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joja [24]
3 years ago
15

How many protons are in this Atom if it has a balanced charge?

Physics
1 answer:
Dmitry [639]3 years ago
6 0

Answer:

the answer is A

Explanation:

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Falling objects drop with an average acceleration of 9.8 m/s2. An arrow is shot with a velocity of 11.76 m/s straight down from
In-s [12.5K]

Answer:

3.8 secs

Explanation:

Parameters given:

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

Initial velocity, u = 11.76 m/s

Final velocity, v = 49 m/s

Using one of Newton's equations of linear motion, we have that:

v = u + gt

where t = time of flight of arrow

The sign is positive because the arrow is moving downward, in the same direction as gravitational force.

Therefore:

49 = 11.76 + 9.8*t\\\\\\\49 - 11.76 = 9.8t\\\\=> 9.8t = 37.24\\\\\\t = \frac{37.24}{9.8} \\\\\\t = 3.8 secs

The arrow was in flight for 3.8 secs

6 0
3 years ago
How much heat energy is produced by 0.5 Wh of electrical energy ?
babymother [125]
1.7 Btu

1 watt = 3.41214 Btu/h

1watt * 1h = 3.41214 Btu/h * h

1 = 3.41214 Btu/ (watt*h)/

0.5 watt * h = 0.5 watt*h * 3.41214 Btu/(watt*h) = 1.706 Btu
6 0
3 years ago
Read 2 more answers
Help Please
hichkok12 [17]
Gravitational potential energy i think
8 0
3 years ago
What is photographic film
Vesnalui [34]

Answer: its a strip of transparent film, one side coated with gelatin emulsion containing microscopically small light-sensitive silver halide crystals

Explanation:

3 0
3 years ago
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
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