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Sholpan [36]
3 years ago
6

A typical 12-V car battery can deliver about 750,000 C of charge before dying. This is not a lot of charge. As a comparison, cal

culate the maximum number of kg of water (at 100 degrees Celsius) that could be boiled into steam (at 100 degrees Celsius) using energy from this battery.
Physics
1 answer:
dimaraw [331]3 years ago
6 0

Answer:

3.982 kg

Explanation:

The latent heat of vaporization = 540 cal/g

= 5.4 ×10⁵ cal/kg

L = 5.4 ×10⁵ × 4.19

  = 2.26 × 10⁶ J/kg

Q = 12 × 750,000

= 9, 000, 000

= 9 × 10⁶ J

the maximum number of kg of water (at 100 degrees Celsius) that could be boiled into steam (at 100 degrees Celsius) is:

= \frac{ 9*10^6}{2.26*10^6 }

= 3.982 kg

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A wire that is 0.65 m long and carrying a current of 8.2 A is at right angles to a uniform magnetic field. The force on the wire
Luda [366]

Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

F=ILB sin \theta

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

\theta is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

\theta=90^{\circ} since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T

3 0
3 years ago
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w
Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

\theta= (0.20 rad/s)(75 s)=15 rad

In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

v=(0.20 rad/s)(13.5 m)=2.7 m/s

6 0
3 years ago
A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction ti
vagabundo [1.1K]

Answer:

 t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

                            v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

d = v. t + \dfrac{v^2}{2a}

t = \dfrac{d}{v} -\dfrac{v}{2a}

t is the reaction time

t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}

 t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.

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