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ElenaW [278]
3 years ago
11

A tuning fork produced four beats per second with a second, 270-Hz tuning fork. What are the two possible frequencies of the fir

st tuning fork?
Physics
1 answer:
EleoNora [17]3 years ago
7 0

Answer:

frequency of the first tuning fork can be 274 Hz or 266 Hz

Explanation:

Given data :

we have number of beats per second = 4

thus, it can also be written as

Beat frequency of the tuning fork, v = 4 Hz

Frequency of the tuning fork, f = 270 Hz

now, the beat frequency is actually the difference of the consecutive frequency.

thus, we have

Frequency of the first tuning fork as = 270 Hz + 4 Hz = 274 Hz

or

Frequency of the first tuning fork as = 270 Hz - 4 Hz = 266 Hz

Therefore, the frequency of the first tuning fork can be 274 Hz or 266 Hz

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The driver of a car slams on the brakes, causing the car to slow down at a rate of
sdas [7]

Answer:

A. The time taken for the car to stop is 3.14 secs

B. The initial velocity is 81.64 ft/s

Explanation:

Data obtained from the question include:

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Final velocity (V) = 0

Time (t) =?

Initial velocity (U) =?

A. Determination of the time taken for the car to stop.

Let us obtain an express for time (t)

Acceleration (a) = Velocity (V)/time(t)

a = V/t

Velocity (V) = distance (s) /time (t)

V = s/t

a = s/t^2

Cross multiply

a x t^2 = s

Divide both side by a

t^2 = s/a

Take the square root of both side

t = √(s/a)

Now we can obtain the time as follow

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Time (t) =..?

t = √(s/a)

t = √(256/26)

t = 3.14 secs

Therefore, the time taken for the car to stop is 3.14 secs

B. Determination of the initial speed of the car.

V = U + at

Final velocity (V) = 0

Deceleration (a) = –26ft/s2

Time (t) = 3.14 sec

Initial velocity (U) =.?

0 = U – 26x3.14

0 = U – 81.64

Collect like terms

U = 81.64 ft/s

Therefore, the initial velocity is 81.64 ft/s

7 0
3 years ago
Guys No one's answering my question so sad! Once again I'm asking the same question –Here
nexus9112 [7]

For the front glass of the car to get wet, V_c \geq 10 \ m/s.

The given parameters:

  • <em>Speed of the car, = Vc</em>
  • <em>Speed of the rain, = 10 m/s</em>

The relative velocity of the car with respect to the falling rain is calculated as;

V_{C/R} = V_C- V_R

  • If the speed of the car equals the speed of the rain, the rain will fall behind the car.
  • If the speed of the rain is greater than speed of the car, the rain will fall far in front of the car.
  • If the speed of the car is greater than speed of the rain, the rain will fall on the car.

Thus, for the front glass of the car to get wet, V_c \geq 10 \ m/s.

Learn more about relative velocity here: brainly.com/question/17228388

8 0
3 years ago
A ball is thrown horizontally from the top of a building 37.5 m high. The ball strikes the ground at a point 80.3 m from the bas
trasher [3.6K]

Answer:

t = 2.77 s

Explanation:

The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:

Equation of movement of the ball in the X axis

X = v₀x*t   Equation  (1)

Equation of movement of the ball in the Y axis

Y = y₀+ v₀y*t -½ g*t² Equation  (2)

Where

X : horizontal position in meters (m)  

Y : vertical  position in meters (m)  

y₀ : initial vertical  position in meters (m)

v₀x :  X-initial speed in m/s  

v₀y :  Y-initial speed in m/s  

g: acceleration due to gravity in m/s²

t : time to position (X,Y)

Data

y₀ = 37.5 m

v₀y = 0

g = 9.8 m/s²

Problem development

The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0

We apply the Equation (2):

Y = y₀+ (v₀y)*t - (½) g*t²

0 =  37.5 +(0)*t- (1/2)*(g)*t²

0 =  37.5 - (1/2)*(9.8)*t²

(1/2)*(9.8)*t²  = 37.5

t² = (2)(37.5)/(9.8)

t= \sqrt{\frac{2*37.5}{9.8} }

t = 2.77 s

5 0
3 years ago
Free coins be quick to get them
olga nikolaevna [1]

Answer:

yeetshhdbdbdhdhdxsudhhdbdjdjd

7 0
3 years ago
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The bones of the forearm (radius and ulna) are hinged to the humerus at the elbow. The biceps muscle connects to the bones of th
sertanlavr [38]

Answer:

The force exerted by the biceps is 143.8 kgf.

Explanation:

To calculate the force exerted by the biceps, we calculate the momentum in the elbow.

This momentum has to be zero so that her forearm remains motionless.

Being:

W: mass weight (6.15 kg)

d_W= distance to the mass weight (0.425 m)

A: weight of the forearm (2.25 kg)

d_A: distance to the center of mass of the forearm (0.425/2=0.2125 m)

H: force exerted by the biceps

d_H: distance to the point of connection of the biceps (0.0215 m)

The momemtum is:

H*d_H-A*d_A-W*d_W=0\\\\H=(A*d_A+W*d_W)/d_H\\\\H=(2.25*0.2125+6.15*0.425)/0.0215\\\\H=(0.478125+2.61375)/0.0215\\\\H=3.091875/0.0215=143.8

The force exerted by the biceps is 143.8 kgf.

8 0
4 years ago
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