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krok68 [10]
3 years ago
10

In 3 seconds a car moving in a straight line increases its speed from 22.4 m/s to 29.1 m/s while a truck increases its speed fro

m 0 mph to 6.7m/s in the same amount of time. What is the acceleration of the truck in m/s? *
SHOW ALL WORK
Physics
1 answer:
ANTONII [103]3 years ago
6 0
Is there a picture?
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8 0
3 years ago
¿A qué velocidad debe circular un auto de carreras para recorrer 87 km en 20 min? (pasar a metros por segundo m/s)
Alexxx [7]

Answer:

Velocidad en m / s = 72,25 m / s

Explanation:

Dado

Distancia a recorrer por el coche de carreras = 87 Km

1 km = 1000 m

Por lo tanto, 87 km = 87000 m

Tiempo necesario para viajar 87 km / 87000 metros = 20 minutos o 20 * 60 = 1200 segundos

Velocidad en m / s = 87000/1200

Velocidad en m / s = 72,25 m / s

7 0
3 years ago
A 1542 kg car has a speed of 13.4 m/s when
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provide formulas to help you more

3 0
3 years ago
A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.
N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2

(c) Moment of inertia :

The net torque acting on it is, \tau=I\alpha, I is the moment of inertia

Also, \tau=Fr

So,

I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2

Hence, this is the required solution.

3 0
3 years ago
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