PHYSICS COLLEGE

A sound wave travels with a velocity of 330 m/s and has a frequency of 500 Hz. What is itswavelength?

Answers

Answer 1

Answer:

Wavelength = 0.66 meters

Explanation:

Given the following data;

Speed = 330 m/s

Frequency = 500 Hz

To find the wavelength;

Mathematically, wavelength is calculated using this formula;

$Wavelength = \frac {speed}{frequency}$

Substituting into the equation, we have;

$Wavelength = \frac {330}{500}$

Wavelength = 0.66 meters

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An object is constrained by a cord to move in a circular path of radius 1-m on a frictionless, horizontal surface. The cord will break if the tension exceeds 26.9-N. The maximum kinetic energy that this object can have is _____ J. Round your answer to the nearest whole number.
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Dogs can hear higher-pitched whistles that humans do. How do you think the sound frequencies that dogs can hear compare to the frequencies that humans can hear?

A toy car is tied to a string and pulled across a table horizontally. Which is thecorrect free-body diagram for this situation?
T
FN
FN
T
FN
EN
T
W
W
W
w
А
B
С
D
Ο Α. Α

Answers

y axis:NandW and also f x axis:T and F T away from car.

Air contains 78.08% nitrogen, 20.095% oxygen, and 0.93% argon. calculate the partial pressure of oxygen if the total pressure of the air sample is 1.7 atm.a.

Answers

partial pressure in a mixture of two or more gases will be given by formula

$P_(partial)$ = mole fraction of gas * total pressure

now here mole fraction is same as percentage of gas in the mixture

Now mole fraction of oxygen is 0.20095 (20.095%)

now here pressure of oxygen in the mixture is given as

$P_(o_2) = 0.20095 * P_(total)$

$P_(o_2) = 0.20095 * 1.7$

$P_(o_2) = 0.342 atm$

so pressure due to oxygen in the mixture will be 0.342 atm

Answer:

20.095

Explanation:

A train moves with a uniform velocity of 36km/hr 10sec. calculate the distance travelled

Answers

Given:-

Speed = 36 km/hr

converting speed into m/s

Speed = 36*5/18

Speed = 10 m/s

t = 10 sec

By using the Formula

Distance = Speed * time

D = 10*10

D = 100 m

Hope it helps....

A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam

Answers

Answer:

the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Explanation:

Given the data in the question;

initial velocity; u = 0 m/s

height; h = 2.5 m

we find the velocity of the ball just before it touches the foam.

using the equation of motion;

v² = u² + 2gh

we know that acceleration due gravity g = 9.81 m/s²

so we substitute

v² = ( 0 )² + ( 2 × 9.81 × 2.5 )

v² = 49.05

v = √49.05

v = 7.00357 m/s

Now as the ball touches the foam

final velocity v₀ = 0 m/s

compresses S = 3 cm = 0.03 m

v₀² = v² + 2as

we substitute

( 0 )² = 49.05 + 0.06a

0.06a = -49.05

a = -49.05 / 0.06

a = -817.5 m/s²

Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

What minimum distance would you have to hit a baseball from the center of the earth so that it would eventually reach the moon? Assume you can hit the ball directly along the line that connects the centers of the earth and moon. The distance between the centers of the earth and moon is ???? = 3.82 × 108 m.

Answers

Answer:

d = 3.44 x 10⁸ m

Explanation:

The minimum distance required will be the distance from the centre of the earth to a point where gravitational intensity due to both earth and moon becomes equal . Once this point is reached , moon will attract the baseball on its own .

Let this distance be d from the centre of the earth

So GM / d² = G m / ( 3.82 x 10⁸ - d )²

M is mass of the earth , m is mass of the moon

M / m = ( d / 3.82 x 10⁸ - d )²

5.972 x 10²⁴ / 7.34 x 10²² = ( d / 3.82 x 10⁸ - d )²

81.36 = ( d / 3.82 x 10⁸ - d )²

9.02 = d / 3.82 x 10⁸ - d

34.45 x 10⁸ - 9.02 d = d

34.45 x 10⁸ = 10.02 d

d = 3.44 x 10⁸ m

A golf ball is dropped from rest from a height of 9.50 m. It hits the pavement, then bounces back up, rising just 9.70 m before falling back down again. A boy catches the ball on the way down when it is 1.20 m above the pavement. Ignoring air resistance calculate the total amount of time the ball is in the air, from drop to catch?

Answers

The ball is in the air for 27.70m because
9.5+9.7=19.2...
9.7-1.2=8.5...
19.2+8.5=27.70...

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