What is the final temperature of 0.3kg of water initially at 20 Celsius after 35 KJ Of heat is added?

Answer:

[tex]\frac{50}{\pi }[/tex]Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f        [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

Now,

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

By comparing equations (i) and (ii) the following holds;

V = 230

ω = 100

Ф = 0

But;

ω = 2 π f = 100

2 π f = 100             [divide both sides by 2]

π f = 50

f = [tex]\frac{50}{\pi }[/tex]Hz

Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz

Answer:

141178.8

Explanation:

use : density x velocity²

1.2 x 343² = 141178.8 pa

Answer:

-1748*10^N/C

Explanation:

See attached file

Answer:

the acceleration during the collision is: - 5  [tex]\frac{m}{s^2}[/tex]

Explanation:

Using the formula:

[tex]a=\frac{\Delta\,v}{\Delta\,t}[/tex]

we get:

[tex]a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}[/tex]

Answer:

Explanation:

given:

radius (r) = 1.88 km

velocity (v) = 136.3 km/hr

required:

banking angle ∡ ?

first:

convert 1.88 km to m = 1.88km * 1000m / 1km

r = 1880 m

convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)

v = 37.86 m/s

now.. calculate the angle

Ф = inv tan (v² / r * g)            we know that gravity = 9.8 m/s²

Ф = inv tan (37.86² / (1880 * 9.8))

Ф = 4.4°

Answer:

[tex]r_f=\frac{1}{6}r_i[/tex]

Explanation:

To find the new separation of the charges, you first take into account the formula for the electric force, when the force are separated a distance of ri.

You use the following expression:

[tex]F_i=k\frac{q_Aq_B}{r_i^2}[/tex]          (1)

k: Coulomb's constant

qA: charge of A particle

qB: charge of B particle

When the charges are separated to a new distance rf, the new force is 36Fi, if the charges have not changed, you have:

[tex]F_f=36F_i=k\frac{q_Aq_B}{r_f^2}[/tex]         (2)

To find the new separation you replace the expression for Fi of the equation (1) into the equation (2) and solve for rf in terms of ri:

[tex]36F_i=36k\frac{q_Aq_B}{r_i^2}=k\frac{q_Aq_B}{r_f^2}\\\\\frac{36}{r_i^2}=\frac{1}{r_f^2}\\\\r_f=\frac{1}{6}r_i[/tex]

The new separation of the charges is 1/6 times of the initial separation

Answer:

[tex]I = 0.083 kg m^2[/tex]

Explanation:

Mass of the bucket, m = 23 kg

Radius of the pulley, r = 0.050 m

The bucket is released from rest, u = 0 m/s

The time taken to fall, t = 2 s

Speed, v = 8.0 m/s

Moment of Inertia of the pulley, I = ?

Using the equation of motion:

v = u + at

8 = 0 + 2a

a = 8/2

a = 4 m/s²

The relationship between the linear and angular accelerations is given by the equation:

[tex]a = \alpha r[/tex]

Angular acceleration, [tex]\alpha = a/r[/tex]

[tex]\alpha = 4/0.050\\\alpha = 80 rad/s^2[/tex]

Since the bucket is falling, it can be modeled by the equation:

mg - T = ma

T = mg - ma = m(g-a)

T = 23(9.8 - 4)

The tension, T = 133.4 N

The equation for the pulley can be modeled by:

[tex]T* r = I * \alpha\\133.4 * 0.050 = I * 80\\6.67 = 80 I\\I = 6.67/80\\I = 0.083 kg m^2[/tex]

Answer:

-i - 7j

Explanation:

The computation of components of the force is shown below:-

torque T = r cross F

T = (4.00 i + 5.00 j + 0 k) X (1.00 i + 7.00 j)

Now we will cross multiplying vectorilly

T = 4 × (iXi) + 28 × (jXi) + 0 + 5 × (iXj)35 × (jXj) + 0

T = 4 × 0 + 28 × (-k) + 5 × (k) + 35 × 0

T = 28 × (-k) + 5 × (k) = -23k

net torque |T| = 23 N - m

direction >> negative k

Or Simply we can do by the below method

r × f

28 - 5 = 23 K

-i - 7j

Answer:

STATIC,  STATIC

KINETIC friction is less than static friction

Explanation:

In this exercise you are asked to complete the sentences with the correct words.

STATIC friction prevents the relative movement of two surfaces in contact.

For moving surfaces the friction is STATIC is greater than the kinetic friction.

For the last two sentences I think they are misspelled, the correct thing is

KINETIC friction is less than static friction

Answer:

208 V

Explanation:

resistor voltage = Vr = 124 V

capacitor voltage Vc = 167 V

the total voltage in the RC circuit is the resultant voltage of the resistor and the capacitor

total voltage i= [tex]\sqrt{Vr^{2} + Vc^{2} }[/tex]

==> [tex]\sqrt{124^{2} + 167^{2} } =[/tex] 208 V

Answer:

Explanation:

1 )

angular velocity of merry go round = 2π n

= 2π  x 3 / 60

ω = .1 x π radian / s

This will be the rotational speed of the whole system including that of Cora and Cameron . It will not depend upon their relative position with respect to

the centre of the merry go round .

So rotational speed of Cora = Rotational speed of Cameron

option ( d ) is correct .

2 )

Tangential speed  v = ω R where R is the distance from the centre of merry go round .

Tangential speed of Cora = .1 x π x 1

= .314 m /s

Tangential speed of Cameron = .1 x π x 2

= .628 m /s .

So tangential speed of Cameron is twice that of Cora .

Option ( e ) is correct .

Answer:

aaaaaaaaaaaaaaaaaaa

Explanation:

Answer:

12.5 rad/s²

Explanation:

Angular Acceleration: This can be defined as the ratio of linear acceleration and radius. The S.I unit is rad/s²

From the question,

a = αr................... Equation 1

Where a = linear acceleration, α = angular acceleration, r = radius.

But,

a = (v-u)/t.............. Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

(v-u)/t = αr

make α the subject of the equation

α = (v-u)/tr................. Equation 3

Given: v = 30 m/s, u = 0 m/s, t = 6 s, r = 0.4 m

Substitute into equation 3

α = (30-0)/(0.4×6)

α = 30/2.4

α = 12.5 rad/s²

Answer:

Explanation:

A converging lens id also known as a convex lens. A convex lens has a positive focal length.

Using the lens formula to determine the image distance from the lens for each object distance.

1/f = 1/u + 1/v

f = focal length

u = object distance

v = image distance

For an object placed at distance of 18.0cm with focal length 14.0cm,

1/v = 1/f-1/u

1/v = 1/14 - 1/18

1/v = 9-7/126

1/v = 2/126

v = 126/2

v = 63cm

Since the image distance is positive for an object 18cm from the lens, the image formed by the object at this distance is a real and inverted image.

The magnification = v/u = 63/18 = 3.5

Similarly for an object placed at distance of 7.0cm with focal length 14.0cm,

v = uf/u-f

v = 7(14)/7-14

v= -14.0 cm

Since the image distance is negative for an object placed 7.0 cm from the lens, the image formed by the object at this distance is a virtual and upright image.

The magnification = v/u = 14/7 = 2

Note that the negative value is not taken into account when calculating magnification. The negative value only tells us the nature of the image formed.

Given that,

Distance = r

Electric field :

Electric field is equal to the multiplication of electric constant and charge divided by square of distance.

In mathematically form,

[tex]E=\dfrac{kq}{r^2}[/tex]

[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}[/tex]

Electric force :

Electric force is equal to the multiplication of electric constant and both charges divided by square of distance.

In mathematically form,

[tex]E=\dfrac{kqQ}{r^2}[/tex]

[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{qQ}{r^2}[/tex]

We know that,

The relation between electric field and electric force is

[tex]E=\dfrac{F}{q}[/tex]

According to question,

If electron is released from rest then electron move towards the source point due to attractive force.

Hence, The electron move towards the source point due to attractive force.

Answer:

speed of puck acc. to the radar gun = 138 km/h

speed of player = 15 km/h

since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,

speed of puck = speed of player + speed of puck acc. to player

138 = 15 + speed of puck acc. to player

speed of puck acc. to player = 138 -15

speed of puck acc. to player = 123 km/h

Brainly this answer if you think it deserves it

Answer:

a) 23.51 m/s

b) 1.07 kg

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

[tex]K = \frac{1}{2} mv^2[/tex]

where m = mass of the body and v = speed of the body

We know that momentum is given as:

p = mv

Therefore:

K = 1/2 * pv

=> v = 2K / p

v = (2 * 295) / 25.1 = 23.51 m/s

The velocity of the body at that instant is 23.51 m/s.

b) Momentum is given as:

p = mv

=> m = p / v

m = 25.1 / 23.51  = 1.07 kg

The mass of the body at that instant is 1.07 kg

Answer:

The potential at the center of the sphere is -924 V

Explanation:

Given;

radius of the sphere, R = 0.22 m

electric field at the surface of the sphere, E = 4200 N/C

Since the electric field is directed towards the center of the sphere, the charge is negative.

The Potential is the same at every point in the sphere, and it is given as;

[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}[/tex] -------equation (1)

The electric field on the sphere is also given as;

[tex]E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}[/tex]

[tex]|q |= 4 \pi \epsilon _o} R^2E[/tex]

Substitute in the value of q in equation (1)

[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V[/tex]

Therefore, the potential at the center of the sphere is -924 V

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

[tex]i_1+i_3=i_2[/tex]                 (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

[tex]i_3=i_2-i_1=0.75A-0.40A=0.35A[/tex]

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

[tex]0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}[/tex]

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

Answer:

The speed of the rod is 1.383 m/s

Explanation:

Given;

length of the conducting rod, L = 31 cm = 0.31 m

induced emf on the rod, emf = 0.75V

magnetic field around the rod, B = 1.75 T

Apply the following Faraday's equation for electromagnetic induction in a moving rod to determine the speed of the rod.

emef = BLv

where;

B is the magnetic field

L is length of the rod

v is the speed of the rod

v = emf / BL

v = (0.75) / (1.75 x 0.31)

v = 1.383 m/s

Therefore, the speed of the rod is 1.383 m/s

Answer:

21 m/s

Explanation:

There are three forces on the pendulum:

Weight force mg pulling down,

Vertical tension component Tᵧ pulling up,

and horizontal tension component Tₓ pulling towards the center of the curve.

Sum of forces in the y direction:

∑F = ma

Tᵧ − mg = 0

T cos 37° = mg

T = mg / cos 37°

Sum of forces in the centripetal direction:

∑F = ma

Tₓ = mv²/r

T sin 37° = mv²/r

(mg / cos 37°) sin 37° = mv²/r

g tan 37° = v²/r

v = √(gr tan 37°)

v = √(9.8 m/s² × 60 m × tan 37°)

v = 21 m/s

Answer:

a. 70 rad/s²

b. 5000 rev

Explanation:

As we know,

[tex]\omega = 20000\frac{rev}{min}\frac{2 \pi rad}{1 \ rev}\frac{1 \ min}{60 \ sec}[/tex]

then,

[tex]\omega=2100 \ rad/s[/tex]

a...

⇒  [tex]\bar{\alpha}=\frac{\omega-\omega_{0}}{\Delta t}[/tex]

On substituting the values, we get

⇒      [tex]=\frac{2100}{30}[/tex]

⇒      [tex]=70 \ rad/s^2[/tex]

b...

⇒  [tex]\theta=\theta_{0}=\omega_{0}t+\frac{1}{2}\alpha t^2[/tex]

       [tex]=\frac{1}{2}\alpha t^2[/tex]

       [tex]=\frac{1}{2}\times 70\times (30)^2[/tex]

       [tex]=31500 \ rad[/tex]

       [tex]=31500 \ rad\frac{1 \ rev}{2\pi rad}[/tex]

       [tex]=5000 \ rev[/tex]

(a) The average angular acceleration will be 70 rad/s².

(b) 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.

Angular acceleration is defined as the pace of change of angular velocity with reference to time. It is denoted by α. Its unit is rad/s².

The given data in the problem is;

n is the revolution of centrifugal rotor =  20000 rpm

t is the time interval= the 30s

(α) is the Angular acceleration=?

n₁ is the revolution when the acceleration is constant =?

(a) The average angular acceleration will be  70 rad/s².

The value of the angular velocity is given by

[tex]\rm \omega_f = \frac{2\pi N}{60} } \\\\ \rm \omega_f = \frac{2 \times 3.14 \times 20000}{60} \\\\ \rm \omega_f= 2100\ rad/sec.[/tex]

The formula for angular acceleration is guven by;

[tex]\rm \alpha =\frac{ \omega_f-\omega_i}{dt} \\\\ \rm \alpha =\frac{ 2100-0}{3}\\\\ \rm \alpha =70\ rad/sec^2[/tex]

Hence the average angular acceleration will be 70 rad/s².

(b 5000 revolutions have the centrifuge rotor turned during its acceleration period.

[tex]\rm \theta= \theta_0+\frac{1}{2} \alpha t^2 \\\\ \rm \theta= \frac{1}{2} \times 70 \times (30)^2 \\\\ \rm \theta=31500\ rad[/tex]

As we know that the angular velocity is given by

[tex]\rm \omega = \frac{\theta}{t} \\\\ \rm \omega = \frac{31500}{30} \\\\ \rm \omega = 1050 \ rad/sec[/tex]

The relation of angular velocity and revolution will be

[tex]\rm n= \frac{ \omega \times 60}{2\pi} \\\\ \rm n= \frac{ 2100 \times 60}{2\times 3.14 } \\\\ \rm n = 20063.69 \ rev[/tex]

Hence 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.

To learn more about angular acceleration refer to the link ;

https://brainly.com/question/408236

Answer:

Time taken to stop = 6 sec

Explanation:

Given:

Rotation of wheel = 1.10 × 10² rev/min

Final velocity (v) = 0

Angular acceleration (a) = - 1.92 rad/s²

Find:

Time taken to stop

Computation:

Initial velocity (u) = (1.10 × 10² × 2π rad) / 60 sec

Initial velocity (u) = 11.52 rad / sec

We know that,

V = U +at

t = (v-u)a

t = (0 - 11.52) / (-1.92)

Time taken to stop = 6 sec

Answer:

[tex]v_i = 1.44\frac{m}{s}[/tex]

Explanation:

The computation of the initial velocity of the fly is shown below:-

But before that we need to do the following calculations

For 4.22 seconds

[tex]\bar v = \frac{-5.80 m}{4.22 s}[/tex]

[tex]= -1.37\frac{m}{s}[/tex]

For uniform acceleration

[tex]\bar v = \frac{v_i +v_f}{2}[/tex]

[tex]= v_i + v_f[/tex]

[tex]= -2.74\frac{m}{s}[/tex]

With initial and final velocities

[tex]= -1.33\frac{m}{s^2}[/tex]

[tex]= \frac{v_i +v_f}{4.22s}[/tex]

[tex]= -v_i + v_f[/tex]

[tex]= -5.61\frac{m}{s}[/tex]

So, the initial velocity is

[tex]v_i = 1.44\frac{m}{s}[/tex]

We simply applied the above steps to reach at the final solution i.e initial velocity

Answer:

The atomic weight of the metal is 58.7 g/mol

Explanation:

Given;

density of the metal sample, ρ = 8.91 g/cm³

edge length of the face centered cubic cell, α = 352.4 pm = 352.4 x 10⁻¹⁰ cm

Volume of the unit cell of the metal;

V = α³

V = (352.4 x 10⁻¹⁰  cm)³

V = 4.376 x 10⁻²³ cm³

Mass of the metal in unit cell

mass = density x volume

mass = 8.91 g/cm³ x 4.376 x 10⁻²³ cm³

mass = 3.899 x 10⁻²² g

Atomic weight, based on 4 atoms per unit cell;

4 atoms = 3.899 x 10⁻²² g

6.022 x 10²³ atoms = ?

= (6.022 x 10²³atoms x 3.899 x 10⁻²² g) / (4 atoms)

= 58.699 g/mol

= 58.7 g/mol (this metal is Nickel)

Theerefore, the atomic weight of the metal is 58.7 g/mol

Answer:

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Explanation:

Given that

s = 9 cos(t) + 9 sin(t), t ≥ 0

Then acceleration and velocity is

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Answer:

Radius r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Explanation:

Area of a circle;

A = πr^2

A = area

r = radius

Making r the subject of formula;

r = √(A/π) ........1

Given;

A = 1300 cm^2

Substituting into the equation 1;

r = √(1300/π)

r = 20.34214472564 cm

r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Answer:

same direction = 80 (n)

opposite direction = 20 (n) going one direction

Explanation:

same direction means they are added to each other

and opposite means acting on eachother

Answer:

Only the horizontal component of a projectile’s momentum is conserved. Where as The vertical component of the momentum is not conserved, because the net vertical force Fy–net is not zero

Explanation:

Answer:

E = - dV/dx

Explanation:

Las superficies equipòtenciales son superficie donde el potencial eléctrico es constante por lo cual nos podemos desplazaren ella sin realizar nigun trabajo.

El campo electrico es el campo que existen algún punto en el espacio creado por alguna ddistribucion de carga.

De los antes expuesto las dos magnitudes están relacionadas

         E = - dV/dx

por lo cual el potenical es el gradiente del potencial eléctrico.

Como el campo eléctrico sobre un superficie equipotenciales constante, podemos colocar una punta de prueba con un potencial dado y seguir la linea que de una diferencia de potencial constar, lo cual permite visualizar las forma de cada linea equipotencial