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Elza [17]
3 years ago
6

The duration between the emission of the sound from the echo reception is: t - 1.5 s. the distance between the observer and the

sound receiver knowing that the speed of sound in the air is worth
V - 340 m/s is:
Physics
2 answers:
ryzh [129]3 years ago
7 0


A sound wave has a frequency of 60.0 Hz. What is its period? If the speed of sound in air is 343 m/s, what is the wavelength of the sound wave? 25. Water waves in ...
algol133 years ago
4 0

Answer:

. tt tt t t t t t t tt t tt t t jk j k j k j k j k jk k j k jk

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During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between
Whitepunk [10]

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

4 0
3 years ago
Read 2 more answers
Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed
olasank [31]

Answer:

Average acceleration is (11.05)g\ m/s^2

Explanation:

It is given that,

Initial velocity, u = 0

Final velocity, v = 6.5 km/s = 6500 m/s

Time taken, t = 60 s

Acceleration, a=\dfrac{v-u}{t}

a=\dfrac{v}{t}  

a=\dfrac{6500\ m/s}{60}  

a=108.33\ m/s^2

Since, g=9.8\ m/s^2

So, a=(11.05)g\ m/s^2

So, the angular acceleration of the missile is (11.05)g\ m/s^2. Hence, this is the required solution.

4 0
3 years ago
A computational model predicts the maximum kinetic energy a roller coaster car can have given its mass, the speed at the highest
Ymorist [56]

Answer:

D

Explanation:

The decrease in potential energy is equal to the increase in kinetic energy.

mgh

250 x 9.8 x 30

=73, 500

3 0
3 years ago
Studen
Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

a

From this motion what is the distance covered

b

What is the magnitude and direction of the displacement

Answer:

a

  D =   72 \  m

b

Magnitude

             d =  - 20 \ m

 Direction  

West

Explanation:

From the question we are told that

The first distance covered westward is d_w_1  =  18 \  m /tex]     The  first distance covered eastward is [tex]d_e1 =  12 \  m /tex]      The second distance covered westward is [tex]d_w_2  =  28 \  m /tex]      The  second distance covered eastward is [tex]d_e2 =  14 \  m /tex]   Generally the distance covered is mathematically represented as       [tex]D =  d_w1 + d_w2 + d_e1 + d_e2

=> D =   18 + 28 + 12 +  14

=> D =   72 \  m

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

d =  -d_w1 -d_w2 + d_e1 + d_e2

=>  d =  -18-28 + 12 + 14

=> d =  - 20 \ m

The direction is west

7 0
3 years ago
The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kineti
ivanzaharov [21]

Answer:

KE = KE (incidental) - KE of emitted photons

or KE = h * f - Wf

So   h * f = KE + Wf = 1.2 + 1.88 = 3.08    incident energy

If you double the frequency then h * f = 6.16

KE = 6.16 - 1.2 = 4.96 eV

7 0
3 years ago
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