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3 years ago

The following are connected in parallel circuit at home EXCEPT:

1 answer:

8 0

Light bulbs, think about parallel circuits as something you plug into, so for example you plug in a tv for it to work, same with a refrigerator and Christmas lights

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An object is moving at 2.50 m/s [E]. At a time 3.00 seconds later the object is traveling at 1.50 m/s

babunello [35]

Given parameters:

First velocity = 2.50m/s

Time of travel = 3s

Second velocity = 1.50m/s

Unknown:

The displacement during the first interval = ?

Velocity is the displacement of a body with time. Displacement is a distance move in a specific direction by a body.

Velocity = $\frac{Displacement}{Time taken}$

So;

Displacement = Velocity x Time taken

Now input the parameter for the first velocity and time of travel;

Displacement = 2.5 x 3 = 7.5m

The displacement id 7.5m

7 0

3 years ago

A boy throws a ball with an initial velocity of 19.6 m/s. What maximum height does the ball reach?

Wewaii [24]

<h2>Hello!</h2>

The answer is: 19.59 m

Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.

When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:

We will use the following formula:

$Vf^2=Vi^2+2*g*s$

Where:

Vf= Final velocity = 0

Vi= Initial velocity = $\frac{19.6m}{s}$

g = Gravity Acceleration = $\frac{9.81m}{s^{2} }$

s = Traveled distance

$0=19.6^2+2*-9.81*s\\s=\frac{19.6^2}{2*9.81}=\frac{384.16}{19.62}=19.59m$

Have a nice day!

8 0

3 years ago

Which chemical equation is balanced?

nika2105 [10]

Answer:

Explanation:

<h2>Mark me me brainliest plz i really need it （⌒▽⌒）</h2>

6 0

3 years ago

Read 2 more answers

Help me please I’ve been stuck on this forever now ☹️

valina [46]

C, velocity I believe this is the answer to your question

7 0

3 years ago

Read 2 more answers

An automobile of mass 1.46 cross times to the power of blank 10 cubed kg rounds a curve of radius 25.0 m with a velocity of 15.0

Vsevolod [243]

The centripetal force exerted on the automobile while rounding the curve is $1.31\times10^4 N$

Explanation:

given that

$Mass\ of\ the\ automobile\ m =1.46\times 10^3 kg\\radius\ of\ the\ curve\ r =25 m\\velocity\ of\ the\ automobile\ v=15m/s\\$

Objects moving around a circular track will experience centripetal force towards the center of the circular track.

$centripetal\ force=mv^2/r\\=1.46\times10^3\times15^2/25\\=1.46\times 10^3\times 225/15\\=1.31\times 10^4 N$

4 0

3 years ago