A pulley in the form of a uniform disk)
with mass 63 kg and a radius 20 cm is at-
tached to the ceiling in a uniform gravita-
tional field and rotates with no friction about
its pivot. The masses are connected by a
massless inextensible cord and T1, T2, and T3
are magnitudes of the tensions.
T3
20 cm
T2
63 kg
1
39 kg
T1
3.8 m
15 kg
2
What is the magnitude of the tension T?
The acceleration due to gravity is 9.8 m/s?.
Assume up is positive.
Answer in units of N.

Answer:

The  value is  [tex]q =  70.8\ C[/tex]

Explanation:

From the question we are told that

   The  area is  [tex]A =  2.0 \ k m^2 = 2000000 \  m[/tex]

   The height is  [tex]h  =  10 \  km  =  10000 \  m[/tex]

    The  electric field is  [tex]E =  4.0*10^{6} \  N/C[/tex]

Generally the electric field is mathematically represented as

      [tex]E =  \frac{q}{A *  \epsilon_o }[/tex]

Here [tex]\epsilon_o }[/tex] is the permitivity of free space with value  

      [tex]\epsilon_o }   =  8.85 *10^{-12} \  N \ m^2 /C^2[/tex]

=>   [tex]4.0*10^{6} =  \frac{q}{2000000 *    8.85 *10^{-12} }[/tex]

=> [tex]q =  70.8\ C[/tex]

Answer:

480

Explanation:

first simplefie 80 ÷10

now

8x 60

480

The main reason Venus has a "uniform" temperature is atmospheric convection.

Atmospheric convection is a term to refer to the difference in temperature, in the atmospheric layer. In short, it refers to the movement of air bodies of different temperatures in the atmosphere of a planet.

In the case of Venus, atmospheric convection contributes to this planet's temperature uniformity because hot air is in constant contact with the planet's surface, preventing it from cooling.

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Answer:

A - ¼ of the force

Explanation:

According to Pascal’s principle, there's an equation that relates the force to area ratios in hydraulic systems, provided that the pistons will be at the same vertical height and that the friction in the given hydraulic system is negligible. This formula is;

F1/A1 = F2/A2

Thus;

F1/F2 = A1/A2

Where F is the applied force and A is area of pad.

Now, we are told that the size of the pad is doubled. Thus;

If the initial radius is r, then the new radius is 2r

Thus;

F1/F2 = (πr²)/(π(2r)²)

F1/F2 = r²/4r²

F1/F2 = ¼

F1 = ¼F2

Thus, the answer is ¼ of the force

Answer: I'm not very good at all. I need help with determine the velocity of an object just before it hits the ground, given the object's mass and the distance it falls. For example, a .060 kg (or 60 g) ball falls from a 12 meter building. Also, assume it is free falling (no air resistance) and the average time it takes to hit the ground it 1.16 seconds.

Explanation:

Explanation:

Crossing over is the swapping of genetic material that occurs in the germ line. During the formation of egg and sperm cells, also known as meiosis, paired chromosomes from each parent align so that similar DNA sequences from the paired chromosomes cross over one another.

Answer:

Explanation:

After cross-over, the chromosomes are randomly shuffled to the daughter cells as the cell divides. This process results in in gametes that with unique combinations of chromosomes and is called independent assortment.

The magnitude of the point charge held stationary at the origin and at given electric potential is determined as 3.42 x 10⁻⁶ C.

The magnitude of the charge held stationary at the origin is calculated as follows;

V = kq/r

where;

Substitute the given parameters into the above equation and solve magnitude of the point charge.

kq = Vr

q = Vr/k

q = (2.8 x 10⁴ x 1.1)/(9 x 10⁹)

q = 3.42 x 10⁻⁶ C

Thus, the magnitude of the point charge held stationary at the origin and at given electric potential is determined as 3.42 x 10⁻⁶ C.

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Answer:

We are given that the student has a weight of 400N

So, the force applied by the boy has to be added to 400 since he has to apply force to keep his body on the stairs

We are given:

u = 0 m/s

v = v m/s

t = 4 s

a = a m/s/s

s = 3 m

w = 400 N

Converting to Mass

w = m * g    (where m is the actual mass and g is the value of gravity)

letting g = 10m/s/s

400  = 10 * m

m = 40 kg

Finding Acceleration:

From the second equation of motion:

s = ut + (a*t*t) / 2

3 = (0)(4) + (a)(4)(4)/2

3 = 8a

a = 3/8 m/s/s

Force Applied by the Student:

Now that we have the acceleration, we can find the force applied by the student to go up the stairs

F = ma (Newton's second law of motion)

F = (40)(3/8)

F = 15 N

Hence, the student is applying 15N to climb up the stairs and another 400 N to support his mass

Total Force applied by the Student:

F(total) = F(climbing) + F(weight)

F(total) = 15 + 400

F(total) = 415 N

Work done by the Student:

Work done = F.S   (dot product of force applied and displacement)

Work done = (415)(3)

Work done = 1245 Joules

Power Output of the Student:

power output is the amount of work done per second

Power = Work Done / t

Power = 1245 / 4

Power = 311.25 Watt

**note that the this is just the power consumed by the student to climb the stairs**

Answer:

F(friction) = μ M g         definition of frictional force

μ  = F / (M g) = 11 N / 50 N = .22

The co-efficient of kinetic friction between the block and the table is 0.22.

To find the co-efficient of friction, the given values are,

Friction force - 11 N

Normal force - 50 N

         μ = 11 N / 50 N

             = 0.22

Thus, the co-efficient of the kinetic friction is 0.22

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Answer:

The answer is non-mineral nutrients.

Answer:

non-mineral nutrients.

Explanation:

Answer:

a.

Explanation:

i did a quizz for it im sorry if im wrong

Answer:

it's because of friction

Explanation:

In slanted one the friction will be less so it would be more difficult to climb

Answer:

explain the meaning of the term work done and give an example of where work is done

The force of attraction between the balloon and the rod is 9 x 10⁻³ N.

The force of attraction between the balloon and the rod is calculated by applying Coulomb's law as shown below;

[tex]F = \frac{kq_1 q_2}{r^2}[/tex]

where;

F = (9 x 10⁹) x (2.0 x 10⁻⁸)²/(0.02)²

F = 9 x 10⁻³ N

Thus, the force of attraction between the balloon and the rod is 9 x 10⁻³ N.

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[tex]\:\:\:\:\:\:\:\:\:\:\star [/tex]We can calculate the force between the balloon and the rod applying "Coulomb's Law". Which states-

[tex]\:\:\:\:\:\:\:\:\:\:\star\:\sf \underline{F_{(air)} = K\times \dfrac{ q_1 q_2}{r^2}} \\ [/tex]

[tex] \:\:\:\:\:\:\:\:\:\:\star\:\sf\underline{F_{(air)} = \dfrac{1}{4\pi \epsilon_{0} } \dfrac{q_1q_2}{r^2} }\\ [/tex]

Where-

As per question, we are given that -

Now that required values are given, so we can substitute the values into the formula and solve for Force -

[tex] \:\:\:\:\:\:\:\:\:\:\star\:\sf\underline{F_{(air)} = \dfrac{1}{4\pi \epsilon_{0} } \dfrac{q_1q_2}{r^2} }\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)}= 9\times 10^9 \times \dfrac{-2\times 10^{-8}\times 2\times 10^{-8} }{(0.02)^2}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)}= -9\times 10^9 \times\dfrac{4\times10^{-16}}{4\times 10^{-4}}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)}= -9\times 10^9 \times\dfrac{\cancel{4}\times10^{-16}}{\cancel{4}\times 10^{-4}}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)} = -9\times 10^9 \times 10^{-16} \times 10^{4}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)} = -9\times 10^9 \times 10^{-16+4} \\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)} = -9\times 10^9 \times 10^{-12} \\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)} = -9\times 10^{9-12}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf Force_{(air)} = -9\times 10^{-3}\\ [/tex]

[tex] \:\:\:\:\:\:\:\longrightarrow \sf \underline{Force_{(air)} = -9\times 10^{-3}\:N}\\ [/tex]

Answer:

perpendicular

Explanation:

The projectile motion of an object can be described by two independent vector components that are horizontal and vertical motion.

The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.

The item is known as a projectile, and the course it takes is known as a trajectory. Falling object motion is a simple one-dimensional kind of projectile motion with no horizontal movement.

The projectile's motion is divided into two parts: horizontal and vertical motion.

Because perpendicular motion components are independent of one another, these two motion components may be treated individually.

Hence two independent vector components are horizontal and vertical motion.

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Answer:

0.65 sec

Explanation:

V = L*ΔI/Δt

so Δt = L*ΔI/V = 1.20 * (7.00 - 1.00)/11.00 = 0.65 sec

Answer: When there is no change in chemical composition of a substance then it is known as physical change. When we say that mass is conserved during a physical change then it means that total mass of the substance or object will remain the same from beginning till the end

Answer:

Sample Response: When a physical change occurs, the mass of the substance is conserved. This means that the total mass of the substance remains the same from beginning to end. The physical properties of the substance, such as size and shape, may change, but the amount of matter in the substance does not change.

Explanation:

Answer:

D.

Explanation:

Budgeting time, avoiding stress, and prioritizing.

Answer:

a)  0.002 cm³

b) 0.005 cm³

Explanation:

density d = 19 kg/cm³

Given mass m, find volume V

The definition of density is

d = m/V

so V = m/d

a) m = 38 g = 0.038 kg

so V = 0.038/19 = 0.002 cm³

b) m = 0.095 kg

V = 0.095/19 = 0.005 cm³

Answer:

Explanation:

Writing out the Newton's Law pf Cooling:

dT/dt = -k * (T - A),

where T is the temperature of the coffee, A is the room temperature, and k is a positive constant.

If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C,

T = 100

A = 25

dT = 100 - 90 = 10

dt = 1

Putting the figures into the equation:

10/1 = -k * (100 - 25)

k = -10/75°C

After 4 minutes, dT/4 = 10/75 (100 - 25) = 10

dT = 40

Temperature after 4 minutes = 100 - 40 = 60°C

The temperature of a cup of coffee varies according to Newton's Law of Cooling, the temperature of the coffee after 4 minutes is approximately 67°C.

To tackle this problem, we can apply Newton's Law of Cooling's differential equation and solve it using variable separation.

dT/dt = -k(T - A)

At t = 0 (initial condition): T = 100°C

At t = 1 minute: T = 90°C

dT/dt = -k(T - A)

At t = 0: dT/dt = -k(100 - 25)

So,

-10 = -k(75)

k = 10/75

Separating variables and integrating, we have:

1/(T - A) dT = -k dt

∫(1/(T - A)) dT = ∫(-k) dt

ln|T - A| = -kt + C

ln|100 - 25| = 0 + C

ln|75| = C

So, the equation will be:

ln|T - A| = -kt + ln|75|

ln|(T - 25)/(75)| = -kt

Now,

ln|((T - 25)/(75))| = -(10/75)(4)

|((T - 25)/(75))| = [tex]e^{(-40/75)[/tex]

T - 25 = ± 75 *  [tex]e^{(-40/75)[/tex]

T = 25 ± 75 *  [tex]e^{(-40/75)[/tex]

T ≈ 25 ± 42.42

Therefore, the temperature of the coffee after 4 minutes is approximately:

T ≈ 25 + 42.42 = 67.42°C

Thus, the temperature of the coffee after 4 minutes is approximately 67°C.

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Answer:

a = 2.5 m/s^2

Explanation:

u = 0

v = 25

t = 10

(using first eq. of motion)

a = (v - u) /t

a = (25 - 0) /10

a = 25/10

a = 2.5 m/s^2

Given parameters:

Volume of metal block = 0.035mL

Mass of block  = 765mg

Unknown:

Volume of metal block = ?

Solution:

Density is the mass per unit volume of a substance.

  Density  = [tex]\frac{mass}{volume}[/tex]

Mass is the amount of matter in a substance

Volume is space occupied by a substance

     Mass of the block in g;

                         765mg =    765 x 10⁻³ g

                                     = 0.765g

 Density = [tex]\frac{0.765}{0.035}[/tex]   = 21.86g/mL

The density of the metal block is 21.86g/mL

Answer:

ززث ع ثذآرثفاقآثہ تجفہجاءآڑجءبیجدءچءاءاڑڑشءڑدءحابڑءحبزیاڑچءازءچزیبزچدزچ،چدز،ڈزیجپا،جڈزیجڈیبزءفحدءڈحضغ دح دفھچءچدجءیچزءچددفحءدفءاڑحعخ حدھغچغھافءاڑاڑہشھدفڈچقڈ،شڈیچڑ

Answer:

Graph #2

Explanation:

Notice, on the position time graph, there is a constant rate of change. This means that there is a constant velocity, meaning it's a straight line. The line is decreasing so the velocity will be negative.

Answer:

neutrons protons electrons

Explanation:

after the Big Bang, the universe was filled with neutrons, protons, electrons, anti-electrons, photons and neutrinos.