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Andreas93 [3]
3 years ago
14

A car battery seems to be malfunctioning (not providing the proper voltage and current to start your car). While the car is off

and you are not trying to start it, a friend uses a voltmeter to measure the voltage and finds that your battery voltage is 12.0 V, just as it is supposed to be. After the voltage check, you try to start your car again to no avail. Can you rule out the battery as the problem?
Physics
1 answer:
Stella [2.4K]3 years ago
4 0

No. You cannot rule out the battery even after the open circuit voltage measurement. The open-circuit voltage may not have changed but the battery's internal resistance may have greatly increased.

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WILL MARK BRAINLIEST PLEASE HELP AND SHOW ALL WORK
jok3333 [9.3K]

Answer:

Same reading.

Explanation:

Assume that after the string breaks the ball falls through the liquid with constant speed. If the mass of the bucket and the liquid is 1.20 kg, and the mass of the ball is 0.150 kg,

A.) Before the string break, the total weight = weight of the can + weight of the water.

According to Archimedes' Principle which state that: “A body immersed in a liquid loses weight by an amount equal to the weight of the liquid displaced.” Archimedes principle also states that: “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it

B.) After the string break.

The scale will have the same reading as before the string break.

6 0
3 years ago
A ball is pushed into a spring-loaded launcher with a force of 20 N, which compresses the spring 0.08 m.
damaskus [11]

Answer:

0.8J

Explanation:

Given parameters:

Force  = 20N

Compression  = 0.08m

Unknown:

Spring constant  = ?

Elastic potential energy  = ?

Solution:

To solve this problem, we use the expression below:

           F = k e

F is the force

k is the spring constant

e is the compression

             20  = k x 0.08

              k  = 250N/m

Elastic potential energy;

       EPE  = \frac{1}{2} k e²    =  \frac{1}{2}  x 250 x 0.08²

 Elastic potential energy = 0.8J

3 0
2 years ago
The table represent the thickness, top density, and bottom density of the different layers of the Earth. In most of the layers,
Nostrana [21]

Answer:inner core?

Explanation:

7 0
3 years ago
Read 2 more answers
A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In
Daniel [21]

Answer:81.57\mu V

Explanation:

Given

radius of circular region r=1.50 mm

A=\pi r^2=7.069\times 10^{-6}\ m^2

Magnetic Field B=1.50\ T

time t=130 ms

Flux is given by

\phi =B\cdot A

change in Flux d\phi =(B_f-B_i)A

Emf induced is e=\frac{\mathrm{d} \phi}{\mathrm{d} t}

e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}

e=81.57 \mu V

3 0
3 years ago
Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea
Setler [38]

Answer:

a)  M = 4,997 10²⁰ kg ,  b)   T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

          v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

         a = (v₀ - v) / t

         a = (15 - (-15)) /9.00 = 30/9

         a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

          F = m a

         G m M / r² = m a

         M = a r² / G

Let's calculate

         M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

         M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

         R = R_{m} + h

         R = 1 10⁵ + 2.00 10⁴

         R = 12 10⁴ m

         F = m a

        G m M / R² = m a

Centripetal acceleration is

         a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

        d = 2π R

        v = d / t

        v = 2π R / T

Let's replace

        G m M / R² = m (2π R / T)² / R

        G M = R³ 4π² / T²

        T² = 4π² R³ / G M

       T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

       T² = 6.82 10¹⁶ / 3.33 10¹⁰

       T = √ (2,048 10⁶)

       T = 1.43 10³ s

3 0
2 years ago
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