Ask question

Ask question

All categories

2 years ago

13

What happens to the velocity of a sound wave in air if the temperature of the air increases?

2 answers:

Yuki888 [10]2 years ago

6 0

The answer is D :) hope this helps

Burka [1]2 years ago

5 0

Decreases ( no B is answer)

You might be interested in

Where do tadpoles live? a. on land c. both on land and water b. in water d. none of the above

Bezzdna [24]

Tadpoles live in water

5 0

2 years ago

Read 2 more answers

Which type of energy is released when a bond between atoms is broken

scoray [572]

Answer:

D

Explanation:

7 0

3 years ago

A ship sets sail from Rotterdam, The Netherlands, heading due north at 8.00 m/s relative to the water. The local ocean current i

Nuetrik [128]

Answer:

The velocity of the ship relative to the earth V = 9.05 $\frac{m}{s}$

Explanation:

The local ocean current is = 1.52 m/s

Direction $\theta$ = 40°

Velocity component in X - direction $V_{x}$ = 1.52 $\cos 40$ °

$V_{x}$ = 1.164 $\frac{m}{s}$

Velocity component in Y - direction $V_{y}$ = 8 + 1.52 $\sin 40$ °

$V_{y}$ = 8.97 $\frac{m}{s}$

The velocity of the ship relative to the earth

$V = \sqrt{V_{x}^{2} + V_{y} ^{2} }$

Put the values of $V_{x}$ and $V_{y}$ we get,

⇒ $V = \sqrt{1.164^{2} + 8.97 ^{2} }$

⇒ V = 9.05 $\frac{m}{s}$

This is the velocity of the ship relative to the earth.

7 0

3 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

2 years ago

A pool and stops at the

Bas_tet [7]

In Linear motion the swimmer swims

8 0

2 years ago