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Olenka [21]
3 years ago
14

On a nice summer day,Kim takes her niece Madison for a walk in her stroller.If they start from rest and accelerate at a rate of

0.5m/s2,what is their velocity after they have traveled 5.0
Physics
1 answer:
11111nata11111 [884]3 years ago
6 0

2.5m/s

Explanation:

Given parameters:

Initial velocity = 0m/s

Acceleration = 0.5m/s²

time of travel = 5s

Solution:

Final velocity = ?

Solution:

Acceleration can be defined as the change in velocity with time:

          Acceleration = \frac{Final velocity - Initial velocity}{time}

  From the equation above, the unknown is final velocity:

Final velocity - initial velocity = Acceleration x time

 since initial velocity = 0

   Final velocity = 0.5 x 5 = 2.5m/s

Learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

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a desktop computer and monitor together draw about 2 A of current they plug into a wall outlet that is 120 V what is the Resista
Delvig [45]

Answer:

60 \Omega

Explanation:

the relation between current, voltage and resistance in an electrical circuit is given by Ohm's law:

V=IR

where V is the voltage, I is the current and R is the resistance. In this problem, the current is I=2 A, the voltage is V=120 V, therefore we can arrange the previous equation and find the resistance:

R=\frac{V}{I}=\frac{120 V}{2 A}=60 \Omega

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A robot is exploring​ charon, the dwarf planet​ pluto's largest moon. Gravity on charon is 0.278 meters per second ​[m divided b
DochEvi [55]

In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up

First we will find the mass of the stone

As it is given that stone is spherical in shape so first we will find its volume

V = \frac{4}{3}\pi r^3

V = \frac{4}{3}\pi *(\frac{0.06}{2})^3

V = 1.13 * 10^{-4} m^3

Now it is given that it's specific gravity is 10.8

So density of rock is

\rho = 10.8 * 10^3 kg/m^3

mass of the stone will be

m = \rho V

m = 10.8* 10^3 * 1.13 * 10^{-4}

m = 1.22 kg

now change in potential energy is given as

\Delta U = mgH

here

g = gravity on planet = 0.278 m/s^2

H = height lifted upwards = 15 cm

\Delta U = 1.22* 0.278 * 0.15

\Delta U = 0.051 J

Now energy supplied by internal circuit of robot is given by

E = Vit

V = voltage supplied = 10 V

i = current = 1.83 mA

t = time = 12 s

E = 10* 1.83 * 10^{-3} * 12

E = 0.22 J

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5 0
3 years ago
Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
dedylja [7]

Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

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1V electrical potential 
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