7. Two capacitors, 5.8 µF and 2.1 µF, are connected in parallel. A 3 V DC voltage is applied across the capacitors. What would be the expected accumilative charge stored in the capacitors

Answer:

Charge, [tex]q=7.68\times 10^{-12}\ C[/tex]

Explanation:

It is given that,

The number of electron in a RBCs, [tex]n=4.8\times 10^7[/tex]

We need to find the total charge of these electrons in the red blood cell. Let it is q. Using the quantization of charge as follows :

q = ne

e is the change on electron

[tex]q=4.8\times 10^7\times 1.6\times 10^{-19}\\\\q=7.68\times 10^{-12}\ C[/tex]

So, the net charge is [tex]7.68\times 10^{-12}\ C[/tex].

Answer:

[tex]\mu_s=0.51[/tex]

Explanation:

Coefficient of friction

The static coefficient of friction is a measure of the force needed to start moving an object from rest along a rough surface.

It's calculated as:

[tex]\displaystyle \mu_s=\frac{F_r}{N}[/tex]

Where Fr is the friction force and N is the normal force exerted by the surface over the object.

In the absence of any other vertical force, the normal is equal to the weight of the object:

[tex]N = W = m.g[/tex]

The box has a mass of m=80 Kg, thus the normal force is:

[tex]N = 80\ Kg * 9.8\ m/s^2[/tex]

N = 784 N

The student needs to push with a force of 400 N to make the box move. That is the force that barely outcomes the friction force. Thus:

[tex]F_r=400\ N[/tex]

Calculating the coefficient:

[tex]\displaystyle \mu_s=\frac{400}{784}[/tex]

[tex]\mathbf{\mu_s=0.51}[/tex]

This question involves the concept of the orbital period.

The period of revolution of the second satellite is "3.95 x 10⁶ s".

First, we will consider the orbital period of the first satellite:

[tex]T_1=\sqrt{\frac{4\pi^2 r_1^3}{GM}}[/tex]

where,

Therefore,

[tex]1\ x\ 10^6\ s=\sqrt{\frac{4\pi^2(8\ x\ 10^6\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(M)}}\\\\M = \frac{4\pi^2(8\ x\ 10^6\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1\ x\ 10^6\ s)^2}\\\\M = 3.03\ x\ 10^{20}\ kg[/tex]

Now, we find out the orbital period of the second satellite:

[tex]T_2=\sqrt{\frac{4\pi^2 r_2^3}{GM}}[/tex]

where,

Therefore,

[tex]T_2=\sqrt{\frac{4\pi^2(2\ x\ 10^7\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(3.03\ x\ 10^{20}\ kg)}}[/tex]

T₂ = 3.95 x 10⁶ s

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The correct option is A i.e.  Van der Waals forces.

what is Van der Waals forces?

Van der Waals forces are weak intermolecular forces that are depending on atom or molecule distance. Interactions between uncharged atoms/molecules produce these forces.

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#SPJ2

Carcinoma refers to a malignant neoplasm of epithelial origin or cancer of the internal or external lining of the body. Carcinomas, malignancies of epithelial tissue, account for 80 to 90 percent of all cancer cases. Epithelial tissue is found throughout the body.

Answer:

A law is defined as a description or a statement given after an observed phenomenon. A theory is a simplification of certain observational data as to how and why it happened.

Explanation:

I hope this helped solve your question.

Answer:

3

Explanation:

Answer:

When the projectile is launched straight up, there isn't a horizontal ... The initial acceleration was 9.8 m/s2 pointing up, so the acceleration at any other point should be the same.

Explanation:

Hope it helped =)

The maximum altitude that the cannonball will reach if fired straight up at a speed of 25 m/s is; 31.86 m

According to the question;

The cannonball is fired straight up at a speed of 25 m/s

Additionally, the cannonball is fired against the force of gravity.

Consequently, the motion is in the opposite direction of the acceleration due to gravity.

From the equation of motion;

At the maximum altitude, V = 0.

The maximum altitude that it will reach is;

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Answer:

He concluded the fact that most alpha particles went straight through the foil is evidence for the atom being mostly empty space. A small number of alpha particles being deflected at large angles suggested that there is a concentration of positive charge in the atom.

Explanation:

Answer:

Explanation:

given

Mass of cart= 150kg

mass of each wheel=45kg

mass of 4 wheels= 180kg

angle of the track=  17 ∘

distance of track= 27m

The height  of the tracl is calculated thus:

sin 17° = h / 27

h = sin 17*27

h=7.89m

"Potential energy at top = kinetic energy of cart and wheels at bottom + rotational energy of wheels at bottom "

1. Potential energy at top= (M+4m)gh

2. kinetic energy of cart and wheels at bottom= 1/2 (M+4m) v²

3. rotational energy of wheels at bottom= 4(1/2 Iω²)

The total is expressed as

(M+4m)gh = 1/2 (M+4m) v² + 4(1/2 Iω²) -------------1

we know that  I = mr² / 2

Put I= mr² / 2

(M+4m)gh = 1/2 (M+4m)v² + 4(1/2 (mr² / 2) ω²)

(M+4m)gh = 1/2 (M+4m)v² +  m r² ω²

we know that  v²= r² ω²

(M+4m)gh = 1/2 (M+4m)v² +  m v²

(M+4m)gh = v² (M/2 + 2m + m)

(M+4m)gh = v² (M/2 + 3m)

v = √[(M+4m)gh / (3m + M/2)]

v = √[(150 + 4*45) * 9.8 m/s² * 7.89 m / (3*45 + 150/2)]

v=√25516.26/142.5

v=√179.06

v = 13.4 m/s

Answer:

C. Lighting is made of atoms

Explanation:

Everything is made of atoms that are, in turn, made out of charged particles. All charged particles come in one of two types: positive and negative (or plus and minus). The minus particles are the electrons, and the plus particles are the much heavier protons which are buried deep in the nucleus.

Answer:

x is vertical and y is horizontal

Explanation:

Answer:

wavelength = 3000 m

Explanation:

We use the formula for the relationship between the speed (c), frequency (v), and wavelength [tex](\lambda)[/tex]:

[tex]c=\lambda\,*\,\nu\\300000000 \,\,m/s = \lambda\,*100000 \,\,1/s\\\lambda \,=\,3000\,\,m[/tex]

Answer:

C

Explanation:

Answer:

Hip-hop Dancing

Explanation:

Aerobic exercises are light (somewhat, a more accurate word would be bearable) workouts that you can endure over long periods of time. Anaerobic workouts describe workouts that require sudden burst of energy, like the ones you seen in most forms of dance, such as Hip-hop dance.

Hope this helps!

Answer:

I believe its C

Explanation:

Answer:

C-An object weighs about two times as much on Jupiter as on Neptune.

Explanation:

on edg

Answer:

0.287 m

Explanation:

The velocity of block A when it reaches block B is:

KE₀ = KE + W

½ mv₀² = ½ mv² + Fd

½ mv₀² = ½ mv² + mg μ d

v₀² = v² + 2g μ d

v² = v₀² − 2g μ d

v² = (10 m/s)² − 2 (9.8 m/s²) (0.3) (4 m)

v = 8.75 m/s

The coefficient of restitution is 0.6, so the relative velocity between A and B after the collision is:

e = |Δv after| / |Δv before|

0.6 = Δv / (8.75 m/s)

Δv = 5.25 m/s

Momentum is conserved, so the speed of block B after the collision is:

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(15 kg) (8.75 m/s) + (10 kg) (0 m/s) = (15 kg) (v) + (10 kg) (v + 5.25 m/s)

131.2 m/s = 25v + 52.5 m/s

25v = 78.7 m/s

v = 3.15 m/s

Energy is conserved, so the compression of the spring is:

KE = EE + W

½ mv² = ½ kd² + Fd

½ mv² = ½ kd² + mg μ d

½ (10 kg) (3.15 m/s)² = ½ (1000 N/m) d² + (10 kg) (9.8 m/s²) (0.3) d

49.6 = 500 d² + 29.4 d

0 = 500 d² + 29.4 d − 49.6

d = [ -29.4 ± √(29.4² − 4(500)(-49.6)) ] / 1000

d = (-29.4 ± 316.2) / 1000

d = 0.287 m

Answer:

F_{players} = 2528.4[N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body must be equal to the product of mass by acceleration.

On the sled act two forces, the force of friction and The Force executed by the football players.

The movement of the football players will be taken as positive, while the friction will be taken as negative, since it is opposed to the movement of the sled.

ΣF = m*a

where:

F = force [N]

m = mass [kg]

a = acceleration = 0.580 [m/s²]

[tex]F_{players} - f_{friction} = m*a[/tex]

The friction force is defected as the coefficient of friction by the normal force, the normal force on a horizontal surface can be calculated as the product of mass by gravitational acceleration.

[tex]f_{friction}=0.8*(300*9.81)\\f_{friction}=2354.4[N][/tex]

Now we can calculate the force exerted by the players.

[tex]F_{players}-2354.5=300*0.580\\F_{players}=2528.4[N][/tex]

You are most likely going through depression. I think you should talk to someone, it could help or try talking to one of us on brainly. I hope you feel better eventually. Your life is valuable try and realize that.

Answer:

You don't need a reason to be sad. It sound as though you are suffering from depression. I understand that. Last year I thought about taking my life, but my parents found out how I was feeling and have been so supportive. Something My mom said that really got to me was, if you killed yourself, who would find you. For me it would have been my little sister. I would have scared her for life. I have come a long way since then, I am doing family therapy, as well as persona, therapy. If you are feeling sad, there is a hotline you can text rather than call. Let them talk you out of it. They won't call anyone if you don't ask them to. Talking to a complete stranger helps, they don't know anything about you so they can't judge you. Also try listening to music, find an artist that you can relate to. For me I like listening to NF when I am feeling down. It helps.

Answer:

A. 7 s ≤ t < 10 s

Explanation:

From Definite Integral Theory and Mechanical Physics we remember that change in position is represented by sum of areas below the curve. According to the graphic, we see that the following condition must be satisfied:

[tex]A_{1}+A_{2} = 0[/tex] (1)

Where:

[tex]A_{1}[/tex] - Change in position from [tex]t = 0\,s[/tex] and [tex]t = 3\,s[/tex], measured in meters.

[tex]A_{2}[/tex] - Change in position from [tex]t = 5\,s[/tex] and [tex]t = T[/tex], measured in meters.

By definitions of triangle, we expand the equation above:

[tex]\frac{1}{2}\cdot (3\,s)\cdot \left(1.2\,\frac{m}{s} \right) -\frac{1}{2}\cdot (T-5\,s)\cdot v = 0\,m[/tex]

[tex]1.8\,m-0.5\cdot T \cdot v +2.5\cdot v = 0[/tex]

And the time is now cleared:

[tex]1.8\,m+2.5\cdot v = 0.5\cdot T\cdot v[/tex]

[tex]T = \frac{1.8+2.5\cdot v}{0.5\cdot v}[/tex]

Where:

[tex]v[/tex] - Final velocity of the cart, measured in meters per second.

[tex]T[/tex] - Time, measured in seconds.

If we know that [tex]v = 1\,\frac{m}{s}[/tex], then the intended instant is:

[tex]T = \frac{1.8+2.5\cdot (1)}{0.5\cdot (1)}[/tex]

[tex]T = 8.6\,s[/tex]

The value does not coincide with the time from the graph.

If we know that [tex]v = 1.5\,\frac{m}{s}[/tex], then the intended instant is:

[tex]T = \frac{1.8+2.5\cdot (1.5)}{0.5\cdot (1.5)}[/tex]

[tex]T = 7.4\,s[/tex]

This value does not coincide with the time from the graph.

If we know that [tex]v = 1.1\,\frac{m}{s}[/tex], then the intended instant is:

[tex]T = \frac{1.8+2.5\cdot (1.1)}{0.5\cdot (1.1)}[/tex]

[tex]T = 8.273\,s[/tex]

This results offer a reasonable approximation, which that the correct answer is A.

The car will return to its position at the time interval of 7 s ≤ t < 10 s. Hence, option (A) is correct.

The change in position is represented by sum of areas below the curve. According to the graphic, we see that the following condition must be satisfied:

A + A' = 0 ..................................................(1)

A is the change in position from t = 0 and t = 3 s, measured in metres.

A' is the change in position from t = 5 and t = T s, measured in metres.

As per the triangle law,

1/2 ×(3 s)(1.2) - 1/2 ×(T - 5)v = 0

1.8 - 0.5T(v) +2.5(v) =0

T = (1.8 + 2.5v)/ (0.5v)

Here, v is the final velocity of cart.

For v = 1 m/s, we have,

T = (1.8 + 2.5(1))/ (0.5(1))

T = 8.6 s

Since, the values do not coincide with the time from the graph, so taking another value, v = 1.5 m/s to obtain the time as,

T = (1.8 + 2.5(1.5))/ (0.5(1.5))

T = 7.4 s

This value does not coincide with the time from the graph.  If we know that , v = 1.1 m/s, then the intended instant is:

T = (1.8 + 2.5(1.1))/ (0.5(1.1))

T = 8.273 s

Thus, we can conclude that the car will return to its position at the time interval of 7 s ≤ t < 10 s. Hence, option (A) is correct.

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Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

[tex]E = hv[/tex]

[tex]E = \frac{hc}{\lambda}[/tex]

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

[tex]E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}[/tex]

[tex]E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})[/tex]

E = 4.96 x 10³ eV

B)

The energy of a particle at rest is given as:

[tex]E = m_{0}c^2[/tex]

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

[tex]E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\[/tex]

[tex]E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\[/tex]

E = 4.19 x 10⁴ eV

C)

The energy of a particle at rest is given as:

[tex]E = m_{0}c^2[/tex]

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

[tex]E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\[/tex]

[tex]E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\[/tex]

E = 3.73 x 10⁹ eV

A) The energy in electron volts for a particle with this wavelength if the particle is a photon is; .E = 4969.5 eV or 4.9695 keV

B) The energy in electron volts for a particle with this wavelength if the particle is an electron is; E = 23.58 eV

C) E = 0.003306 eV

A) The formula for the energy here is;

E = hc/λ

where;

h is planck's constant = 6.626 × 10⁻³⁴ J.s

c is speed of light = 3 × 10⁸ m/s

λ is wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Thus;

E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(0.25 x 10⁻⁹)

79.512 × 10⁻¹⁷ J

converting to eV gives;

E = (79.512 × 10⁻¹⁷)/(1.6 × 10⁻¹⁹)

E = 4969.5 eV or 4.9695 keV

B) Formula for the energy if the particle is an electron is;

E = h²/(2mλ²)

where m = 9.31 × 10⁻³¹ kg

E = (6.626 × 10⁻³⁴)²/(2 × 9.31 × 10⁻³¹ × (0.25 x 10⁻⁹)²)

E = 37.726 × 10⁻¹⁹ J

Converting to eV gives;

E = (37.726 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)

E = 23.58 eV

C) Mass of alpha particle is; m = 6.64 × 10⁻²⁷ kg

E = h²/(2mλ²)

where m = 6.64 × 10⁻²⁷ kg

E = (6.626 × 10⁻³⁴)²/(2 × 6.64 × 10⁻²⁷ × (0.25 x 10⁻⁹)²)

E = 52.896 × 10⁻²³ J

Converting to eV gives;

E = (52.896 × 10⁻²³)/(1.6 × 10⁻¹⁹)

E = 0.003306 eV

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Answer:

[tex] \frac{1}{2} \times m \times {v}^{2} \\ \frac{1}{2} \times m \times {2}^{2} \\ 2 \times m[/tex]

I dont the child mass...you should substitute that value to (m)

then you can get your answer

Answer:

720 V

Explanation:

Given that,

The number of turns in primary coil, N₁ = 60

The number of turns in secondary coil, N₂ = 360

The input rms voltage, V₁ = 120 V

We need to find the output rms voltage of the secondary coil . The relation between number of turns in primary coil - secondary coil to the input rms voltage to the output rms voltage is given by :

[tex]\dfrac{N_1}{N_2}=\dfrac{V_1}{V_2}\\\\V_2=\dfrac{N_2V_1}{N_2}\\\\V_2=\dfrac{360\times 120}{60}\\\\V_2=720\ V[/tex]

Answer:

61.6886 kg

Explanation:

Answer:

61 kg

Explanation:

there ya go hope this was useful

Answer:Well toasting it makes it a solid so that’s a chemical change that happens! Hope this helped!

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

[tex]y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (1)

Where:

[tex]y(t)[/tex] - Position of the mass as a function of time, measured in meters.

[tex]A[/tex] - Amplitude, measured in meters.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]m[/tex] - Mass of the block, measured in kilograms.

[tex]t[/tex] - Time, measured in seconds.

[tex]\phi[/tex] - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

[tex]k = \frac{m\cdot g}{\Delta y}[/tex] (2)

Where:

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]\Delta y[/tex] - Deformation of the spring due to gravity, measured in meters.

If we know that [tex]m=1.65\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta y = 0.260\,m[/tex], then the spring constant is:

[tex]k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}[/tex]

[tex]k = 62.237\,\frac{N}{m}[/tex]

If we know that [tex]A = 0.130\,m[/tex], [tex]k = 62.237\,\frac{N}{m}[/tex], [tex]m=1.65\,kg[/tex], [tex]x(t) = 0\,m[/tex] and [tex]\phi = 0\,rad[/tex], then (1) is reduced into this form:

[tex]0.130\cdot \cos (6.142\cdot t)=0[/tex] (1)

And now we solve for [tex]t[/tex]. Given that cosine is a periodic function, we are only interested in the least value of [tex]t[/tex] such that mass reaches equilibrium position. Then:

[tex]\cos (6.142\cdot t) = 0[/tex]

[tex]6.142\cdot t = \cos^{-1} 0[/tex]

[tex]t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s[/tex]

[tex]t \approx 0.255\,s[/tex]

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

The time taken for the spring to reach new equilibrium position is 0.26 s.

The given parameters;

The general wave equation is given as;

[tex]y(t) = A\ cos(\omega t \ + \phi)[/tex]

The angular frequency is given as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\[/tex]

The spring constant is calculated as;

[tex]F = mg\\\\kx = mg\\\\k = \frac{mg}{x} \\\\k = \frac{1.65 \times 9.8}{0.26} \\\\k = 62.2 \ N/m[/tex]

The angular frequency is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{62.2}{1.65} }\\\\\omega = 6.14 \ rad/s[/tex]

The time taken for the spring to reach new equilibrium position is calculated as follows;

[tex]y(t) = A \ cos(\omega t)\\\\0 = 0.13 \times cos(6.14t)\\\\6.14t = cos^{-1}(0)\\\\6.14t = 1.57 \ rad\\\\t = \frac{1.57 \ rad}{6.14 \ rad/s} \\\\t = 0.26 \ s[/tex]

Thus, the time taken for the spring to reach new equilibrium position is 0.26 s.

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Answer:

The time for the rock to reach the bottom is 4.02 seconds.

Explanation:

Given;

mass of the rock, m = 5 kg

height of the rock fall, h = 79 meters

The time to drop to the given height is given by;

[tex]t = \sqrt{\frac{2h}{g} }[/tex]

where;

t is the time to fall to the bottom

g is acceleration due to gravity = 9.8 m/s²

[tex]t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*79}{9.8} }\\\\t = 4.02 \ s[/tex]

Therefore, the time for the rock to reach the bottom is 4.02 seconds.

Answer:

The linear (tangential) speed of a passenger is 0.4188 m/s

The distance traveled by the person in 5 minutes ride is 125.64 m

Explanation:

Given;

radius of the Ferris, r = 12 m

1 revolution per 3 minutes, [tex]\omega = \frac{2\pi (radian)}{3\ (minutes)} *\frac{1\ minute}{60 \ seconds} = 0.0349 \ rad/s[/tex]

The linear (tangential) speed of a passenger is given by;

v = ωr

v = (0.0349)(12)

v = 0.4188 m/s

The distance traveled by the person in 5 minutes ride is given by;

d = vt

d = (0.4188)(5 x 60)

d = 125.64 m

Answer:

a. 186.2 J b. 8.63 m/s c. 190 m d. 43.2 s e. 186.2 J

Explanation:

a. From conservation of energy, the potential energy loss of block = kinetic energy gain of the block.

So, U + K = U' + K' where U = initial potential energy of block = mgh, K = initial kinetic energy of block = 0, U' = final potential energy of block at bottom of curve = 0 and K' = final kinetic energy of block at bottom of curve.

So, mgh + 0 = 0 + K'

K' = mgh where m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s², h = initial height above the ground of block = radius of curve = 3.8 m

So, K' = 5 kg × 9.8 m/s² × 3.8 m = 186.2 J

b. Since the kinetic energy of the block K = 1/2mv²  where m = mass of block = 5 kg, v = velocity of block at bottom of curve

So, v = √(2K/m)

= √(2 × 186.2 J/5 kg)

= √(372.4 J/5 kg)

= √(74.48 J/kg)

= 8.63 m/s

c. To find the stopping distance, from work-kinetic energy principles,

work done by friction = kinetic energy change of block.

So ΔK = -fd where ΔK = K" - K' where K" = final kinetic energy = 0 J (since the block stops)and K' = initial kinetic energy = 186.2 J, f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance

ΔK = -fd

K" - K' = - μmgd

d = -(K" - K')/μmg

Substituting the values of the variables, we have

d = -(0 J - 186.2 J)/(0.02 × 5 kg × 9.8 m/s²)

d = -(- 186.2 J)/(0.98 kg m/s²)

d = 190 m

d. Using v² = u² + 2ad where u =initial speed of block = 8.63 m/s, v = final speed of block = 0 m/s (since it stops), a = acceleration of block and d = stopping distance = 190 m

So, a = (v² - u²)/2d

substituting the values of the variables, we have

a = (0² - (8.63 m/s)²)/(2 × 190 m)

a = -74.4769 m²/s²/380 m

a = -0.2 m/s²

Using v = u + at, we find the time t that elapsed while the block is moving on the horizontal track.

t = (v - u)/a

t =(0 m/s - 8.63 m/s)/-0.2 m/s²

t = - 8.63 m/s/-0.2 m/s²

t = 43.2 s

e. The work done by friction W = fd where

= μmgd where f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance = 190 m

W = 0.02 × 5 kg × 9.8 m/s² × 190 m

W = 186.2 J

The potential energy of the loss of the block will be equal to the kinetic energy gain. The kinetic energy of the block is 186.2 J at the bottom of the curved surface.

The potential energy of the loss of the block will be equal to the kinetic energy gain.

So,

[tex]U = mgh[/tex]

Where,

[tex]U[/tex] - potential energy

[tex]m[/tex] - mass of block =  5 kg

[tex]g[/tex] - gravitational  acceleration = 9.8 m/s²

[tex]h[/tex] = height = radius of curve = 3.8 m

Put the values in the formula,

[tex]U = 5 \times 9.8 \times 3.8 \\\\ U = 186.2 \rm \ J[/tex]

Therefore, the kinetic energy of the block is 186.2 J at the bottom of the curved surface.

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Answer:

yess

Explanation:

Answer:

yes me