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3 years ago

13

Why must you use temperature to specify how hot or cold something is?

2 answers:

podryga [215]3 years ago

7 0

At others time we need to k ow how hot or cold something is when we need to measure temperature correctly we need to use an instrument called a thermometer this measure temperature in degrees Celsius or degrees Fahrenheit

neonofarm [45]3 years ago

3 0

Explanation:

Well to come to think because using the terms hot and cold can be confusing

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A ball that is attached to a string travels in a horizontal, circular path, as shown in Figure 1. At time t0 , the ball has a sp

Levart [38]

Answer:

F₁ = 4 F₀

Explanation:

The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:

F₀ = mv₀²/r --------------- equation (1)

where,

F₀ = Force on string at t₀

m = mass of ball

v₀ = speed of ball at t₀

r = radius of circular path

Now, at time t₁:

v₁ = 2v₀

F₁ = mv₁²/r

F₁ = m(2v₀)²/r

F₁ = 4 mv₀²/r

using equation (1):

<u>F₁ = 4 F₀</u>

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4 years ago

What is the equivalent resistance of a circuit that contains two 50.0 0

ruslelena [56]

Answer:

the correct answer is A :)

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3 years ago

Atoms of most elements are less likely to react when they have

sergejj [24]

Answer:

when they have 8 valence electrons

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3 years ago

How are scientists studying the effects of space on the human body?

Vsevolod [243]

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3 years ago

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A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

4 years ago