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scZoUnD [109]
3 years ago
8

Three cars with identical engines and tires start from rest, and accelerate at their maximum rate. Car X is the most massive, an

d car Z is the least massive. Which car needs to travel the farthest before reaching a speed of 60 mi/h?
A. Car Z
B. Car Y
C. Car X
D. All cars need to travel the same distance, although some cars will take longer than other cars.
Physics
1 answer:
aksik [14]3 years ago
6 0

Answer:C) car X

Explanation:

Given

All the cars have identical Engine thus Force Produced by car X will be equal to  Y and Z

and Force=mass\times acceleration

Since Car X is most massive so acceleration associated with it will be minimum

acceleration of car X is minimum thus it will travel farthest  

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Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thi
77julia77 [94]

Answer:

k1 + k2

Explanation:

Spring 1 has spring constant k1

Spring 2 has spring constant k2

After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.

x1 = x2

Since the force exerted to each spring might be different, let's assume F1 for spring 1 and F2 for spring 2. Hence the equations of spring constant for both springs are

k1 = F1/x -> F1 =k1*x

k2 = F2/x -> F2 =k2*x

While F = F1 + F2

Substitute equation of F1 and F2 into the equation of sum of forces

F = F1 + F2

F = k1*x + k2*x

= x(k1 + k2)

Note that this is applicable because both spring have the same extension of x (I repeat, EXTENTION, not length of the spring)

Considering the general equation of spring forces (Hooke's Law) F = kx,

The effective spring constant for the system is k1 + k2

3 0
3 years ago
You're staring at a traditional clock for 1 hour. after 1 hour, what is the angular displacement of the second hand? the minute
Anvisha [2.4K]
<span>During the hour, the second hand will have traveled a full 360 degrees times 60 for a total displacement of 21600 degrees or 21600*pi/180 = 120 pi radians, or approximately 376.99 radians. Also during that hour, the minute hand will have gone around a full 360 degrees, or 360*pi/180 = 2 pi radians, or approximately 6.28 radians. And finally, the hour hand will have gone 360/12 = 30 degrees or 30*pi/180 = pi/6 radians, or approximately 0.52 radians.</span>
6 0
3 years ago
Dispositivo que muestra tanto la dirección como la magnitud de la corriente eléctrica.
slamgirl [31]

Answer:

<em>Galvanómetro</em>

Explanation:

Un galvanómetro es un dispositivo eléctrico utilizado para <em>detectar la presencia de corriente y voltaje pequeños o para medir su magnitud.</em> Los galvanómetros se utilizan principalmente en puentes y potenciómetros.

5 0
3 years ago
The potential energy between two atoms in a particular molecule has the form U(x) = 2.1 x 8 − 5.2 x4 where the units of x are le
a_sh-v [17]

Answer:

x\approx 0.948

Explanation:

The correct formula for the potential energy between two atoms in a particular molecule is:

U(x) = \frac{2.1}{x^{8}}-\frac{5.2}{x^{4}}

Where x is the distance.

According to the definitions of potential energy and work, as well as the Work-Energy Theorem and the Principle of Energy Conservation. The relation between that and related force is:

F = -\frac{dU}{dx}

The function is derived in terms of distance:

F (x) = \frac{84}{5\cdot x^{9}} -\frac{104}{5\cdot x^{5}}

Then, it is needed to find at least of x so that F(x) equals to 0.

\frac{84}{5\cdot x^{9}}-\frac{104}{5\cdot x^{5}}=0

\frac{84}{x^{4}}-104 = 0

84-104\cdot x^{4} = 0

x=\sqrt[4]{\frac{84}{104} }

x\approx 0.948

7 0
3 years ago
A test charge of +4 µC is placed halfway between a charge of +6 µC and another of +2 µC separated by 20 cm. (a) What is the magn
S_A_V [24]

Answer:

(a) Magnitude: 14.4 N

(b) Away from the +6 µC charge

Explanation:

As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:

F_e = K\frac{qq_{test}}{r^2}

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.

Let's say that a force that goes toward the +6 µC charge is positive, then:

F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N

F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N

The magnitude will be:

F_e = -21.6 + 7.2 = -14.4 N, away from the +6 µC charge

3 0
4 years ago
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