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3 years ago

What is the kinetic energy of a 0.5 kg puppy that is running 1.5m/s

Physics

1 answer:

bija089 [108]3 years ago

3 0

Answer: Well the answer is KE = 5.625E-7 i just don't know the units for it...

Hope this helps....... Stay safe and have a Merry Christmas!!!!!!!!!! :D

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A runner with a mass of 80kg accelerates from 0 to 9 m/s in 3 s find the net force

frosja888 [35]

So, first you find your acceleration which is 3m/s^2, using the acceleration formula.
Now set up your equation, F=ma, so put in the stuff, F=80kg·3m/s^2. Then solve your equation by multiplying, and you get F=240N, since newtons are your measurement.
Hope this helps

4 0

4 years ago

What is the hidden gene called

vagabundo [1.1K]

The hidden gene is called recessive

3 0

3 years ago

Read 2 more answers

(a) You short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. If the current

erica [24]

(a) $1.11 \Omega$

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

$r=\frac{V}{I}$

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

$r=\frac{20 V}{18 A}=1.11\Omega$

(b) 360 W

The power generated by the battery is given by the equation

$P=VI$

where

V = 20 V is the voltage of the battery

I = 18 A is the current

Substituting into the formula,

$P=(20 V)(18 A)=360 W$

(c) 360 J

The energy dissipated by the internal resistance is given by

$E=Pt$

where

P = 360 W is the power generated

t = 1 s is the time

Solving the equation, we find

$E=(360 W)(1 s)=360 J$

(d) 1.65 A

The battery is now connected to a $R=11 \Omega$ resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

$R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega$

And so, the current flowing through the circuit is

$I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A$

(e) 29.9 W

The power dissipated in the external resistor is given by

$P=I^2 R$

where

I = 1.65 A is the current

$R=11 \Omega$ is the resistance

Solving the equation, we find

$P=(1.65 A)^2(11 \Omega)=29.9 W$

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

$V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V$

4 0

3 years ago

A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits t

Dovator [93]

Answer:

The location of helicopter is behind the packet.

Explanation:

As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.

The horizontal velocity remains same as there is no force in the horizontal direction. The vertical velocity goes on increasing as acceleration due to gravity acts.

So, the helicopter is behind the packet.

7 0

3 years ago

The damage caused by electric shock depends on the current flowing through the body; 1 ma can be felt and 5 ma is painful. above

WARRIOR [948]

We can solve the problem by using Ohm's law.

The resistance of the person, with dry skin, is $R=60000 \Omega$ . In order to be felt, the current must be at least
$I=1 mA=0.001 A$

Ohm's law gives us the relationship between current and voltage:
$V=IR$
where
I is the current
R the resistance
V the voltage

Using the data of the problem, we find that the minimum voltage needed is
$V=IR=(0.001 A)(60000 \Omega)=60 V$

3 0

3 years ago