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2 years ago

15

A bunny hopping along the road can travel at an average of 0.6 m/s. Calculate far the bunny can travel in 11 seconds.

1 answer:

galben [10]2 years ago

3 0

Answer:

6.6 m

Explanation:

.6 m/s * 11 s = 6.6 m ( see how the 's' cancels out and you are left with 'm' as the answer ?)

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2 years ago

Read 2 more answers

Suppose a straight wire with a length of 2.0 m runs perpendicular to a magnetic field with a magnitude of 38 T. What current wou

iragen [17]

Answer:

9.67 A

Explanation:

The weight of a student with a mass of m = 75 kg is:

$W=mg=(75 kg)(9.8 m/s^2)=735 N$

where g=9.8 m/s^2 is the acceleration due to gravity.

We want the magnetic force on the wire to be equal to this weight. The magnetic force on the wire is

$F=ILB sin \theta$

where

I is the current in the wire

L = 2.0 m is the length of the wire

B = 38 T is the magnetic field

$\theta=90^{\circ}$ is the angle between the direction of B and L

Since we want W=F, we can write

$ILB sin \theta=W$

And we can solve it to find the current I:

$I=\frac{W}{BLsin\theta}=\frac{735 N}{(38 T)(2.0 m)(sin 90^{\circ})}=9.67 A$

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3 years ago

A 4.50-kg wheel that is 34.5 cm in diametet rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1

lisabon 2012 [21]

Answer:

$\alpha =10.93radian/sec^2$

Explanation:

We have given given the final angular velocity $\omega _{final}=13.5rad/sec$

And $\omega _{initial}=22rad/sec$

Displacement $\Theta =13.8radian$

We have to find the angular acceleration $\alpha$

According to law of motion $\omega _{final}^2=\omega _{initial}^2+2\alpha \Theta$

So $13.5^2=22^2+2\times \alpha \times 13.8$

$\alpha =-10.93radian/sec^2$

In question we have tell about magnitude only so $\alpha =10.93radian/sec^2$

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2 years ago