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3 years ago

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What is the speed of a wave that has a frequency of 200 Hz and a

1 answer:

Luba_88 [7]3 years ago

8 0

The wave speed to this question is 400 meters

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The practice of playing top hits several times a day on radio stations is called _____. blocking rotation circulation format dri

spin [16.1K]

The practice of playing top hits several times a day on radio stations is called rotation.

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2 years ago

Read 2 more answers

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is:

$7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}$

$p_{B,2} = 3\,\frac{kg\cdot m}{s}$

The magnitude of the large object's momentum change is:

$p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}$

$p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}$

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

4 0

3 years ago

Mr.Z holds 200 kg above his head for 5 seconds. What is the work done on the weights?

Alenkasestr [34]

m = mass held by mr. Z above his head = 200 kg

g = acceleration due to gravity = 9.8 m/s²

F = force applied by mr. Z to hold the mass

Using equilibrium of force , force equation is given as

F = mg

F = (200) (9.8)

F = 1960 N

Since the mass is not moved,

d = displacement of the mass = 0 m

we know that , work done is given as

W = F d

inserting the values

W = (1960) (0)

W = 0 J

8 0

3 years ago

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a 200 WW electric im

ivolga24 [154]

Answer:

The heat is 115478.4 J.

Explanation:

Given that,

Mass of water = 0.400 kg

Power = 200 W

Suppose, we determine how much heat must be added to the water to raise its temperature from 20.0°C to 89.0°C?

We need to calculate the heat

Using formula of heat

$Q=mc\Delta T$

Where, m = mass of water

c = specific heat

Put the value into the formula

$Q=400\times4.184\times(89-20)$

$Q=115478.4\ J$

Hence, The heat is 115478.4 J.

7 0

3 years ago

An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi

Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

$a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2$

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

7 0

3 years ago