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Luba_88 [7]
3 years ago
13

anice is watching her granddaughter drive a Barbie Jeep with a 6 V battery and an electric motor with 5 ohm of resistance. How m

uch power is the motor using to drive the Jeep?
Physics
2 answers:
Scrat [10]3 years ago
5 0

Answer:

Power required will be 7.2 watt

Explanation:

We have given battery of Barble jeep is 6 volt

So potential difference V = 6 volt

Resistance of electric motor R = 5 ohm '

We have to find the power motor using to drive the jeep

Power is given by P=\frac{V^2}{R}, here V is voltage and R is resistance

So P=\frac{V^2}{R}=\frac{6^2}{5}=7.2watt

katrin2010 [14]3 years ago
5 0

Answer:

7 watts esentially

Explanation:

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The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?
vitfil [10]

The expression of the electric flux is

\Phi = \frac{Q}{\epsilon_0}

Here,

Q = Total charge enclosed in the closed surface

\epsilon_0 = Permittivity due to free space

Rearranging to find the charge,

Q = \epsilon_0 \Phi

Replacing with our values we have finally

Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)

Q = 1.6284*10^{-8} C (\frac{10^9nC}{1C})

Q = 0.1684nC

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

7 0
3 years ago
I need homework help
KIM [24]
1) the weight of an object at Earth's surface is given by F=mg, where m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is 
F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N

2) On Mars, the value of the gravitational acceleration is different:g=3.7 m/s^2. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus: 
g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg: 
g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as 
g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2

<span>6) On Earth, the gravity acceleration is </span>g=9.81 m/s^2<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is 
</span>F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N<span>
</span>
5 0
3 years ago
Help pls lol and tysm
Zanzabum

Answer:

umm section 2 and 4

Explanation:

because at section 2 it starts and at section 4 it moves again and stops at 3.

7 0
2 years ago
TRUE OR FALSE: In single-slit diffraction, the central band gets thinner as the width of the slit increases.
olganol [36]

Answer:

Explanation:

True

4 0
3 years ago
Read 2 more answers
A man's higher initial acceleration means that a man can outrun a horse over a very short race. A simple - but plausible - model
Lera25 [3.4K]

Answer:

 t_man = 10.16 s,   t_horse = 10.73 s,  the winner is the man

Explanation:

To solve this problem we are going to find the time of each one separately.

Man we look for distance and time during acceleration

         x₁ =  v₀ t₁ + ½ a₁ t₁²

as it comes out of rest its initial velocity is zero

        x₁ = ½ a₁ t₁²

        x₁ = ½ 6.0 1.8²

        x₁ = 9.72 m

at this point its speed is

        v₁ = v₀ + a t

        v₁ = 0 + 6  1.8

        v₁ = 10.8 m / s

From here on it continues at constant speed, the distance that the man needs to travel from the point where the man leaves at 100m is

        x₂ = 100 - x₁

        x₂ = 100- 9.72

        x₂ = 90.28 m

the time for this part is

        v₁ = x₂ / t₂

         t₂ = x₂ / v₂

         t₂ = 90.28 / 10.8

         t₂ = 8.36 s

the total time for the man is

        t_man = t₁ + t₂

        t _man = 1.8 + 8.36

        t_man = 10.16 s

We repeat the calculation for the horse

distance traveled during the acceleration period

         x₃ = v₀ t + ½ a₂ t₃²

as part of rest its initial velocity is zero

        x₃ = ½  a₂ t₃²

        x₃ = ½  5.0  4.8²

        x₃ = 57.6 m

the velocity at this point is

         v₃ = v₀ + a₂ t₃

         v₃ = 0 + 5  4.8

         v₃ = 24 m / s

the rest of the route is at constant speed, the remaining distance

         x₄ = 200 - x₃

         x₄ = 200 - 57.6

         x₄ = 142.4 m

the time to go through it is

         t₄ = x₄ / v₃

         t₄ = 142.4 / 24

         t₄ = 5.93 s

the total time for the horse is

         t_horse = t₃ + t₄

         t_horse = 4.8 + 5.93

         t_horse = 10.73 s

when we compare the times we see that the man arrives a little before the horse, the winner is the man

5 0
3 years ago
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