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2 years ago

I am looking to buy a new car this summer. I want to buy the fastest car I can and in my

Physics

1 answer:

Mashcka [7]2 years ago

7 0

Answer

Se togli 15 mph da 95 e 15, capisci quanto tempo la macchina 2 fa da 0 mph a 70 mph. La prima macchina fa da 0 mph a 60 mph in 5 secondi, e la seconda da 0 mph a 70 mph in 5 secondi. Risulta essere più veloce la seconda macchina. Spero di essere stato utile :)

Explanation:

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A particle with charge 8 µC is located on the x-axis at the point −10 cm , and a second particle with charge 3 µC is placed on t

bixtya [17]

Answer:Force on -7 uC charge due to charge placed at x = - 10cm

now we will have

towards left

similarly force due to -5 uC charge placed at x = 6 cm

now we will have

towards left

Now net force on 7 uC charge is given as

towards left

Explanation:

6 0

3 years ago

How fast can the 140 a current through a 0.200 h inductor be shut off if the induced emf cannot exceed 80.0 v?

Vesna [10]

Recall that to compute for the emf of a circuit given current and inductance, we must recall that

$emf = - M \frac{\Delta I }{\Delta t}$

where I is the current (A), M is the mutual inductance (h), and t is the time (ms). Since the current must not exceed 80.0 V, we have

$80.0 \geq 0.200(\frac{140}{t})$
$t \geq \frac{28.0}{80}$
$t \geq 0.35$

From this, we see that it must take at least 0.35 ms so it doesn't exceed 80 V.
Answer: 0.35 ms

7 0

2 years ago

An automobile battery has an emf of 12.6 V and an internal resistance of 0.0600 . The headlights together present equivalent res

murzikaleks [220]

Answer:

(a) $V=11.86\ V$

(b) $V=9.76\ V$

Explanation:

<u>Electric Circuits</u>

Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

$V=R.I$

(a) The electromagnetic force of the battery is $\varepsilon =12.6\ V$ and its internal resistance is $R_i=0.06\ \Omega$ . Knowing the equivalent resistance of the headlights is $R_e=5.2\ \Omega$ , we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

$\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)$

Solving for i

$\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A$

i=2.28\ A

The potential difference across the headlight bulbs is

$V=\varepsilon -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V$

$V=11.86\ V$

(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is

$V=\varepsilon -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V$

5 0

3 years ago

What is a supernova?

kozerog [31]

A supernova is a star that suddenly increases greatly in brightness because of a catastrophic explosion that ejects most of its mass.

8 0

2 years ago

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number = 2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2 m

and when x = x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is 0.0549 m

5 0

3 years ago