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denpristay [2]
3 years ago
5

A 2.43ug particle moves at 1.97 x 108 m/s. What is its momentum?

Physics
1 answer:
Ira Lisetskai [31]3 years ago
4 0

Answer:

0.48 kgm/s

Explanation:

m = mass of the particle = 2.43 μg = 2.43 x 10⁻⁶ x 10⁻³ kg = 2.43 x 10⁻⁹ kg

v  = velocity of the particle = 1.97 x 10⁸ m/s

p = momentum of the particle

momentum of the particle is given as

p = m v

inserting the values

p = (2.43\times 10^{-9})(1.97\times 10^{8})

p = 0.48 kgm/s

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The voltage across the terminals of a generator is 5.7 v when it supplies a current of 0.3 A. It becomes 5.1 V when I=0.9A. Find
snow_tiger [21]

Answer:

  • The emf of the generator is 6V
  • The internal resistance of the generator is 1 Ω

Explanation:

Given;

terminal voltage, V = 5.7 V, when the current, I = 0.3 A

terminal voltage, V = 5.1 V, when the current, I = 0.9 A

The emf of the generator is calculated as;

E = V + Ir

where;

E is the emf of the generator

r is the internal resistance

First case:

E = 5.7   + 0.3r -------- (1)

Second case:

E = 5.1 + 0.9r -------- (2)

Since the emf E, is constant in both equations, we will have the following;

5.1 + 0.9r = 5.7   + 0.3r  

collect similar terms together;

0.9r - 0.3r = 5.7 - 5.1

0.6r = 0.6

r = 0.6/0.6

r = 1 Ω

Now, determine the emf of the generator;

E = V + Ir

E = 5.1 + 0.9x1

E = 5.1 + 0.9

E = 6 V

6 0
3 years ago
1) How is the temperature of a gas related to the kinetic energy of its particles?
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3 years ago
A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman,
lina2011 [118]

Answer:

44.1 m

Explanation:

<u>Given:</u>

  • V_a = speed of sound in air = 343 m/s
  • V_r = speed of sound in the rod = 15V_a
  • \Delta t = times interval between the hearing the sound twice = 0.12 s

<u>Assumptions:</u>

  • l = length of the rod
  • t = time taken by the sound to travel through the rod
  • T = time taken by the sound to travel to through air to the same point = t+\Delta t = t+0.12\ s

We know that the distance traveled by the sound in a particular medium is equal to the product of the speed of sound in that medium and the time taken.

For traveling sound through the rod, we have

l=V_r t\\\Rightarrow t = \dfrac{l}{V_r}..........eqn(1)

For traveling sound through the air to the women ear for traveling the same distance, we have

l=V_aT\\\Rightarrow l=V_a(t+0.12)\\\Rightarrow l=V_a(\dfrac{l}{V_r}+0.12)\,\,\,\,\,\,(\textrm{From eqn (1)})\\\Rightarrow l=V_a(\dfrac{l}{15V_a}+0.12)\\\Rightarrow l=\dfrac{l}{15}+0.12V_a\\\Rightarrow l-\dfrac{l}{15}=0.12V_a\\\Rightarrow \dfrac{14l}{15}=0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = 44.1\ m

Hence, the length of the rod is 44.1 m.

4 0
3 years ago
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