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Furkat [3]
3 years ago
12

A doctor counts 68 heartbeats in 1.0 minute. What are the corresponding period and frequency of the heart rhythm

Physics
1 answer:
sergejj [24]3 years ago
8 0

Answer:

f=1.13s^{-1}=1.13Hz

Explanation:

Hello,

In this case, a frequency stands for a rate in which some action is done per unit of time. In this case, for the heartbeat, since 68 actions (heartbeats) occur in 1.0, the frequency turns out:

f=\frac{68}{1.0min}=68min^{-1}

Or as most commonly used in Hz (s^{-1}):

f=68\frac{1}{min} *\frac{1min}{60s}=1.13s^{-1}=1.13Hz

Best regards.

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The electric motor of a model train accelerates the train from rest to 0.540 m/s in 27.0 ms. The total mass of the train is 610
Gnesinka [82]

Answer:

The value  is  P  =3.294 \  W

Explanation:

From the question we are that

     The velocity  v  =  0.540  \  m/s

      The  time taken is  t =  27.0 ms  = 27.0 *10^{-3} \ s

     The  total mass of the train is  m  =  610 \ g  =  0.610 \ kg

Generally the average power delivered is mathematically represented as

     P  =\frac{KE }{t}    

     P  =\frac{  \frac{1}{2}  *  m  *  v^2 }{t}    

=>  P  =\frac{  \frac{1}{2}  *  0.610   *   0.540 ^2 }{ 27.0 *10^{-3}}    

=>  P  =3.294 \  W  

5 0
2 years ago
Read 2 more answers
One solution to minimize resonance with buildings is to ______ the width to span ratio.
Shtirlitz [24]

Answer:

Increase

Explanation:

Resonance is a phenomenon which occurs when a body A in motion set another body B into motion of it own natural frequency. So for resonance to be minimize in a body is to increase the width to span ratio. So as to reduce the overall vibration which affects directly building resonance, the stiffness or trusses and girders should be increase. The increase in this aspect helps to reinforce building structure and support.

3 0
3 years ago
Swinging a golf club or baseball bat are examples of ______________ stretching.
kari74 [83]

Answer:

Static stretching is the answer.

Explanation:

Static stretching is the most common form that greatly improves flexibility. However, static stretches does little to contract the muscles needed to generate powerful golf swings. Dynamic stretches help improve your range of motion while reducing muscle stiffness.

6 0
3 years ago
What are the names of the 4 types of fronts? How are they created?
jeka57 [31]

Answer:

Stationary Front, warm front, cold front, Occluded Front.

Explanation:

Stationary Front. When the surface position of a front does not change (when two air masses are unable to push against each other; a draw), a stationary front is formed.

cold front is the leading edge of a cooler mass of air at ground level that replaces a warmer mass of air and lies within a pronounced surface trough of low pressure. It often forms behind an extratropical cyclone (to the west in the Northern Hemisphere, to the east in the Southern), at the leading edge of its cold air advection pattern—known as the cyclone's dry "conveyor belt" flow. Temperature differences across the boundary can exceed 30 °C (86 °F) from one side to the other. When enough moisture is present, rain can occur along the boundary. If there is significant instability along the boundary, a narrow line of thunderstorms can form along the frontal zone. If instability is weak, a broad shield of rain can move in behind the front, and evaporative cooling of the rain can increase the temperature difference across the front. Cold fronts are stronger in the fall and spring transition seasons and weakest during the summer.

A warm front is a density discontinuity located at the leading edge of a homogeneous warm air mass, and is typically located on the equator-facing edge of an isotherm gradient. Warm fronts lie within broader troughs of low pressure than cold fronts, and move more slowly than the cold fronts which usually follow because cold air is denser and less easy to remove from the Earth's surface. This also forces temperature differences across warm fronts to be broader in scale. Clouds ahead of the warm front are mostly stratiform, and rainfall gradually increases as the front approaches. Fog can also occur preceding a warm frontal passage. Clearing and warming is usually rapid after frontal passage. If the warm air mass is unstable, thunderstorms may be embedded among the stratiform clouds ahead of the front, and after frontal passage thundershowers may continue. On weather maps, the surface location of a warm front is marked with a red line of semicircles pointing in the direction of travel.

In meteorology, an occluded front is a weather front formed during the process of cyclogenesis. The classical view of an occluded front is that they are formed when a cold front overtakes a warm front, such that the warm air is separated (occluded) from the cyclone center at the surface. The point where the warm front becomes the occluded front is called the triple point; a new area of low-pressure that develops at this point is called a triple-point low. A more modern view of the formation process suggests that occluded fronts form directly during the wrap-up of the baroclinic zone during cyclogenesis, and then lengthen due to flow deformation and rotation around the cyclone.

3 0
2 years ago
Read 2 more answers
You hear a sound with a frequency of 256 Hz. The amplitude of the sound increases and decreases periodically: it takes 2 seconds
german

To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.

The relationship that defines both agreements is given by the equation,

f_{beat}=\frac{1}{T}

Then the frequency for the previous period given (2sec) is

f_{beat}=\frac{1}{2}

f_{beat} = 0.5Hz

The beat frequency of two frequencies is equal to the difference between the two frequencies, then

f_{beat} = |f_1-f_2|\\f_{beat} = |256Hz-2Hz|\\f_{beat} = 254Hz

<em>Hence option A is incorrect.</em>

We can do this process for 254Hz as f_1 and 258 Hz for f_2 , then

f_{beat} =|254Hz-258Hz|

f_{beat} = 4Hz

<em>Hence option B is incorrect. </em>

We can also do this process for 255Hz as  f_1 and 257 Hz for f_2 , then

f_{beat} =|255Hz-257Hz|

f_{beat} = 2Hz

<em>Hence option C is incorrect. </em>

We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then

f_{beat} =|255.5Hz-256.5Hz|\\f_{beat} = 1Hz

<em>Hence option D is incorrect. </em>

We can also do this process for 255.75Hz as f_1 and 256.25 Hz for f_2, then

f_{beat} =|255.75Hz-256.25Hz|\\f_{beat} = 0.5Hz

<em>Hence option E is incorrect. </em>

Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz

3 0
3 years ago
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