An atom of tin has an atomic number of 50 and a mass number of 119. How many protons, electrons, and neutrons are found in one neutral atom of tin?
O 50 protons, 69 electrons, 50 neutrons
O 50 protons, 50 electrons, 69 neutrons
69 protons, 50 electrons, 69 neutrons
69 protons, 69 electrons, 50 neutrons

Answer:

bb kettle

Explanation:

it transfres electricsl to kinetic

Answer:

they are right it is a new moon

Explanation:

took the test

Answer:

256 N

Explanation:

formula of centripetal force = mv²/r

m= 2kg

v= 8m/s

r= 0.5m

mv²/r = 2×8²/0.5 = 256N

Answer:

B It is always increasing.

Explanation:

In Physics, entropy can be defined as the tendency or ability of a substance to reach maximum disorder i.e to be randomly distributed.

This ultimately implies that, entropy is a thermodynamic quantity that measures the degree of maximum disorder or randomness of a system.

The S.I unit used for the measurement of the degree of maximum order or randomness of a system is Joules per Kelvin (JK¯¹). An example of entropy is the mixing of ideal gases.

Generally, the entropy in an irreversible process always increases and as such the change in entropy has a positive value.

Hence, the entropy of the universe is always increasing because its energy flow is considered to be in a downward direction rather than upward i.e from a hot region to a cold region; making the energy to be evenly distributed.

Answer:

a place where two bones come together is known as a join

Answer:

Hey mate.....

Explanation:

This is ur answer....

Joints – A place in the body where bones come together. Non-Moveable Joints (sometimes called fixed or fibrous) – A place in the body where two or more bones come together but do not move.

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Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ  = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

I don't know the answer but I just want points sorry

Answer:

(a) The volume of water is 100 cm³

(b) The volume of the rock is 20 cm³

(c) The density of the rock is 30 g/cm³

Explanation:

The given parameters of the perspex box are;

The area of the base of the box, A = 10 cm²

The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the  water, V, is given as follows;

[tex]V_T[/tex] = [tex]V_r[/tex] + V

[tex]V_T[/tex] = A × h₂

∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³

The total volume, [tex]V_T[/tex] = 120 cm³

The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V

∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, [tex]V_r[/tex] = 20 cm³

(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)

∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³

Answer:

37.5 m/s

Explanation:

Using,

Formula

v = ωr....................... Equation 1

Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.

From the question,

Given: ω = 30 rad/s, r = 1.25 m

Substitute these values into equation 1

v = 30(1.25)

v = 37.5 m/s.

Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s

Answer:

Products is the result. Reactants produce the result

Explanation:

,,

Answer:

r =[tex]\frac{ 1 \pm \sqrt{ \frac{m}{M} } }{1 - \frac{m}{M} }[/tex]

Explanation:

Let's apply the universal gravitation law to the body (c), we use the indications 1 for the planet and 2 for the moon

          ∑ F = 0

           -F_{1c} + F_{2c} = 0

             F_{1c} = F_{2c}

let's write the force equations

             [tex]G \frac{m_c M}{r^2} = G \frac{m_c m}{(d-r)^2}[/tex]

where d is the distance between the planet and the moon.

              [tex]\frac{M}{r^2} = \frac{m}{(d-r)^2}[/tex]

             (d-r)² = [tex]\frac{m}{M} \ \ r^2[/tex]  

              d² - 2rd + r² = \frac{m}{M} \ \ r^2

              d² - 2rd + r² (1 - [tex]\frac{m}{M}[/tex]) = 0

              (1 - [tex]\frac{m}{M}[/tex])  r² - 2d r + d² = 0

we solve the second degree equation

              r = [2d ± [tex]\sqrt{ 4d^2 - 4 ( 1 - \frac{m}{M} ) }[/tex] ] / 2 (1- [tex]\frac{m}{M}[/tex])

              r = [2d ±  2d [tex]\sqrt{ \frac{m}{M} }[/tex]] / 2d (1- [tex]\frac{m}{M}[/tex])

              r =[tex]\frac{ 1 \pm \sqrt{ \frac{m}{M} } }{1 - \frac{m}{M} }[/tex]

there are two points for which the gravitational force is zero

The distance between object from planet will be "[tex]\frac{R}{[1+\sqrt{\frac{m}{M} } ]}[/tex]".

According to the question,

Let,

As we know,

→ [tex]Prerequisites-Gravitational \ force = \frac{GMm}{r^2}[/tex]

Now,

→ [tex]\frac{GMm_1}{x^2} = \frac{Gmm_1}{(R-x)^2}[/tex]

→ [tex]\frac{(R-x)^2}{x^2} = \frac{m}{M}[/tex]

→     [tex]\frac{R-x}{x} =\sqrt{\frac{m}{M} }[/tex]

→          [tex]x = \frac{R}{[1+ \sqrt{\frac{m}{M} } ]}[/tex]

Thus the answer above is appropriate.          

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The dependent variable is academic performance while the independent variable is presence/absence of tutorial support.

The correct results are:

In every experiment, there is a dependent and independent variable as well as an experimental and a control group.

The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.

The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.

The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).

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Answer: 2000 J

Explanation: work W = F s

?.............................

Answer:

Explanation:

From the information given;

mass of the crate m = 41 kg

constant horizontal force = 135 N

where;

[tex]s_1 = 15.0 \ m \\ \\ s_2 = 12.0 \ m[/tex]

coefficient of kinetic friction [tex]u_k[/tex] = 0.28

a)

To start with the work done by the applied force [tex](W_f)[/tex]

[tex]W_F = F\times (s_1 +s_2) \times cos(0) \ J[/tex]

[tex]W_F = 135 \times (12 +15) \times cos(0) \ J \\ \\ W_F = (135 \times 37 )J \\ \\ W_F =4995 \ J[/tex]

Work done by friction:

[tex]W_{ff} = -\mu\_k\times m \times g \times s_2 \\ \\ W_{ff} = -0.320 \times 41 \times 9.81 \times 12 \ J \\ \\ W_{ff} = -1544.49 \ J[/tex]

Work done  by gravity:

[tex]W_g = mg \times (s_1+s_2) \times cos (90)} \ J \\ \\ W_g = 0 \ j[/tex]

Work done by normal force;

[tex]W_n = N \times (s_1 + s_2) \times cos (90) \ J[/tex]

[tex]W_n = 0 \ J[/tex]

b)

total work by all forces:

[tex]W = F \times (s_1 + s_2) + \mu_k \times m \times g \times s_2 \times 180 \\ \\ W = 135 \times (15+12) \ J - 0.320 \times 41 \times 9.81 \times 12[/tex]

W = 2100.5  J

c) By applying the work-energy theorem;

total work done = ΔK.E

[tex]W = \dfrac{1}{2}\times m \times (v^2 - u^2)[/tex]

[tex]2100.5 = 0.5 \times 41 \times v^2[/tex]

[tex]v^2 = \dfrac{2100.5}{ 0.5 \times 41 }[/tex]

[tex]v^2 = 102.46 \\ \\ v = \sqrt{102.46} \\ \\ \mathbf{v = 10.1 \ m/s}[/tex]

Answer:

The correct answer is A) It is upright and 1.5m behind the mirror

Explanation:

Your reflection must be upright, meaning vertical/erect, and the distance will be the exact same. Also, the reflected ray appears as if it had traveled from an object located behind the mirror.

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

Answer:

B. kinetic energy

Explanation:

Thermal energy (It’s a low form of energy ) is a form of kinetic energy as it is produced as a result of motion of particles either if they vibrate at their position or they move along longer paths.

Answer:

Δθ = 15747.37 rad.

Explanation:

       [tex]\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)[/tex]

       [tex]\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)[/tex]

       [tex]\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)[/tex]

       [tex]\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)[/tex]

       [tex]\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)[/tex]

      [tex]\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)[/tex]

Answer:

75.4

Explanation:

r= 2

h= 6

v= 22/7 *r*r*h

v= 75.42

Answer:

534.8 meters

Explanation:

Use T=(2*pi*r)/v

560=(2*pi*r)/6

3360=2*pi*r

1680=pi*r

534.8 meters=radius

It takes 560s for a runner to complete one circular lap, moving at a speed of 6.00 m/s. The radius of a track is 534.7 m.

Mathematically, it can be calculated as follows :

distance = speed × time

The formula relating distance (d), speed (s), and time (t) is

d = st

First, Calculating the distance,

d = 560 s × 6 m·s⁻¹

  = 3360 m

When, Calculating the track radius,

The distance travelled is the circumference of a circle,

C = 2пr

r = 3360/2п

 = 534.7 m

The radius of the track is 534.7 m.

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Answer:

a)  Wapp = 520 N

b)  ΔUf = 447 N

c) Wn = 0

d) Wg = 0

e) ΔK = 73 J

f) v = 1.96 m/s

Explanation:

a)

       [tex]W_{app} = F_{app} * \Delta X = 130 N * 4.0 m = 520 N (1)[/tex]

b)

       [tex]W_{ffr} = F_{fr} * \Delta X * cos (180) (2)[/tex]

       [tex]W_{ffr} = F_{fr} * \Delta X * cos (180) = -\mu_{k} * m* g* \Delta X = \\ -0.300*38.0kg9.8 m/s2*4.0m = -447 J (3)[/tex]

c)

d)

e)

        [tex]\Delta K = W_{app} + W_{ffr} = 520 J - 447 J = 73 J (4)[/tex]

f)

       [tex]\Delta K = 73 J = \frac{1}{2}*m*v^{2} (5)[/tex]

a) The work done by the applied force   [tex]W_{AP}=520\ J[/tex]

b) The change in the internal energy [tex]\Delta U=447\ J[/tex]

c) Work done by normal force  [tex]W_n=0[/tex]

d) Work done by gravitation   [tex]W_g=0[/tex]  

e) The change in KE will be [tex]\Delta KE=73\ J[/tex]

f) The final speed v = 1.96 m/s

The work done on any object can be defined as the force applied on the object and its displacement due the effect of the force.

If the object achieve movement due to the work then the energy in the object will be kinetic energy.

If the object attains some height  against the gravity then the energy in the object will be potential energy.

Now it is given in the question that

The horizontal force   [tex]F_h=130\N[/tex]

mass of the object  m= 38 kg

Coefficient of friction [tex]\mu=0.3[/tex]

Displacement of the object [tex]\delta x=4\ m[/tex]

(a) The work done will be

[tex]W=F_h\tines \Delta x[/tex]

[tex]W=130\times 4=520\ J[/tex]

(b) The increase in the internal energy

The increase in the internal energy of the box is due to the energy generated by the force of friction so

[tex]W_f=F_f\times \Delta x\times Cos(180)[/tex]

here  [tex]F_f[/tex] is the frictional force and is given as

[tex]\mu=\dfrac{F_f}{R}[/tex]

Here R is the normal reaction and its value will be weight of the box in opposite direction.

[tex]\mu=\dfrac{F_f}{-mg}[/tex]

[tex]F_f=-mg\times \mu[/tex]

[tex]W_f=F_f\times \Delta x\times Cos180=-mg\times\mu \times cos180[/tex]

[tex]W_f=-38\times 9.81\times 0.3\times4=-447\J\ J[/tex]

(c) The work done by the normal force will be zero because the displacement is horizontal against the normal work so the work done will be zero.

(d) The work done by the gravitational force will also be zero. Because the displacement is horizontal and the gravitational force acts downward.

(e) The change in the KE of the box.

The change in the KE of the box will be the net energy of the box so from the work energy theorem the net energy will be

[tex]\Delta KE =W_{AP}-W_f=520-447=73\ J[/tex]

(f) The speed of the box

The KE of the box will be

[tex]KE=\dfrac{1}{2} mv^2[/tex]

[tex]v=\sqrt{ \dfrac{2\times KE}{m}[/tex]

[tex]v=\sqrt{\dfrac{2\times73}{38} }=1.96\ \dfrac{m}{s}[/tex]

Thus

a) The work done by the applied force   [tex]W_{AP}=520\ J[/tex]

b) The change in the internal energy [tex]\Delta U=447\ J[/tex]

c) Work done by normal force  [tex]W_n=0[/tex]

d) Work done by gravitation   [tex]W_g=0[/tex]  

e) The change in KE will be [tex]\Delta KE=73\ J[/tex]

f) The final speed v = 1.96 m/s

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Answer:

J = -0.522 m/s

Explanation:

Given that,

The mass of the baseball, m = 0.145 kg

Initial velocity, u = 14.5 m/s

Final velocity, v = 10.9 m/s

aWe need to find the direction of the impulse that the glass imparts to the baseball. Impulse is equal to the change in momentum such that,

[tex]J=m(v-u)[/tex]

Substitute all the values,

[tex]J=0.145\times (10.9-14.5)\\\\=-0.522\ kg-m/s[/tex]

The direction of impulse is opposite to the direction of velocity.

Answer:

v = 2.57 m / s

Explanation:

For this exercise let's use conservation of energy

starting point. When it is at an angle of 30º

          Em₀ = K + U = ½ m v₁² + m g y₁

final point. Lowest position

          Em_f = K = ½ m v²

as there is no friction, the energy is conserved

          Em₀ = Em_f

          ½ m v₁² + m g y₁ = ½ m v²

Let's find the height(y₁), which is the length of the thread minus the projection (L ') of the 30º angle

         cos 30 = L ’/ L

         L ’= L cos 30

         y₁ = L -L '

          y₁ = L- L cos 30

we substitute

          ½ m v₁² + m g L (1- cos 30) = ½ m v²

           v = [tex]\sqrt{ v_1^2 +2gL(1-cos30 )}[/tex]

let's calculate

           v = [tex]\sqrt{ 2^2 + 2 \ 9.8 \ 1.0 (1- cos 30)}[/tex]

           v = 2.57 m / s

Answer:

c.

Equal to 200 N..........