A painter sits on a scaffold that is connected to a rope passing over a pulley. The other end of the rope rests in the hands of the painter who wants to lift the scaffold. She plans to pull downward on the loose end of the rope, thinking that the scaffold will then rise vertically with her along for the ride. The scaffold has a mass of 52 kg, and her mass is 63 kg. The painter pulls downward on the rope with a force of 600.0 N, while she and the scaffold are hanging from the other end above the ground.

Required:
a. What is the net acceleration on the system consisting of the painter and the scaffold?
b. What is the magnitude of the normal force exerted on the painter by the scaffold?

Answer:

d. 110 MW

Explanation:

The computation of the power loss is given below:

= I^2 × R

= 1000^2 × 110

= 110 × 10^6 W

= 110 MW

Here I represent the current through the resistance

ANd, R represent the resistance

So, the correct option is d

Answer:

The warm seas create a large humid air mass. The warm air rises and forms a low pressure cell, known as a tropical depression.

Explanation:

Hurricanes arise in the tropical latitudes (between 10 degrees and 25 degrees N) in summer and autumn when sea surface temperature are 28 degrees C (82 degrees F) or higher.

Answer:

air

Explanation:

Answer:

warm front .,

Explanation:

...........

Answer:

B

Explanation:

The crest of all three waves are of equal height

If a boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water, then the horizontal distance traveled by the boy would be 2.58 meters.

There are three equations of motion given by Newton,

v = u + at

S = ut + 1/2 × a × t²

v² - u² = 2 × a × s

As given in the problem if a boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water,

3 = ut + 1/2 × a × t²

3 = 0 + 0.5 × 9.8 × t²

t = 3 / 4.9

t = 0.7824

The horizontal distance traveled by the boy = 3.3 ×  0.7824

                                                                         = 2.58 meters

Thus, the horizontal distance traveled by the boy would be 2.58 meters.

To learn more about equations of motion here, refer to the link;

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Answer:

A. Coefficient of kinetic friction, μ = 0.34

B. The box moves a distance of 9.64 m before coming to a stop

C. The heavier box will stop after 2.1 seconds

Explanation:

a. The coefficient of kinetic or sliding friction is given as: μ = F/R

where applied force, F = m × (∆v)/t

∆v = v - u

∆v = 0 - 8 m/s = -8 m/s; t = 2.4 s

R = normal reaction = m×g

where g = 9.8 m/s²

Substituting in the kinetic friction formula; μ = m∆v/t ÷ 1/m×g

μ = ∆v/g×t

μ = 8 / 9.8 × 2.4

μ = 0.34

b. Using the equation v² = u² + 2as to calculate the distance travelled by the box

where v = 0 m/s; u = 8.0 m/s; a = ? s = ?

From F = ma = μR

a = μR/m = (μ × m × g)/m

a = μg

a = 0.34 × 9.8

a = 3.32 m/s²

This is negative acceleration or deceleration

Substituting in the equation of motion

8² + 2 × -3.32 × s = 0

-6.64s = -64

s = 9.64 m

Therefore, the box moves a distance of 9.64 m before coming to a stop

c. The coefficient of friction is independent of mass.

Using the formula in (a): μ = ∆v/g×t

t = ∆v/μg

t = 7/0.34 × 9.8

t = 2.10 s

Therefore, the heavier box will stop after 2.1 seconds

The  coefficient of kinetic friction of the box is 0.34.

The distance traveled by the box is 9.6 m.

The time taken for the heavier box to stop is 2.1 s.

The  coefficient of kinetic friction of the box is calculated as follows;

[tex]\mu mg = ma\\\\\mu g = a\\\\\mu g = \frac{v}{t} \\\\\mu = \frac{v}{gt} \\\\\mu = \frac{8}{9.8 \times 2.4} \\\\\mu = 0.34[/tex]

The distance traveled by the box is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\2as = -u^2\\\\s = \frac{-u^2}{2a} \\\\s = \frac{-u^2}{2\mu g} \\\\s = \frac{-(8)^2}{2\times 0.34 \times 9.8} \\\\s = -9.6 \ m\\\\|s| = 9.6 \ m[/tex]

The time taken for the heavier box to stop is calculated as;

[tex]\mu = \frac{v}{gt} \\\\t = \frac{v}{\mu g} \\\\t = \frac{7}{0.34 \times 9.8}\\\\ t = 2.1 \ s[/tex]

Learn more about coefficient of kinetic friction here: https://brainly.com/question/20241845

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh   (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT   (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C    (3)

ΔT = 5.39x10⁹ / 4200 * 75000

Hope this helps

Answer:

A plant

Explanation:

because animals don't have cell walls, and fungus and bacteria dont have chloroplasts

Answer:

A. v = √2gh

B. No! The final velocity does not depend on the mass of the car.

C. Yes! the final velocity depends on the steepness of the hill

D. 3.28 m/s

Explanation:

A. Determination of the final velocity.

½mv² = mgh

Cancel out m

½v² = gh

Cross multiply

v² = 2gh

Take the square root of both side

v = √2gh

B. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is no mass (m) in the formula.

Thus, the final velocity does not depend on the mass of the car.

C. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is height (h) in the formula.

Thus, the final velocity depends on the steepness of the hill

D. Determination of the final velocity.

Height (h) = 0.55 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = √2gh

v = √(2 × 9.8 × 0.55)

v = √10.78

v = 3.28 m/s

Answer:

Check Newton's Rings:

d = height of air film

s = distance from center to ring being considered

R = radius of circle considered

The approximate formula is:

d = s^2 / (2 R)   or s = (2 R d)^1/2

If we just use 4000 mi for R and 1/4 mi for d the height

we get s = (2 * 4000 * 1/4)^1/2 = 2000^1/2 mi = 45 mi

Answer:

The first choice, one block per minute.

Answer:

I. 100 Bg

II. 50 Bg

III. 25 Bg

Explanation:

I. Determination of the activity after 26 hours.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 13 hours

Time (t) = 26 hours

Number of half-lives (n) =?

n = t / t½

n = 26 / 13

n = 2

Finally, we shall determine remaining activity

Original activity (N₀) = 400 Bg

Number of half-lives (n) = 2

Activity remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

N = 1/4 × 400

N = 100 Bg

II. Determination of the activity after 39 hours.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 13 hours

Time (t) = 39 hours

Number of half-lives (n) =?

n = t / t½

n = 39 / 13

n = 3

Finally, we shall determine remaining activity.

Original activity (N₀) = 400 Bg

Number of half-lives (n) = 3

Activity remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2³ × 400

N = 1/8 × 400

N = 50 Bg

III. Determination of the activity after 52 hours.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 13 hours

Time (t) = 52 hours

Number of half-lives (n) =?

n = t / t½

n = 52 / 13

n = 4

Finally, we shall determine remaining activity

Original activity (N₀) = 400 Bg

Number of half-lives (n) = 4

Activity remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2⁴ × 400

N = 1/16 × 400

N = 25 Bg

Answer:

is there a. b.  c  or d?

Explanation:

Answer: d = st

Explanation:

We know that the distance is equal to the rate (speed) times the time

d = st

Answer:

The height of the hill is 0.46 m.

Explanation:

Given;

mass of the child and sled, m = 50 kg

initial velocity of the sled, u = 0

final velocity of the sled, v = 3 m/s

The height of the high is calculated from the law of conservation of energy;

P.E at top = K.E at bottom

mgh = ¹/₂mv²

gh = ¹/₂v²

[tex]h = \frac{v^2}{2g} \\\\h = \frac{3^2}{2\times 9.8} \\\\h = 0.46 \ m[/tex]

Therefore, the height of the hill is 0.46 m.

This question is incomplete, the missing image is uploaded along this answer below;

Answer:

a) the required angular acceleration magnitude α is  π rad/s² or 3.14 rad/s²

b) the time at which the contact occur is 8 seconds

Explanation:

Given the data in the question;

first we convert the given angular velocity to rad/s

angular velocity = 240 rpm = ((240/60) × 2π ) = 8π rad/s

so

ωA = 8π rad/s

next we determine angular acceleration at point A; so

ωA = at

8π rad/s = at -------let this be equation

thus, angular acceleration of disk A is ωA and rotates in counter clockwise direction.

Next we determine the velocity of point C;

Vc = rA × ωA

where Vc is velocity at point C, rA is radius of A ( 150/1000)m,  { from the diagram }

so we substitute

Vc = 0.15m × 8π

Vc = 1.2π m/s

for angular velocity at point B;

Vc = rB × ωB

where rB is the radius of B ( 200/1000)m

we substitute

1.2π = 0.2 × ωB

ωB = 1.2π / 0.2

ωB = 6π rad/s

Thus, the wheel B rotates with an angular velocity of 6π rad/s in clock wise direction.

Now,

a) Determine the required angular acceleration magnitude α

we find the the angular acceleration of disk B after 2 seconds, using the expression;

ωB = at

where angular acceleration is a and t is time ( t - 2)

we substitute

ωB = at

6π = a( t - 2) -------- let this be equation 2

now, lets substract equation 1 form equation 2

(6π = a( t - 2)) - (8π = at)

(6π = at - 2a) - ( 8π = at)

-2π = 0 + -2a

2π = 2a

a = 2π/2

a = π rad/s² or 3.14 rad/s²

Therefore, the required angular acceleration magnitude α is  π rad/s² or 3.14 rad/s²

b) determine the time at which the contact occurs;

from equation 1

8π = at

we substitute in the value of a

8π = π × t

t = 8π / π

t = 8 seconds

Therefore, the time at which the contact occur is 8 seconds

Answer:

47947.52 J.

Explanation:

From the question,

Amount of heat given of (Q) = mc(t₁–t₂).................... Equation 1

Where m = mass of water, c = specific heat capacity of water, t₁ = initial temperature, t₂ = final temperature.

Given, m = 640 g = 640 g, c = 4.816 J/g°C, t₁ = 76 °C, t₂ = 28 °c.

Substitute these values into equation 1 above

Q = 640×4.816(48)

Q = 147947.52 J.

Hence the amount of heat given off is 47947.52 J.

Answer:

10 mm

Explanation:

We'll begin by calculating the spring constant of the spring. This can be obtained as follow:

Extention (e) = 5 mm

Force (F) = 125 N

Spring constant (K) =?

F = Ke

125 = K × 5

Divide both side by 5

K = 125 / 5

K = 25 N/mm

Finally, we shall determine how much the spring will stretch when a 250 N force is applied. This can be obtained as follow:

Force (F) = 250 N

Spring constant (K) = 25 N/mm

Extention (e) =?

F = Ke

250 = 25 × e

Divide both side by 25

e = 250 / 25

e = 10 mm

Thus, the spring will stretch 10 mm when a 250 N force is applied.

Answer:

Displacement = 30 miles due east.

Explanation:

Let the distance due west be A

Let the distance due east be B

Given the following data;

A = 50 miles

B = 80 miles

To find the displacement;

Displacement can be defined as the change in the position of a body or an object. It is a vector quantity because it has both magnitude and direction.

Thus, the displacement would be calculated by subtracting distance A from distance B because the rider rode in opposite directions.

Displacement = B - A

Displacement = 80 - 50

Displacement = 30 miles due east.

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

Answer:

Explanation:

a ) The volume of blood flowing per second throughout the vessel is constant .

a₁ v₁ = a₂ v₂

a₁ and a₂ are cross sectional area at two places of vessel and v₁ and v₂ are velocity of blood at these places .

2A x v₁ = A x .40

v₁ = .20 m /s

b )

Let normal pressure be P₁ when cross sectional area is 2A and at cross sectional area A , pressure is P₂

Applying Bernoulli's theorem

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

P₁ - P₂ = 1/2  ρ(v₂² - v₁² )

= .5 x 1060 ( .4² - .2² )

= 63.6 Pa .

Answer:

The cutoff wavelength is the minimum wavelength in which a particular fiber still acts as a single mode fiber. Above the cutoff wavelength, the fiber will only allow the LP01 mode to propagate through the fiber (fiber is a single mode fiber at this wavelength).

Explanation:

Answer:

The mass of the fish is 18.1 kg.

Explanation:

Given;

weight of the fish, F = 108 N

extension of the spring by the given weight, x = 14 cm = 0.14 m

First, determine the elastic constant of the spring by applying Hook's law;

F = kx

where;

k is the spring constant

k = F/x

k = 108 / 0.14

k = 771.43 N/m

When the spring is stretched to 23cm, the mass of the fish is calculated as follows;

[tex]F = mg = Kx\\\\m = \frac{Kx}{g} \\\\m = \frac{771.43\times 0.23}{9.8} \\\\m = 18.1 \ kg[/tex]

Therefore, the mass of the fish is 18.1 kg.

Answer:

the  frequency of the oscillation is 2.5 Hz.

Explanation:

Given;

time to complete the oscillation, t = 0.4 s

number of oscillations, n = 1

The frequency of the oscillation is calculated as;

[tex]F = \frac{n}{t} \\\\F = \frac{1}{0.4} \\\\F = 2.5 \ Hz[/tex]

Therefore, the  frequency of the oscillation is 2.5 Hz.

Answer:

A. shin guards is that answer

Answer:

Shin Guards Brainlyiest please

Explanation:

Answer:

Depending, on how much it's push against together.

Explanation:

Since, the two objects are getting in contact. But, if it's a type of item/thing there's a different frictions, but I know it's normal friction when two objects comes in contact. But, its depending on how much you push it against the two items.

Answer:b

Explanation:

Guess

Answer:

In a scientific experiment, a constant error -- also known as a systematic error -- is a source of error that causes measurements to deviate consistently from their true value.

Explanation:

Answer:

 ρ= 1.378 10⁴ Ω / m

Explanation:

Let's use ohm's law

          V = i R

           R = V / i

let's calculate

           R = 9.0 / 230 10⁻⁶

           R = 3.9 10⁴ Ω

now we can use the definite of resistance

           R = ρ [tex]\frac{L}{A}[/tex]

the area of ​​a circle is

           A = π r² = π (d/2)²

           ρ = R A / L

           ρ = π R [tex]\frac{d^2}{4L}[/tex]

let's calculate

          ρ = π 3.9 10⁴  [tex]\frac{(1.5 \ 10^{-3}^2 }{4 \ 5 \ 10^{-2}}[/tex]

          ρ= 1.378 10⁴ Ω / m