Q1 what is the direction of current?​

Answer:

x = 0.056 m

ΔKE = 0.489 J

Explanation:

Given that

Angle, θ = 38°

Length, L = 1.7 m

Mass, m = 0.09 kg

Spring constant, K = 590 N/m

If we use the Work-Energy theorem, then we know that Potential Energy, PE = Kinetic Energy, KE

This is mathematically written as

1/2kx² = mgH

The height, H we can get by using the relation

H = L.Sinθ

H = 1.7 * Sin 38

H = 1.7 * 0.6157

H = 1.047 m

Next, we use the Work-Energy theorem

1/2kx² = mgH

1/2 * 590 * x² = 0.09 * 9.8 * 1.047

295 * x² = 0.9234

x² = 0.9235 / 295

x² = 0.00313

x = √0.00313

x = 0.056 m

If the spring loses contact at x = 0.056, definitely, it will also lose contact at x = 0.8

Then we use the formula

ΔKE = mg(H - H1)

ΔKE = mg(xsinθ - x2.sinθ)

Where, x = 1.7 , x2 = 0.8

ΔKE = 0.09 * 9.8 (1.7 * sin 38 - 0.8 * sin 38)

ΔKE = 0.882(1.047 - 0.493)

ΔKE = 0.882 * 0.554

ΔKE = 0.489 J

Answer:

9 hr

Explanation:

Speed = distance/time

Let the time taken by the train be t.

=> 70 = 630/t

=> t = 630/70

=> t = 9

it take 9 hr to complete its journey

Answer:

9 hours

Explanation:

Using the formula to solve motion problems: d=r*t (Distance = Rate * Time)

Given:

Distance = 630 km

Rate (Speed) = 70 km/h

Time = Distance / Time

Time = 630/70

               = 9

One similarity is the use of physical body whereas one difference is that one is exercise and the other is a sport.

One similarity between a set and a bump in volleyball is the movement and use of legs and hands.

Whereas one difference between a set and a bump in volleyball is that completing several reps of a particular exercise in a row is called a set while on the other hand,  the basic pass in volleyball is known as bump.

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Answer:

3 m/s

Explanation:

A= F/m

12,000/ 4000 = 3

Answer:

3 m/s^2

Explanation:

The equation you have to use is F=ma because the problem is a Newton's 2nd law problem.

Our known values are:

F ( Force ) = 12,000 N

m ( mass ) = 4,000 kg

a ( acceleration ) = ?

Now we plug in the known values into the equation and solve

F=ma

12,000=4,000a

We have to divide 4,000 by both sides to isolate the a value

12,000/4,000=4,000/4,000a

The 4,000s on the right of the equation cancel.

And 12,000 divided by 4,000 equals 3

The acceleration (a) is 3 meters per second squared (m/s^2)

Next, check to make sure 3 does work by plugging it back into the equation.

12,000=4,000*3

12,000=12,000 ✔

As you can see, the acceleration will be 3 m/s^2

Answer:

Both.

Explanation:

Given that a block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. 1)Which one makes it furthest up the ramp ?

Since both of them have the same mass and the same initial velocity, then, they will both have the same kinetic energy.

That is,

K.E = 1/2mv^2

Friction is a force that opposes motion. And since the frictional force is zero,

Both of them will accelerate from Newton's law.

F = ma

We can therefore conclude that both of them will make it further up the ramp.

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

          x = v₀² / 2a

let's calculate

          x = 2²/(2 0.8)

         x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

Answer:

B or 2

Explanation:

Answer:

fem = 7.58 10⁻⁵ V

Explanation:

For this exercise we use Faraday's law

          fem =   [tex]- \frac{d \Phi _B}{dt}[/tex]

the magnetic flux is

          Ф_B = B. A = B A cos θ

Tje bold are vectros.  Suppose the case where the normal to the surface of the red blood cell is parallel to the field, therefore the angle is zero and the cos 0 = 1

The red blood cell area is

          A =π r²

indicate that the diameter is r = 8.00 mm = 8.00 10⁻³ m

the magnetic field has a frequency of f=60 Hz, and B₀ = 1.00 10⁻³T,  therefore we can write it

          B = B₀ sin (wt) = B₀ sin( 2π f t)

we substitute

           fem = - A dB / dt

           fem = - A B₀  [tex]\frac{ d (sin ( 2\pi f t)}{dt}[/tex]

           fem = - π r² Bo (2πf  cos 2πft)

the maximum electromotive force occurs when the function is ±1

           fem = 2 π² r² B₀ f

let's calculate

           fem = 2π²  (8.00 10⁻³)²  1.00 10⁻³ 60

           fem = 7.58 10⁻⁵ V

Answer:

geiger counter with geiger mueller

Answer:

Explanation:

Angular momentum = [tex]I*w[/tex]

I = Moment of inertia

w = Angular velocity

Hollow pipe can be considered as hollow cylinder

Moment of Inertia of Hollow cylinder = [tex]Mr^2[/tex]

Moment of Inertia of solid cylinder= [tex]\frac{Mr^2}{2}[/tex]

Clearly Moment of inertia of hollow pipe is greater which states that Angular momentum of it will be greater.

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Answer:

It's plastic.

trust me it's plastic, i've rad it somewhere.

All of them have something that's not like the others.

-- Rubber is the only one on the list that has two repeated letters.

-- Plastic is the only one on the list thagt has no repeated letters.

-- Plastic is the only one on the list that has no 'r' in its name.

-- Copper is the only one on the list that is an element, not a compound.

-- Copper is the only good electrical conductor on the list.

-- Plastic is the only one on the list with more than six letters in its name.

-- Rubber is the only one on the list with no 'p' in its name.

-- Plastic is the only one on the list that doesn't end in "-er".

Answer:

117/60*50

Explanation:

Trust me bc I'm smart

Answer:

50N

Explanation:

W=Fd

150=F(3)

50N=F

Answer:

Explanation:

you can say the law of superpoition can tell us that each rock layer is older than the one above it. So, the relative age of the rock or fossil in the rock or fossil in the rock is older if it is farther down in the rock layers. hope it helps

Answer:

(a)

W_friction = 98.1 J

W_net = 550.9 J

(b)

W_friction = 98.1 J

W_net = 463.95 J

Explanation:

(a)

First, we will calculate the work done by friction:

[tex]W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\[/tex]

W_friction = 98.1 J

Now, the work done by Eskimo will be:

[tex]W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 0^o\\[/tex]

W_Eskimo = 649 J

So, the net work will be:

W_net = W_{Eskimo} - W_{friction}

W_net = 649 J - 98.1 J

W_net = 550.9 J

(b)

First, we will calculate the work done by friction:

[tex]W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\[/tex]

W_friction = 98.1 J

Now, the work done by Eskimo will be:

[tex]W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 30^o\\[/tex]

W_Eskimo = 562.05 J

So, the net work will be:

W_net = W_{Eskimo} - W_{friction}

W_net = 562.05 J - 98.1 J

W_net = 463.95 J

Answer: Q= QH - QL

= W + QL - QL

= W

= 1 kj/8

Explanation:

Since the net rate is positive the room will be heated.

Answer:

a

Explanation:

How long would it take a machine to do 5.000

joules of work if the power rating of the machine

is 100 watts?

The given machine will take 50 s to complete the work. the power is the rate of performing work.

It can be defined as the rate of performing work. It can also be written as the amount of work divided by the time it takes to complete the work.

[tex]p = \dfrac wt[/tex]

So

[tex]t = \dfrac w p[/tex]

Where,

[tex]p[/tex] - power = 100 watt  = 100 J/s

[tex]t[/tex] - time = ?

[tex]w[/tex]- work = 5000 J

Put the values in the formula,

[tex]t = \dfrac{ 5000 \rm \ J} {100 \rm \ J/s}\\\\t = 50 \rm \ s[/tex]

Therefore, the given machine will take 50 s to complete the work.

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Answer :

Last choice

Answer:

its frequency is too low

Answer:

1.61 N/kg

Explanation:

Take the universal gravitational constant G as 6.67 × 10^(-11) Nm²/kg²

The required gravitational field strength

= 6.67 × 10^(-11) × 7.3 × 10^(22) / (1738000)²

= 1.61194103 N/kg

= 1.61 N/kg (corr. to 3 sig. fig.)

Answer:

200 meters

Explanation:

20 x 10 = 200

Answer:

c

Explanation:

Answer:

When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.

Explanation:

Answer: True

Explanation: "The lens is a transparent biconvex structure in the eye that, along with the cornea, helps to refract light to be focused on the retina."

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

The image of this hollow sphere and uniform rod is missing, so i have attached it.

Answer:

A) J = 0.7443 kg•m²

B) T = 1.9169 N•m CCW

C) α = 2.5754 rad/s²

D) a = 3.966 m/s²

Explanation:

A) The moment of inertia J of the contraption around the fulcrum is given by the formula;

J = Jℓ + Jr

Let's calculate Jℓ

Jℓ = [((0.34²/3) × 0.50 × 0.34)/1.78] + (0.48 × (0.34 + 0.64)²)

Jℓ = 0.4647 kg•m²

Now, let's Calculate Jr

Jr = ((1.78 - 0.34)²/3) × ((1.78 - 0.34)/1.78) × 0.50

Jr = 0.2796 kg•m²

Thus;

J = 0.4647 + 0.2796

J = 0.7443 kg•m²

(b) Using CCW as positive, Torque in Nm is calculated as;

T = Tℓ - Tr

Let's calculate Tℓ

Tℓ = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81

Tℓ = 4.7739 N•m CCW

Now, let's Calculate Tr;

Tr = [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

Tr = 2.857 N•m CW

Thus;

T = 4.7739 - 2.857

T = 1.9169 N•m CCW

(c) The angular acceleration α of the contraption, using CCW is gotten from the formula;

α = T/J

α = 1.9169/0.7443

α = 2.5754 rad/s²

(d) The linear acceleration a of the right end of the rod, using up as positive is given by;

a = α*(1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

A) the moment of inertia of the contraption is 0.7443 kgm²

B) The torque about the fulcrum is 1.9169 Nm

C) Angular acceleration of the contraption is 2.5754 rad/s²

D) The linear acceleration of the contraption is 3.966 m/s²

(A) The moment of inertia I of the contraption around the fulcrum is given by :

[tex]I = [(0.34^2/3) \times 0.50 \times 0.34)/1.78 + (0.48 \times (0.34 + 0.64)^2)] + [(1.78 - 0.34)^2/3) \times (1.78 - 0.34)/1.78) \times 0.50][/tex]

I = 0.4647 + 0.2796

I = 0.7443 kgm²

(B) Using CCW as positive, Torque in Nm is given by;

T = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81 - [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

T = 4.7739 - 2.857

T = 1.9169 Nm

(C) The angular acceleration (α) of the contraption is given by:

α = T/I

since, torque is defined as T = Iα

α = 1.9169/0.7443

α = 2.5754 rad/s²

(D) The linear acceleration (a) of the right end of the rod

a = αr

where r is the distance from the pivot

a = α × (1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

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Answer:

A Inertia

Explanation:

Answer:

The magnification of an image is equal to the ratio of the image height to the object height.

Answer:

a) the torque required is  10.53 N-m

b) The magnitude force applied tangentially is 12.33 N

Explanation:

Given the data in the question;

mass m = 1.765 kg

radius r = 0.854 m

first we calculate the moment of inertia;

[tex]I[/tex] = [tex]\frac{2}{5}[/tex]mr²

we substitute

[tex]I[/tex] = [tex]\frac{2}{5}[/tex] × 1.765 × (0.854)²

[tex]I[/tex] = 0.514897 kg.m²

a)

Find the torque required to bring the sphere from rest to an angular velocity of 317 rad/s, clockwise, in 15.5 s

ω[tex]_{initial[/tex] = 0

ω[tex]_{final[/tex] = 317 rad/s

t = 15.5 s

we know that; ω[tex]_{final[/tex] = ω[tex]_{initial[/tex] + ∝t

so we substitute

317 = 0 + ∝(15.5)

∝ = 317 / 15.5

∝ = 20.4514 rad/s²

so

ζ = [tex]I[/tex] × ∝

we substitute

ζ = 0.514897 × 20.4514

ζ = 10.53 N-m

Therefore, the torque required is  10.53 N-m

b)

What magnitude force applied tangentially at the equator would provide the needed torque.

ζ = F × r

we substitute

10.53  = F × 0.854

F = 10.53 / 0.854

F = 12.33 N

Therefore, magnitude force applied tangentially is 12.33 N