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4 years ago

one us dollar = 0.73 euros (as of 8-07). Which is worth more, one dollar or one euro? How many dollars is one euro?

Physics

1 answer:

Rom4ik [11]4 years ago

8 0

1 Euro equals<span>1.17 </span>US Dollar

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To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi

abruzzese [7]

Answer:

(a) 5.73 * 10¹³ m/s²

(b) 4.99 * 10⁵ m/s

Explanation:

(a) We know that Electrical force is the product of electric charge and electric field:

F = qE

We also know that Force is the product of mass and acceleration:

F = ma

Therefore, equating both of them:

ma = qE

Acceleration, a, will be:

a = (qE) / m

Electric charge of a positron = 1.602 * 10^(-19) C

Mass of a positron = 9.11 * 10^(-31) kg

Acceleration will be:

a = (1.602 * 10^(-19)) * 9.11 * 10^(-31)) / 326

a = 5.73 * 10¹³ m/s²

(b) Acceleration is the time rate of change of velocity. It is given as:

a = v/t

Velocity, v, will then be:

v = a * t

Time, t = 8.7 * 10^(-9) s

Therefore, velocity will be:

v = 5.73 * 10^(13) * 8.7 * 10^(-9)

v = 49.85 * 10⁴ m/s = 4.99 * 10⁵ m/s

7 0

3 years ago

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ololo11 [35]

Answer:

asd

Explanation:

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3 years ago

An equation for the motion of an object is given as 2vt-4x=Bt^2 where B is a constant. The variable v indicates velocity, in met

Jlenok [28]

Answer:

acceleration m/s²

Explanation:

The units on the left side of the equation are units of length. The units on the right side of the equation must be units of length also. Acceleration times time squared gives length.

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3 years ago

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Which of the following shows the abbreviation for an SI unit of density? kg/m2 g/cm g/cm3 lb/mL

Ymorist [56]

the answer is g/cm^3

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3 years ago

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How much electrical energy is used by the river cooker oven rated 800 W ,230 V is switched on for 30 minutes

mariarad [96]

Answer:

E = 1440 kJ

Explanation:

It is given that,

Power of a cooker oven is 800 W

Voltage at which it is operated is 230 V

Time, t = 30 minutes = 1800 seconds

We need to find the electrical energy used by the cooker oven. The product of power and time is equal to the energy consumed. So,

$E=P\times t\\\\E=800\ W\times 1800\ s\\\\E=1440000\ J\\\\\text{or}\\\\E=1440\ kJ$

So, electrical energy of 1440 kJ is consumed by the cooker oven.

8 0

3 years ago