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Amiraneli [1.4K]
3 years ago
14

What device is used to measure the current in a circuit

Physics
1 answer:
Natalka [10]3 years ago
8 0
The device is called the ammeter 
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What does the stomach acid have in protein digestion
castortr0y [4]
It contains protease which is the enzyme that breaks down protein
5 0
3 years ago
A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum "is 2.51 s". The temp
Artyom0805 [142]

Answer:

0.0034 sec

Explanation:

L = initial length

T = initial time period = 2.51 s

Time period is given as

T = 2\pi \sqrt{\frac{L}{g}}

2.51 = 2\pi \sqrt{\frac{L}{9.8}}

L = 1.56392 m

L' = new length

ΔT = Rise in temperature = 142 °C

α = coefficient of linear expansion = 19 x 10⁻⁶ °C

New length due to rise of temperature is given as

L' = L + LαΔT

L' = 1.56392 + (1.56392) (19 x 10⁻⁶) (142)

L' = 1.56814 m

T' = New time period

New time period is given as

T' = 2\pi \sqrt{\frac{L'}{g}}

T' = 2\pi \sqrt{\frac{1.56814}{9.8}}

T' = 2.5134 sec

Change in time period is given as

ΔT = T' - T

ΔT = 2.5134 - 2.51

ΔT = 0.0034 sec

5 0
3 years ago
A 1050 W carbon-dioxide laser emits light with a wavelength of 10μm into a 3.0-mm-diameter laser beam. What force does the laser
avanturin [10]

The force exerted by the laser beam on a completely absorbing target is 3.5 \times 10^{-6} \ N.

The given parameters;

  • <em>power of the laser light, P = 1050 W</em>
  • <em>wavelength of the emitted light, λ = 10 μm </em>

The speed of the emitted laser light is given as;

v = 3 x 10⁸ m/s

The force exerted by the laser beam on a completely absorbing target is calculated as follows;

P = Fv

F = \frac{P}{v} \\\\F = \frac{1050}{3\times 10^8} \\\\F = 3.5 \times 10^{-6} \ N

Thus, the force exerted by the laser beam on a completely absorbing target is 3.5 \times 10^{-6} \ N.

Learn more here:brainly.com/question/17328266

3 0
2 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
Consider a situation where the acceleration of an object is always directed perpendicular to its velocity. This means that
Bond [772]

Answer:

this situation would not be physically possible

7 0
3 years ago
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