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Ira Lisetskai [31]
2 years ago
8

By what percent is the torque of a motor decreased if its permanent magnets lose 6.5 % of their strength

Physics
1 answer:
djyliett [7]2 years ago
4 0
6 is the answer I remember the answer from when I took this and it was easy
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Erase all the trajectories, and fire the pumpkin vertically again with an initial speed of 14 m/s. As you found earlier, the max
yanalaym [24]

Answer:

\theta=39.49^{\circ}

Explanation:

Maximum height of the pumpkin, H_{max}=9.99\ m

Initial speed, v = 22 m/s

We need to find the angle with which the pumpkin is fired. the maximum height of the projectile is given by :

H_{max}=\dfrac{v^2\ sin^2\theta}{2g}

On rearranging the above equation, to find the angle as :

\theta=sin^{-1}(\dfrac{\sqrt{2gH_{max}}}{v})

\theta=sin^{-1}(\dfrac{\sqrt{2\times 9.8\times 9.99}}{22})

\theta=39.49^{\circ}

So, the angle with which the pumpkin is fired is 39.49 degrees. Hence, this is the required solution.

8 0
3 years ago
What type of chemical reaction is shown here? *
8090 [49]

Answer:

single replacement reaction

4 0
3 years ago
A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Ass
Vlada [557]
1) 15 / 12 = 1.25 ratio
2) to increase acceleration  1.25 times (with same F, or same engine) you have to lower mass 1.25 times
3) 1515/1.25 = 1212 kg

choose A

6 0
3 years ago
Paki po plss po plssssss​
slega [8]

Answer:A language is a structured system of communication. Language, in a broader sense, is the method of communication that involves the use of – particularly human – languages. The scientific study of language is called linguistics

Explanation:

7 0
3 years ago
A spaceprobe has an 29.0 m length when measured at rest. What length
elixir [45]

Answer:

The observer sees the space-probe 9.055m long.

Explanation:

Let L_0 be the length of the space-probe when measured at rest, and L be its length as observed by an observer moving at velocity v, then

(1).\: \: L = L_0\sqrt{1-\dfrac{v^2}{c^2} }

Now, we know that L_0 = 29.0m and v = 0.95c, and putting these into (1) we get:

L = 29\sqrt{1-\dfrac{(0.95c)^2}{c^2} }

L = 29\sqrt{1-0.95^2 }

\boxed{L = 9.055m}

Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.

3 0
3 years ago
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