A 87 kg man has a total mechanical energy of 1780 J .If he is swinging downward and is currently 1.4 m above the ground, what is his speed? Use g = 10 m/s^2

Answer:

Explanation:

Maximum value of force will be possible when both the sphere will have same charge . In that case charge on each sphere = Q / 2 =.5Q

F( max ) = k .5Q x .5Q / R²

=.25kQ² /R²

For the second case

F = k q ( Q-q)/  R²

F = .25kQ² /3R²

.25kQ² /3R² =  k q ( Q-q)/  R²

.25 Q² = 3qQ - 3q²

3q² - 3qQ + .25 Q² = 0

q =

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).

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:

carbon.

Explanation:

the reaction would then create c02 as a product

Answer:

0.01

Explanation:

Given the data:

10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90

True value = 9.81

Mean value :

Σx / n

Sample size, n = 9

(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9

= 88.69 / 9

= 9.854

Standard deviation (σ) :

Sqrt (Σ(X - m)² / n)

[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9

Sqrt(0.113824 / 9)

Sqrt(0.0126471)

σ = 0.1124593

Standard Error = σ / sqrt(n)

Standard Error = 0.1124593 / 9

Standard Error = 0.0124954

Standard Error = 0.01 ( 1 significant digit)

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:

the correct answer is B

Explanation:

Answer:

.B

Explanation:

This is the answer on edge 2021

Have a good day!

Answer:

1. 14 g of chocolate mixture.

2. 24 fl oz of chocolate milk

3. 10 cups of chocolate milk.

4. 12½ cups.

Explanation:

From the question given above, the following data were obtained:

1 TBSP = 7 g

1 Cup = 8 fl oz

2 Table spoons (TBSP) for 1 cup (8 fl oz) of milk.

1. Determination of the mass of chocolate mixture in 1 cup of chocolate milk.

From the question given above,

1 Cup required 2 Table spoons (TBSP)

But

1 TBSP = 7 g

Therefore,

2 TBSP = 2 × 7 = 14 g

Thus, 1 Cup required 14 g of chocolate mixture.

2. Determination of the number fl oz of chocolate milk in 3 cups

1 Cup = 8 fl oz

Therefore,

3 Cups = 3 × 8

3 Cups = 24 fl oz

Thus, 24 fl oz of chocolate milk are in 3 cups.

3. Determination of the number of cups of chocolate milk produce from 20 TBSP.

2 TBSP is required to produce 1 cup.

Therefore,

20 TBSP will produce = 20/2 = 10 Cups.

Thus, 10 cups of chocolate milk produce from 20 TBSP.

4. Determination of the number of cups obtained from 100 fl oz chocolate milk.

8 fl oz is required to produce 1 cup.

Therefore,

100 fl oz will produce = 100 / 8 = 12½ cups.

Thus, 12½ cups is obtained from 100 fl oz chocolate milk.

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:

most likely be included in the analysis section of a lab report

Explanation:

Answer:

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

Explanation:

your welcome :)

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:

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:

They both have an equal temperature.

Explanation:

Any material or object that allow the conduction (transfer) of electric charge or thermal energy is generally referred to as a conductor. Some examples of a conductor are metals, copper, aluminum, graphite, etc.

In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

Hence, if two objects have different temperatures when they come in contact, heat will flow from the warmer object to the cooler one until they both have an equal temperature.

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:

[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:

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:

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:

v= -107.8 m/s

Explanation:

       [tex]v_{f} = v_{o} + a*t = a*t = -g*t = 9.8m/s2*11s = -107.8 m/s (1)[/tex]

Answer:

S = 16.3125m

Explanation:

Given the following data;

Acceleration, a = 14.5m/s²

Time, t = 2.25secs

Since the bottle rocket starts from rest, its initial velocity is 0m/s.

To find the distance S, we would use the second equation of motion.

S = ut + ½at²

Substituting into the equation, we have

S = 0(2.25) + ½*14.5*2.25

S = 0 + 7.25*2.25

S = 16.3125m

Therefore, the bottle rocket covered a distance of 16.3125 meters.

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:

☝

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.