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Effectus [21]
2 years ago
13

What are two ways an engineer can build a car in order for it to accelerate faster

Physics
1 answer:
Ket [755]2 years ago
4 0

Explanation:

Take F=ma

a = F/m

For a higher, F higher or m lower

Means higher horse power for engine or lower mass for the car

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4 0
3 years ago
In one of the original Doppler experiments, a tuba was played at a frequency of 64.0 Hz on a moving flat train car, and a second
wolverine [178]

Answer:

 f_{beat} = 1.64\ Hz

Explanation:

given,

frequency of tuba.f = 64 Hz

Speed of train approaching, v = 8.50 m/s

beat frequency = ?

using Doppler's effect formula

 f' = f(\dfrac{v}{v-v_s})

v_s is the velocity of the source

v is the speed of sound, v = 340 m/s

now,

 f' = 64\times (\dfrac{340}{340 - 8.50})

       f' = 65.64 Hz

now, beat frequency is equal to

 f_{beat} = f' - f

 f_{beat} = 65.64 - 64

 f_{beat} = 1.64\ Hz

hence, beat frequency is equal to 1.64 Hz

3 0
3 years ago
How does the law of conservation of energy apply to machines?
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4 0
3 years ago
Read 2 more answers
A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The
S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

\sum F_x = 0

f-N_w = 0

N_w = f

The forces in the horizontal direction would be,

\sum F_y = 0

N_f -W =0

N_f = W

The sum of Torques at equilibrium,

\sum \tau = 0

Wdcos\theta - N_wLsin\theta = 0

WdCos\theta = fLSin\theta

f = \frac{Wd}{Ltan\theta}

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

f_{max} = \mu W=\frac{Wd}{Ltan\theta}

\theta = tan^{-1} (\frac{d}{\mu L})

Replacing,

\theta = tan^{-1} (\frac{0.9}{0.42*2})

\theta = 46.975\°

Therefore the minimum angle that the person can reach is 46.9°

8 0
3 years ago
Q2) The position of an artillery, with a speed of projectile 650, which can fire in any direction above the horizontal plane and
Wittaler [7]

Answer:

The minimum time to reach the target is 2156s

Explanation:

Check attachment

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