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Butoxors [25]
3 years ago
12

A 9­v battery is connected to a light bulb with a resistance of 3?. what is the current in the circuit

Physics
1 answer:
anygoal [31]3 years ago
7 0
9/3= 3 ampere.. that would be the aswer
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A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor
finlep [7]

Answer:

The initial velocity is 38.46 m/s.

Explanation:

The horizontal distance travel by the tennis ball = 13 m  

The height at which the tennis ball dropped = 56 cm

Now calculate the initial speed of tennis ball.

The vertical velocity is zero.

Below is the calculation. Here, first convert centimetre into kilometre. So, height at which ball dropped is 0.56 km.

v = \sqrt{2 \times 9.8 \times 0.56} = 3.32 m/s \\

t = \frac{3.32}{9.8} = 0.338s \\

Ux \times t = 13 \\

Ux = \frac{13}{0.338} = 38.46 m/s = Initial velocity.

8 0
3 years ago
Define electric current and drift velocity.
Blizzard [7]

Answer:

Current- the flow of free charges, such as electrons and ions

Drift velocity- the average speed at which these charges move

3 0
3 years ago
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An object is launched with a vertical velocity of 44.3 m/s upward, and a horizontal velocity of 12 m/s. What is the maximum heig
Brrunno [24]

b is the answer there you go if you need the answer

7 0
3 years ago
Why do plane mirrors and convex mirrors form only virtual images?
Alekssandra [29.7K]
<span>B. because they cannot make light rays converge</span>
7 0
3 years ago
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A cyclist coasts up a 10.5° slope, traveling 19.0 m along the road to the top of the hill. If the cyclist's initial speed is 9.5
never [62]

Answer:

Final speed, v = 12.57 m/s

Explanation:

Given that,

Slope with respect to horizontal, \theta=10.5^{\circ}

Distance travelled, d = 19 m

Initial speed of the cyclist, u = 9.5 m/s

We need to find the final speed of the cyclist. Let h is the height of the sloping surface such that,

h=d\times sin\theta

h=19\times sin(10.5)  

h = 3.46 m

Let v is the final speed of the cyclist. It can be calculated using work energy theorem as :

\dfrac{1}{2}m(v^2-u^2)=mgh

\dfrac{1}{2}(v^2-u^2)=gh

\dfrac{1}{2}\times (v^2-(9.5)^2)=9.8\times 3.46

v = 12.57 m/s

So, the final speed of the 12.57 m/s. Hence, this is the required solution.

5 0
3 years ago
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