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3 years ago

When measuring current, always place any type of meter in

2 answers:

5 0

In series with the circuit, so for it pass the current to be mensured.

Letter A

If you notice any mistake in my english, please let me know, because i am not native.

Delicious77 [7]3 years ago

4 0

Answer:

Series with the circuit

Explanation:

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What is the order of the events for the water cycle on a typical warm day?

Rudik [331]

B precipitation,condensation,precipitation

6 0

3 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an

a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, $\mu=6\times 10^{-2}$

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

$KE=PE+W$

$\dfrac{1}{2}mv^2=mgh+W$

W is the work done by the friction.

$W=f\times d$

$f=\mu mg\ cos\theta$

$W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}$

$W=\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}$

$\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}$

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0

3 years ago

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of

Aleonysh [2.5K]

Answer:

r = 0.0548 m

Explanation:

Given that,

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.

We need to find the radius of their circular path. The formula for the radius of path is given by :

$r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$