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lina2011 [118]
3 years ago
15

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

ately 57.0 m . If the track is completely flat and the race car is traveling at a constant 30.5 m/s (about 68 mph ) around the turn, what is the race car's centripetal (radial) acceleration
Physics
1 answer:
umka21 [38]3 years ago
7 0

Answer:

a=16.32\ m/s^2

Explanation:

Given that,

The radius of the track, r = 57 m

The speed of a race car, v = 30.5 m/s

We need to find the centripetal acceleration of the car. Its formula that is use to find it is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(30.5)^2}{57}\\\\=16.32\ m/s^2

So, the car's centripetal acceleration is 16.32\ m/s^2.

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¿cual es la velocidad de un auto que recorre 4566 metros en 4 minutos? expresar en km\h
Svetlanka [38]

Explanation:

(4566 m / 4 min) × (1 km / 1000 m) × (60 min / h) = 68.49 km/h

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3 years ago
Suppose the line on a distance-versus-time graph and the line on a speed-versus-time graph are both slanted straight lines going
lesya [120]
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3 years ago
g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl
Fynjy0 [20]

Answer:

v=1.5m/s

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}. Since v=r\omega (the ball rolls without slipping) and for a solid sphere I=\frac{2mr^2}{5}, we have:

mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}

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v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s

6 0
3 years ago
You are a crewman on a pirate ship. You are in a battle with the Royal Navy of England. You have loaded your cannon, which is no
Brut [27]

Answer:

Conservation of momentum - causes a recoil of cannon frame after launch

Explanation:

- It is simple, we can apply the conservation of momentum on the system. Initial momentum P_i before shooting is zero. Then after you strike the cannon up, the ball will go to the target however for the momentum to be conserved the cannon frame must move backwards with the same momentum with which the cannon ball is launched.

Hence, the recoil of the cannon frame would be large enough to dislodge you off the ground and throw you a few feet back.

6 0
3 years ago
Why does a wave undergo a 180° phase change at a fixed boundary?
Leno4ka [110]

Answer:

The phase change of 180^{o} can be theoretically understood as follows:

For transmission or propagation of waves between media the wave motion should maintain a principle of continuity meaning that the wave function at the interface should be continuous and diffrentiable at the interface.

At the point of incidence there are 2 types of waves reflected wave and the incident wave. Now the principle of continuity dictates that the sum of the phases of the above 2 waves should be same as that of transmitted wave. If we use these relations we notice that the reflected wave shall either change it's phase by 180^{o} or will not change it's phase depending on the relationship between the refractive indices of the incident and the reflecting medium. For a solid boundary a phase change of 180^{o} occurs.

6 0
3 years ago
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