement is upward

So, W = 9.8 x 2 x Cos 180°

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Thus, option (A) is correct.

4 0
3 years ago
A convex lens with a focal length of 10 cm is located 30 cm away from a
zloy xaker [14]

Answer: 15 cm

Explanation:

According to the Lens Equation we have the following:

\frac{1}{f}=\frac{1}{v}+\frac{1}{u} (1)

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v is the distance between the image and the lens

Isolating v:

v=\frac{uf}{u-f}} (2)

Solving:

v=\frac{(30 cm)(10 cm)}{30 cm-10 cm}} (3)

Finally:

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8 0
3 years ago
Are photons causing the electrons to flow off of the target? ( the target is calcium)
Sliva [168]

Answer:

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daser333 [38]

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

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so we get as,

T = 50/9.8

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5 0
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MakcuM [25]

Answer: The answer is C.) 25 m/s^2.

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8 0
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