How many miles per day can you walk at a MODERATE Intensity level and your heart rate is 170?

Answer:

Resistance is the opposing of the flow of current through a conductor.

Answer:

[tex]500 \: \mathrm{kg} \cdot \mathrm{m/s}[/tex]

Explanation:

The momentum of an object is given as [tex]p=mv[/tex]. Since Amy has a mass of 50 kg and is travelling 10 m/s, her momentum is [tex]p=mv=50\cdot 10 =\fbox{$500\: \mathrm{kg\cdot m/s}$}[/tex].

Answer:

500

Explanation:

Answer:

Ribosomes are organelles the make protein for the cell.

Answer:

3.85s

Explanation:

Given parameters:

Wavelength = 25m

Velocity  = 6.5m/s

Frequency  = 0.26Hz

Unknown:

Period of the wave = ?

Solution:

The period of a wave is the inverse of the frequency of the wave.

    Period  = [tex]\frac{1}{frequency}[/tex]  

  Period = [tex]\frac{1}{0.26}[/tex]   = 3.85s

Answer:

7.01yard/sec

Explanation:

Given parameters:

Initial position  = 50yard

Final position = 12yard

Time  = 5.42s

Unknown:

Average speed of runner  = ?

Solution:

To solve this problem;

        Speed  = [tex]\frac{distance}{time}[/tex]  

Distance covered  = Initial position  - final position  = 50 - 12  = 38yards

So;

       Speed  = [tex]\frac{38}{5.42}[/tex]   = 7.01yard/sec

Answer:

the answer is B

Explanation:

I think its B because on the top it shows the molecule speed and A looks like the water is cold, C shows that the hot

water is cooler, and D shows that both are cold

Answer:

B.

Explanation:

Answer:

the total distance is 5km and the displacement is 1km

Explanation:

The total distance would be the addition of John running both ways so 3 km, 2 km.

However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.

Think about 2 km as a positive value for the first part of the question and a negative value for the second part.

Answer: B- a doctor, by law, must treat every patient who shows up in a medical facility

Explanation:

Answer:

the correct answer is B

Explanation:

Answer:

.B

Explanation:

This is the answer on edge 2021

Have a good day!

Answer:

[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]

Explanation:

Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].

However, it's also given as change in momentum (impulse-momentum theorem).

Therefore, we can set the change in momentum equal to the former formula for impulse:

[tex]\Delta p=F\Delta t[/tex].

Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:

[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].

Now we plug in our values and solve:

[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).

The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.  

Closeness to the Sound Source: When a rocket is fired on the launchpad, it creates a tremendous amount of noise in proximity to the nearby equipment and structures.

When a rocket is launched, concentrated sound waves are created that can seriously harm neighboring structures, especially if such structures are not built to handle such strong vibrations. In contrast, once a rocket is in the air, the sound waves spread out and become less forceful as they travel through the atmosphere, decreasing the possibility that they may cause harm.

Reflection and Amplification: The launchpad environment can serve as an echo chamber for sound waves because of its huge, solid structures.

Hence, The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.  

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Answer:

because they are the rocks that line the surface of our planet ​

Explanation:

We see sedimentary rocks more than other rock types because they are the rocks that line the surface of our planet.

Sedimentary rocks typically form the earth cover due to the way they are formed.

Answer:

every number to 3 sf = 1) 45.0  2) 250  3) 1.30

Explanation:

your welcome :)

Answer:

C. both have same K.E

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

[tex] Momentum = Mass * Velocity [/tex]

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

[tex] K.E = \frac{1}{2}MV^{2}[/tex]

Where, K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Both momentum and kinetic energy are related to the velocity of an object or a body.

Since the horse and a dog have same momentum. Thus, they both have same kinetic energy.

Answer:

The solution to this question can be defined as follows:

Explanation:

In point (a):

[tex]v_i= 100 \ mV\\\\ i_I = 100 \mu \ A\\\\ v_O = 10 \ V\\\\ R_L = 100 \ \Omega \\\\i_L = \frac{V_0}{R_L} = \frac{10}{100} = 100 \ MA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{100 \times 10^{-3}} =100 \\\\A_v(db) = 20 \lag (100) =40 \ db \\\\ A_i= \frac{i_L}{i_i} = \frac{100 \times 10^{-3}}{100 \times 10^{-6}} =1000 \\\\A_i(db) = 20 \lag (100) =60 \ db \\\\[/tex]

[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_L}{v_i i_i} = \frac{ 10(100 \times 10^{-3})}{100 \times 10^{-6} \times 100 \times 10^{-3}} =100000\\\\ A_p(db) =10 \log (100000) =50 \ db \\\\[/tex]

In point (b):

[tex]v_i = 10 \mu V\\\\ i_i = 100 \ nA \\\\ v_O =1 \ V \\\\ R_L= 10 \ k \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{1}{10 \ K} = 100 \ \muA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{10 \times 10^{-6}} =100000 \\\\A_v(db) = 20 \lag (100000) =100 \ db \\\\ A_i= \frac{i_0}{i_i} = \frac{100 \times 10^{-6}}{100 \times 10^{-9}} =1000 \\\\A_i(db) = 20 \lag (1000) =60 \ db \\\\[/tex]

[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 1 \times 100 \times 10^{-6})}{10 \times 10^{-6} \times 100 \times 10^{-9}} =100000000\\\\A_p(db) =10 \log (100000000) =80 \ db \\\\[/tex]

In point (C):

[tex]v_i =1\ V\\\\ i_I = 1 \ mA\\\\ v_O =5\ V \\\\ R_L = 10 \ \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{5}{10 } = 0.5 \ A \\\\A_v = \frac{V_0}{V_i} = \frac{5}{1} =5 \\\\A_v(db) = 20 \log 5 =13.97 \ db = 14 \db \\\\ A_i= \frac{i_0}{i_i} = \frac{0.5}{1\times 10^{-3}} =500 \\\\A_i(db) = 20 \log (500) =53.97 \ db = 54 \db \\\\[/tex]

[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 5 \times 0.5 }{1 \times 1 \times 10^{-3}} =2500\\\\A_p(db) =10 \log (2500) = 33.97 \ db = 34 \db\\\\[/tex]

Answer:

Twice.

Explanation:

The momentum of an object is given by :

p = mv

Where

m is mass and v is the velocity

If the mass of the ball were doubled, m'=2m and v'=v=3 m/s

New momentum,

p'=m'v'

p'=2m × v

p'=2mv

or

p'=2p

So, the new momentum becomes twice the initial momentum.

The change in internal energy of the engine is -50 joule. Hence, option (B) is correct.

Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.

The absorb energy: Q = 600 Joule

Work done: W = 650 Joule.

Let, the change in internal energy of the engine= dU.

According to conservation of energy:

The absorb energy =  change in internal energy  + Work done

Q = dU + W

dU = Q - W

= 600 joule - 650 joule

= - 50 joule.

Hence, the change in internal energy of the engine is -50 joule.

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Answer:

[tex]10\: \mathrm{J}[/tex]

Explanation:

The kinetic energy of an object is [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.

The toy car's initial kinetic energy is [tex]KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}[/tex].

After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.

To find the final velocity of the toy car, we can use kinematic equation [tex]v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}[/tex]

We are given [tex]v_i=5\: \mathrm{m/s}[/tex] and [tex]\Delta x = 2\: \mathrm{m}[/tex].

To find acceleration:

[tex]F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}[/tex].

Now substitute [tex]v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}[/tex] into [tex]v_f=\sqrt{v_i^2+2a\Delta x}[/tex] to get [tex]v_f\approx 5.92\: \mathrm{m/s}[/tex].

Using this, we can find the final kinetic energy of the toy car is [tex]KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2[/tex].

Thus, the change in kinetic energy is [tex]KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{$10\: \mathrm{J}$}[/tex] (one significant figure).

The change in the kinetic energy of the car is 10 J.

The given parameters;

The acceleration of the car is calculated as follows;

[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{2} \\\\a = 2.5 \ m/s^2[/tex]

The final velocity of the car is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\v = \sqrt{u^2 + 2as} \\\\v = \sqrt{5^2 \ + \ 2(2.5)(2)} \\\\v = 5.92 \ m/s[/tex]

The change in the kinetic energy of the car is calculated as follows;

[tex]\Delta K.E = \frac{1}{2} m(v^2 - u^2)\\\\\Delta K.E = \frac{1}{2} \times 2 \times (5.92^2\ - \ 5^2)\\\\\Delta K.E = 10 \ J[/tex]

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Answer:

1/4πε₀[Hx/ (√x² + b²)^3/2]i.

Explanation:

So, without mincing words let's dive straight into the solution to the question above. There is need to determine the electric field on-axis of a uniformly-charged disk sitting.

The electric field in the x-component, dεₓ = 1/4πε₀[H/ x² + b²] cos .

Thus, the total electric field in the x-component, εₓ = 1/4πε₀ [ xdH/ (x^2 + b^2)^3/2.

Therefore, the electric field = 1/4πε₀ [ xdH/ (x^2 + b^2)^3/2i.

Where x=0

Answer: The displacement is 1 block.

Explanation:

Let's define:

The right is the positive side.

The left is the negative side.

Then if you start at position A, and you walk N blocks to the right, the new position is:

A + N

And if you start at position A, and you walk M blocks to the left, the new position is:

A - M.

In this case, we know that Kayla starts at -3 and she walks 5 blocks to the right.

Then her new position is:

-3 + 5 = 2

Now she walks 3 blocks to the left, then her new position is:

2 - 3 = -1

The displacement will be equal to the difference between the final position (-1) and the initial position (-2)

Then the displacement is:

D = -1 - (-2) = -1 +2 = 1

The displacement is 1 block.

Answer:

TRUE

Explanation:

IM SMART

Answer:

Stay  Safe! ,God bless you . The answer is false ,

Explanation:

Answer:

The earth, The sun, the solar system and the milky way.

Answer:

Explanation:

Let t represent the time for Tina to catch David.

Hence, considering the equation of linear motion S = ut + 1/2at^2..... 1

For David u = 28.0 m/s where 'a' is set to nought

S = ut

S = 28t.......2

For Tina consider equation 1

Where acceleration = 2.90m/s^2 and u is set at nought

S = 1/2×2.90 m/s×t^2.......3

Equate 2 and 3

28t = 1.45t^2

Divide through by t

28 = 1.45t

t = 28/1.45

t = 19.31seconds

Now put the value of t into equation 3

S = 1/2×2.90 m/s×t^2.......3

= 1.45×20×20

= 580m

Tina must have driven 580meters before passing David

Considering the equation of linear motion : V^2 = U^2+2as

Where u is set at nought

V^2 = 2as

V^2 = 2×2.9×580

V^2 = 3364

V = √3364

V = 58m/s

Her speed will be 58m/s

(a) Tina should drive for 580 m, before passing the David.

(b) The speed of Tina during her passage through the David is 58 m/s.

Given data:

The initial velocity of the David is, u = 28.0 m/s.

The magnitude of acceleration is, [tex]a = 2.90 \;\rm m/s^{2}[/tex].

(a)

We can use the second kinematic equations of motion to obtain the distance covered by Tina, before passing the David. As per the second kinematic equation of motion,

[tex]s= u't + \dfrac{1}{2}at^{2}[/tex]

Here, u' is the initial speed of Tina and t is the time interval. Then,

Let t represent the time for Tina to catch David.

Hence, considering the equation of linear motion as,

S = ut + 1/2at²...............................................................(1)

Also,

S = ut

S = 28t ...........................................................................(2)

For Tina consider equation 1

S = 1/2×2.90t²................................................................(3)

Equate 2 and 3

28t = 1.45t²

 28 = 1.45t

t = 28/1.45

t = 19.31 seconds

Now put the value of t into equation (3)

S = 1/2×2.90 t².

   = 1.45×20×20

   = 580m

Thus, we can conclude that Tina should drive for 580 m, before passing the David.

(b)

Now, using the third kinematic equation of motion to obtain the speed of Tina during her passage through David as,

v² = u²+2as

Solving as,

v² = 28.0² + 2(2.90)(580)

v = √3364

v = 58m/s

Thus, we can conclude that the speed of Tina during her passage through the David is 58 m/s.

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Answer:

Explanation:

Expression for fundamental  frequency of tone produced in a wire under tension of T and length L is given as follows

[tex]f=\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]

m is mass per unit length .

We shall apply this formula for given wires .

For shorter wire

[tex]60 =\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]

For longer wire for second harmonic

length of wire is 2L , tension is 4T ,

[tex]f =\frac{2}{4L} \times \sqrt{\frac{4T}{ m} }[/tex]

[tex]f =\frac{2\times 2}{4L} \times \sqrt{\frac{T}{ m} }[/tex]

f = 2 x 60 = 120 Hz .

Answer:

Explanation:

The change in momentum is in the negative direction.