The specific heat of a certain type of cooking oil is 1.75J/g°C. How much heat energy is needed to raise the temperature of 2.63kg of this oil from g

Answers

Answer 1

Complete question:

The specific heat of a certain type of cooking oil is 1.75J/g°C. How much heat energy is needed to raise the temperature of 2.63 kg of this oil from 23 °C to 191 °C?

Answer:

The heat energy required to raise the temperature of this cooking oil is 773220 J

Explanation:

Given;

mass of the cooking oil, m = 2.63 kg = 2630 g

specific heat capacity of the cooking oil, C = 1.75J/g°C

initial temperature of the cooking oil, T₁ = 23° C

final temperature of the cooking oil, T₂ = 191° C

The heat energy required to raise the temperature of this cooking oil;

Q = mcΔT

Q = 2630 x 1.75 x (191 - 23)

Q = 773220 J

Therefore, the heat energy required to raise the temperature of this cooking oil is 773220 J


Related Questions

Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is four times greater than that of wire B. What is the ratio of the radius of A to that of B

Answers

Answer:

r₁/r₂ = 1/2 = 0.5

Explanation:

The resistance of a wire is given by the following formula:

R = ρL/A

where,

R = Resistance of wire

ρ = resistivity of the material of wire

L = Length of wire

A = Cross-sectional area of wire = πr²

r = radius of wire

Therefore,

R = ρL/πr²

FOR WIRE A:

R₁ = ρ₁L₁/πr₁²   -------- equation 1

FOR WIRE B:

R₂ = ρ₂L₂/πr₂²   -------- equation 2

It is given that resistance of wire A is four times greater than the resistance of wire B.

R₁ = 4 R₂

using values from equation 1 and equation 2:

ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²

since, the material and length of both wires are same.

ρ₁ = ρ₂ = ρ

L₁ = L₂ = L

Therefore,

ρL/πr₁² = 4ρL/πr₂²

1/r₁² = 4/r₂²

r₁²/r₂² = 1/4

taking square root on both sides:

r₁/r₂ = 1/2 = 0.5

A woman is listening to her radio, which is 174 m from the radio station transmitter. (a) How many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1540 kHz

Answers

Explanation:

It is given that,

The distance between the radio and the radio station is 174 m

We need to find how many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1540 kHz.

f = 1540 kHz

Wavelength,

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{1540\times 10^3}\\\\\lambda=194.8\ m[/tex]

Let there are n wavelengths of the radio waves. So,

[tex]n=\dfrac{d}{\lambda}\\\\n=\dfrac{174}{194.8}\\\\n=0.89\ \text{wavelengths}[/tex]

There are 0.89 wavelengths.

the capacitor is initially unchanged. immediatley after the switches closes, the capacitor voltage is

Answers

Answer:

Immediately after the switches closes, the capacitor voltage is zero.

Explanation:

Charge on capacitor is given as;

[tex]q_o = CV_{battery}[/tex]

after the switches closes, the charge on the capacitor is the same as before, therefore, the voltage drop in the capacitor is zero.

Apply Kirchoff's voltage rule for short term;

[tex]q = q_o[/tex] = 0 (since it is uncharged)

[tex]V_{battery} + IR =0[/tex]

Where;

q₀ is the charge on the capacitor before

q is the charge on the capacitor after

Therefore, immediately after the switches closes, the capacitor voltage is zero.

A grandfather clock (with a pendulum) keeps perfect time on Earth. If you were to transport this clock to the Moon, would its period of oscillation increase, decrease, or stay the same? Would its frequency of oscillation increase, decrease, or stay the same? Explain.

Answers

Answer:

Frequency of oscillation will increase

Explanation:

Because The period of a pendulum is inversely proportional to the square root of the gravitational acceleration, so as. Gravity increases on the moon period of oscillation decreases whereas period of oscillating pendulum is inversely proportional to frequency thus frequency is directly proportional to gravity so as gravity increases on the moon the frequency will increase

Answer:

a) Therefore the period of oscillation in the moon will increase.

[tex]T_{E}<T_{M}[/tex]

b) The frequency in the moon will decrease

[tex]\omega_{E}>\omega_{M}[/tex]

Explanation:

The period of the pendulum is given by this equation:

[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex] (1)

As we can see, T is proportional to L and inversely proportional to the gravity vector. We know that g in the earth is grader than g in the moon.

[tex]g_{E}>g_{M}[/tex] (2)

Therefore the period of oscillation in the moon will increase.

[tex]T_{E}<T_{M}[/tex] (3)              

Now, the frequency of oscillation is:

[tex]\omega=\sqrt{\frac{g}{L}}[/tex]

In this case, ω is proportional to g, then using (1) we can conclude that:

[tex]\omega_{E}>\omega_{M}[/tex]

Therefore the frequency in the moon will decrease.

I hope it helps you!

           

A small truck has a mass of 2065 kg. How much work is required to decrease the speed of the vehicle from 23.0 m/s to 12.0 m/s on a level road

Answers

Answer:

397.51 kJ

Explanation:

Since the change in velocity is done on a level road, there is no change in the potential energy. The workdone is the workdone on reducing the kinetic energy.

Workdone W = change in kinetic energy ∆K.E

W = ∆K.E

K.E = 0.5mv^2

∆K.E = 0.5m(m1^2 - m2^2)

Given;

Mass m = 2065 kg

Initial velocity v1 = 23.0 m/s

Final velocity v2 = 12.0 m/s

W = ∆K.E = 0.5m(m1^2 - m2^2)

Substituting the given values;

W = ∆K.E = 0.5×2065(23^2 - 12^2)

W = 397512.5J

W = 397.51 kJ

A body of mass 10 kg moving on a circular path of radius 1 m completes a rotation in 2 pi sec the centripetal acceleration is A 1m/s 2 B 2m/s 2 C 4m/s 2 D none

Answers

Answer:

1 m/s²

Explanation:

a = v² / r

a = ω² r

a = (2π rad / 2π s)² (1 m)

a = 1 m/s²

Using the image below, identify the numbered parts of the wave

Answers

Answer:

1) Wavelength

2) Peak of Wave

3) Trough of Wave

4) Amplitude of Wave

(1)  Wavelength

(2) Peak of Wave

(3) Trough of Wave

(4) Amplitude of Wave

What are the terms shown in the image of wave?

Wavelength -The distance between two successive crests or troughs of the wave is called as wavelength

Amplitude- Maximum distance between highest point to the axix of the wave is amplitude.

Crest - Positive amplitude of the wave is crest

Trough- Negative amplitude of the wave is called trough.

Hence  

(1)  Wavelength

(2) Peak of Wave

(3) Trough of Wave

(4) Amplitude of Wave

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An evacuated tube uses an accelerating voltage of 50 kV to accelerate electrons to hit a copper plate and produce X-rays. Non-relativistically, what would be the maximum speed (in m/s) of these electrons

Answers

Answer:

Explanation:

Accelerating voltage = 50 x 10³ V

energy created = q V where q is charge on electron and V is accelerating potential

So qV = 1/2 m v² , m is mass of electron and v is velocity

Putting values

1.6 x 10⁻¹⁹ x 50 x 10³ = 1/2 x 9.1 x 10⁻³¹ x v²

v² = 175.8 x 10¹⁴

v = 13.25 x 10⁷ m /s

A brass rod with a mass of 0.300 kg slides on parallel horizontal iron rails, 0.440 m apart, and carries a current of 15.0 A. The coefficient of friction between the rod and rails is 0.300. What vertical, uniform magnetic field is needed to keep the rod moving at a constant speed

Answers

Answer:

The magnitude of the magnetic field is  [tex]B = 0.0890 \ T[/tex]

Explanation:

From the question we are told that

   The mass of the rod is  [tex]m =0.300 \ kg[/tex]

    The distance of separation is  [tex]d = 0.440 \ m[/tex]

     The current is  [tex]I = 15.0 \ A[/tex]

     The coefficient of friction is   [tex]\mu = 0.300[/tex]

     

Generally for the rod the rod to continue moving at a constant speed

   The frictional force must equal to the magnetic field force so

    [tex]F_m = F_f[/tex]

Where  [tex]F_m = B* I * d[/tex]

and     [tex]F_f = \mu * m * g[/tex]

   [tex]B*I *d = \mu * m * g[/tex]

=>    [tex]B = \frac{\mu * m * g }{I * d }[/tex]

substituting values

       [tex]B = \frac{0.2 * 0.300 * 9.8 }{ 15 * 0.440 }[/tex]

       [tex]B = 0.0890 \ T[/tex]

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s. The dipole is initially oriented so that the charge Q is located in the plane that bisects the dipole. Assume that r>>s
Immediately after the dipole is released:
a. What is the magnitude of the force on the dipole?
b. What is the magnitude of the torque on the dipole?

Answers

Answer:

a) the magnitude of the force is

F= Q([tex]\frac{kqs}{r^3}[/tex]) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b)  the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

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

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = [tex]\frac{kp}{r^{3} }[/tex] , where p = q × s

E(r) = [tex]\frac{kqs}{r^{3} }[/tex] where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q([tex]\frac{kqs}{r^3}[/tex])

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall  F = Qqs/4πε₀r³

τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

Part A: The expression of the force on the dipole is [tex]F = \dfrac {Qqs}{4 \pi \epsilon_0 r^3}[/tex].

Part B: The expression of the torque on the dipole is [tex]\tau = \dfrac {Qqs}{4\pi\epsilon_0 r^2}[/tex].

How do you calculate the force and torque on the dipole?

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s. Also, r>>s.

Part A

The electric field due to dipole at a distance r is given below.

[tex]E_r = \dfrac {qs}{4\pi \epsilon_0 r^3}[/tex]

The magnitude of the force can be given as below.

[tex]F = QE_r[/tex]

[tex]F = \dfrac {Qqs}{4 \pi \epsilon_0 r^3}[/tex]

Hence the expression of the force on the dipole is [tex]F = \dfrac {Qqs}{4 \pi \epsilon_0 r^3}[/tex].

Part B

The torque on the dipole will dependent on the perpendicular forces on the dipole. The expression of the torque is given below.

[tex]\tau = F \times r \times sin \theta[/tex]

For the perpendicular forces, θ = 90°. Hence the torque is given below.

[tex]\tau = F\times r[/tex]

[tex]\tau = \dfrac {Qqs}{4 \pi \epsilon_0 r^3} \times r[/tex]

[tex]\tau = \dfrac {Qqs}{4\pi\epsilon_0 r^2}[/tex]

Hence the expression of the torque on the dipole is [tex]\tau = \dfrac {Qqs}{4\pi\epsilon_0 r^2}[/tex].

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4. The capacitance of a capacitor is increased by a factor of 1.5 when it is completely filled
with a certain dielectric material. Find the dielectric constant of the material and its
electric susceptibility​

Answers

Answer:

a. Dielectric constant, ε = 1.5 b. Electric susceptibility, χ = 0.5

Explanation:

a. Dielectric constant

Since the capacitance of the capacitor is increased by a factor of 1.5. Let its initial capacitance be C and its final capacitance after adding the material be C'.

Since C' = εC where ε = relative permittivity,

Also, C' = 1.5C

Comparing both equations for C', ε = 1.5.

Since ε = relative permittivity = dielectric constant,

dielectric constant = 1.5

So, the dielectric constant = 1.5

b. Electric susceptibility

The electric susceptibility χ is given by

χ = ε - 1 where ε = dielectric constant

Since ε = 1.5,

χ = ε - 1

χ = 1.5 - 1

χ = 0.5

So the electric susceptibility χ = 0.5

An eagle is flying horizontally at a speed of 4.6 m/s when the fish in her talons wiggles loose and falls into the lake 4.2 m below. Calculate the vertical component of the velocity of the fish relative to the water when it hits the water.

Answers

Answer:

v = 9.07 m/s

the vertical component of the velocity of the fish relative to the water when it hits the water is 9.07 m/s

Explanation:

Given;

An eagle is flying horizontally at a speed of 4.6 m/s

Initial horizontal velocity uh = 4.6 m/s

Initial vertical velocity uy = 0

Height to fall d = 4.2 m

Acceleration due to gravity g = 9.8 m/s^2

The final vertical velocity of the fish when it hits the water can be calculated using the equation of motion;

v^2 = u^2 + 2as

v^2 = uy^2 + 2gd

uy = 0

v^2 = 2gd

v = √(2gd)

Substituting the given values;

v = √(2×9.8×4.2)

v = 9.073036977771 m/s

v = 9.07 m/s

the vertical component of the velocity of the fish relative to the water when it hits the water is 9.07 m/s

A spring with k = 35.5 N/m has a mass of 5.50 kg attached to it. An external force F = (4.40 N)sin[(6.80 s−1)t] drives the spring mass system so that it oscillates without any resistive forces. What is the amplitude of the oscillatory motion of the spring-mass system?

Answers

Answer:

A = 0.02 m

Explanation:

The spring constant, k = 35.5 N/m

The attached mass, m = 5.50 kg

The expression for the external force, F =  (4.40 N)sin[(6.80 s⁻¹)t].....(1)

The general expression for the external force, F = F₀ sin (wt).............(2)

Comparing equations (1) and (2):

The forced frequency, [tex]\omega = 6.80 rad/s[/tex]

F₀ = 4.40 N

The natural frequency can be calculated using the formula:

[tex]\omega_0 = \sqrt{\frac{k}{m} } \\\\\omega_0 = \sqrt{\frac{35.5}{5.5} } \\\\\omega_0 = 2.54 rad/s[/tex]

The amplitude of oscillation of a spring-mass system in the steady state:

[tex]A = \frac{F_0}{m(\omega^2 - \omega_o^2)} \\\\A = \frac{4.4}{5.5(6.8^2 - 2.54^2)} \\\\A = 0.02 m[/tex]

John coasts on his skateboard at a speed of 1 m/s. He uses his foot to
increase his speed to 3 m/s. If he has a mass of 95 kg, what impulse does he
apply?
A. 285 kg•m/s
B. 95 kg•m/s
c. 190 kg•m/s
D. 145 kg•m/s

Answers

Answer:

C

Explanation:

The value of the impulse of his skateboard will be 190 kg-m/s. Then the correct option is C.

What is an impulse?

The change in momentum is equal to the product of impact force applied while colliding and the time for that impact.

The impulse is given as the product of the mass and the difference between the final velocity and initial velocity.

Impulse  = Mass x (Final velocity - initial velocity)

John coasts on his skateboard at a speed of 1 m/s.

He uses his foot to increase his speed to 3 m/s. If he has a mass of 95 kg.

Then the value of the impulse of his skateboard will be given as,

Impulse = 95 x (3 - 1)

Impulse = 95 x 2

Impulse = 190 kg-m/s

Thus, the value of the impulse of his skateboard will be 190 kg-m/s.

Then the correct option is C.

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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge at point P, which is on the x-axis, 9.83 mm from the origin

Answers

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

[tex]\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}][/tex]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

[tex]\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°[/tex]

The distance r is:

[tex]r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m[/tex]

You replace the values of all parameters in the equation (1):

[tex]\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}[/tex]

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

"Index of refraction is defined as the speed of light in a vacuum divided by the speed of light in the medium. What is the index of refraction for a glass sample

Answers

Complete Question

Index of refraction is defined as the speed of light in a vacuum divided by the speed of light in the medium. What is the index of refraction for a glass sample? Speed of light in a vacuum c = 3.0 x 108 m/s. Speed of light in the glass = 1.50 x 108 m/s.

Answer:

The  index of refraction is  [tex]n_g = 2[/tex]

Explanation:

From the question we are told that

    The  speed of light in a vacuum is  [tex]c = 3.0 *10^{8} \ m/s[/tex]

      The speed of the in the glass is  [tex]c_g = 1.50 *10^{8} \ m/s[/tex]

From the question the index of refraction for a glass is mathematically evaluated as

        [tex]n_g = \frac{c}{c_g }[/tex]

       [tex]n_g = \frac{3.0 *10^{8}}{1.50 *10^{8} }[/tex]

       [tex]n_g = 2[/tex]

     

If the diameter of a radar dish is doubled, what happens to its resolving power, assuming that all other factors remain unchanged? Its resolving power

a. is reduced to one-half its original value.
b. Quadruples.
c. is reduced to one-quartet its original value.
d. Halves.
e. Doubles.

Answers

Answer:

e. Doubles.

Explanation:

Resolving power is given by the formula as follows :

[tex]\dfrac{1}{d\theta}=\dfrac{D}{1.22\lambda}[/tex]

Here, [tex]d\theta[/tex] is the angle subtended by two distant objects

D is diameter of the telescope

Here, the diameter of a radar dish is doubled, assuming all other factors remain unchanged, then the resolving power gets doubled. Hence, the correct option is (e).

The amount of heat necessary to change 1 kg of a solid into a liquid at the same temperature is called the Group of answer choices

Answers

Answer:

Latent heat of fusion

Explanation:

The amount of heat required to change the state of a unit mass (say 1kg) of a substance is called latent heat. This change typically occurs without a significant change in temperature. The latent heat occurs as a result of the work required to break the forces that hold atoms or molecules in a given matter (solid, liquid or gas). Now, since the transformation is from solid into liquid, the amount of heat required to do this is called the latent heat of fusion. This transformation takes place at the melting point of the solid.

An object of mass 4kg is moving along a horizontal plane. If the coefficient of kinetic friction is 0.2 find the friction force acting on the object.

Answers

Answer:

The friction force acting on the object is 7.84 N

Explanation:

Given;

mass of object, m = 4 kg

coefficient of kinetic friction, μk = 0.2

The friction force acting on the object is calculated as;

F = μkN

F = μkmg

where;

F is the frictional force

m is the mass of the object

g is the acceleration due to gravity

F = 0.2 x 4 x 9.8

F = 7.84 N

Therefore, the friction force acting on the object is 7.84 N

Each charge is equidistant from the origin. In which direction is the net electric field at the point P on the y-axis?

Answers

Answer:

"Upwards and towards the left" is the right answer.

Explanation:

The magnitude of the field will be:

⇒  [tex]E=\frac{kq}{r^2}[/tex]

And direction -> for negative charges, to positive charges, except charges.  

Charging across the y-axis. It would be up to the aggregate field.  Because the x-axis needs to charge. Total production is to the west.  

Thus the net field is upwards as well as to the left.

A wheel 2.40 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.40 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following.
A. What is the tangential speed?
B. Total acceleration
C. Angular position of point P.

Answers

Answer:

Explanation:

Radius of wheel = 1.2 m

A )

To know angular speed after t sec , we use the formula

ω = ω₀ + α t  , where ω₀ is initial velocity , α is angular acceleration

ω = 0 + 4.4 x 2

= 8.8 rad / s

v= ωR , v is tangential speed , ω is angular speed , R is radius of wheel .

= 8.8 x 1.2 = 10.56 m /s

B )

radial acceleration

Ar = v² / R

= 10.56² / 1.2

= 92.93 m /s²

Tangential acceleration

At = angular acceleration x radius

= 4.4 x 1.2 = 5.28 m /s²

Total acceleration

=  √ ( At² + Ar² )

=√ (5.28² +92.93²)

= 93 m /s²

C )

θ = ωt + 1/2 α t²     where θ is angular position after time t .

= 0 + .5 x 4.4 x 2²

= 8.8 rad

= 180x 8.8/ 3.14  = 504.45 degree

initial position = 57.3°

final position = 504 .45 + 57.3

= 561.75 °

= 561.75 - 360

= 201.75 ° .

Position of radius vector of point P will be at angle of 201.75 from horizontal axis .

Total energy of a particle executing S.H.M of amplitude A is proportional to:
(a)A?
(b) A-2
(c) A
(d) A-​

Answers

Answer:

If x = A sin w t    where w is the angular frequency

then v = w A cos w t

Since KE = 1/2 m Vmax^2 and Vmax = w A     maximum KE

the total energy is proportional to A^2

Also, since the maximum potential energy is

PEmax = 1/2 K A^2    where the KE is zero (maximum amplitude)

one can again see that the total energy is proportional to A^2

The net torque acting on an object is zero. What can you say about the angular momentum (about the same axis) of the object?

Answers

Answer:

The angular momentum (about the same axis) of the object is constant

Explanation:

Torque is the time time derivative of angular momentum.

τ = dL / dt

where;

dL  is the change in  angular momentum

τ is torque acting on the object

dt is change in time

when net torque acting on an object is zero, then

dL / dt = 0

Change in angular momentum (ΔL) is zero, therefore we can say that the angular momentum (L) is constant.

Thus, the angular momentum (about the same axis) of the object is constant.

To water the yard, you use a hose with a diameter of 3.0 cm. Water flows from the hose with a speed of 2.2 m/s. If you partially block the end of the hose so the effective diameter is now 0.50 cm, with what speed does water spray from the hose

Answers

Answer:

v₂ = 79.69 m/s

Explanation:

The initial diameter of the hose, d₁ = 3.0 cm = 0.03 m

Initial Cross Sectional Area, A₁ = πd₁²/4

A₁ = (π* 0.03²)/4

A₁ = 0.00071 m²

The initial speed of water from the hose, v₁ = 2.2 m/s

The diameter of the hose after blocking the end, d₂ = 0.50 cm = 0.005 m

Cross Sectional Area of the hose after blocking the end, A₂ = πd₂²/4

A₂ = (π* 0.005²)/4

A₂ = 0.0000196 m²

To get the speed, v₂, at which the water spray from the hose after blocking the end, we will use the continuity equation:

A₁v₁ = A₂v₂

0.00071 * 2.2 = 0.0000196 v₂

0.001562 = 0.0000196 v₂

v₂ = 0.001562/0.0000196

v₂ = 79.69 m/s

Wind gusts create ripples on the ocean that have a wavelength of 3.03 cm and propagate at 3.37 m/s. What is their frequency (in Hz)?

Answers

Answer:

Their frequency is 111.22 Hz

Explanation:

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration and is expressed in units of length (m).

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

The propagation speed of a wave is the quantity that measures the speed at which the wave's disturbance propagates throughout its displacement. The speed at which the wave propagates depends on both the type of wave and the medium through which it propagates. Relate wavelength (λ) and frequency (f) inversely proportional using the following equation:

v = f * λ.

Then the frequency can be calculated as: f=v÷λ

In this case:

λ=3.03 cm=0.0303 m (1m=100 cm)v= 3.37 m/s

Replacing:

[tex]f=\frac{3.37 \frac{m}{s} }{0.0303 m}[/tex]

Solving:

f=111.22 Hz

Their frequency is 111.22 Hz

a bird experiences an acceleration of -1.80m/s^2 for 3.81 s and ends up 26.5 m to the left of its starting point. what is its final velocity?

Answers

For a questions which involves constant acceleration, you can use the SUVAT equations.

S - displacement - -26.5m ( If you to take the right as a positive value - since displacement is a vector quantity involving direction and magnitude)

U

V - Final Velocity - unknown

A - Acceleration - -1.8m/s^2

T - Time - 3.81s

s = vt - 1/2at^2

-26.5m = 3.81v - 1/2 • -1.8 • 3.81^2

Rearrange

-26.5m/-1/2•-1.8•3.81^2=3.81v

3.81v = -2.028~

v = -0.532~ (3sf)

79. An example of an electrical insulator is
a. graphite.
b. glass.
c. aluminum
d. tungsten

Answers

Answer:

B. Glass

Explanation:

An electrical insulator is a substance that does not conduct electricity.

Glass has tightly bounded electrons, that is why it is an insulator of electricity.

Hey there! The answer to your question is below.

The correct answer would be B.GLASS

Glass is a insulator and rubber too

Glass, wood, plastic and more are all insulators

So the answer would be B. Glass

Hope this helps!

By: xBrainly

A small object with mass 3.80 kg moves counterclockwise with constant speed 1.65 rad/s in a circle of radius 2.70 m centered at the origin. It starts at the point with position vector 2.70 m. Then it undergoes an angular displacement of 8.70 rad.
(a) What is its new position vector?
in meters
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis?
(c) What is its velocity?
in m/s
(d) In what direction is it moving?
_____° from the +x direction.
(e) What is its acceleration?
in m/s2
(f) What total force is exerted on the object?
in N

Answers

Answer:

Explanation:

angular velocity

ω = 1.65 rad /s

radius R = 2.70 m

angular displacement = 8.70 rad

a )

New position vector in vector form

= R cos8.7 i + R sin8.7 j

= 2.7 cos8.7 i + 2.7 sin8.7 j

= 2.7 x .748 i + 2.7 x .663 j

= 2.01 i + 1.79 j

b )

8.7 radian = 180/π x 8.7 degree

= 498.72 degree

= 498.72 - 360

= 138.72 degree

It will be in second quadrant .

angle made with positive x - axis

= 138.72 degree .

c )

velocity

v = ω R

= 1.65 x 2.7

= 4.455 m /s

d )

It is moving in a direction making 138.72° with positive x direction .

e )

acceleration will be centripetal acceleration

= v²/ R  

= 4.455² / 2.7

= 7.35 m /s²

f ) force = mass x acceleration

= 3.8 x 7.35

= 27.93 N .

A chain lying on the ground is 10 m long and its mass is 70 kg. How much work (in J) is required to raise one end of the chain to a height of 6 m

Answers

Answer:

The work done in moving the chain is 4116 J

Explanation:

Given;

mass of the chain, m = 70 kg

length of the chain, l = 10 m

vertical height through which one end of the chain was raised, h = 6 m

Work done in raising this chain to this height is equal to potential energy due to this vertical height

W = PE

W = mgh

where;

m is mass of the chain

g is acceleration due to gravity

h is the vertical height through which the chain was raised

W = 70 x 9.8 x 6

W = 4116 J

Therefore, the work done in moving the chain is 4116 J

Which one of the following is the shortest length?
A)
100 meters
C)
104 millimeters
E)
10 nanometers
B)
10² centimeters
D)
105 micrometers

Answers

Convert all lengths to metres

A) 100 meters B) 1 meters C) 0.104 meters D) 0.000105 meters E) 0.00000001 meters

Therefore D is the answer

Answer:

Option E (10 nanometers) is the shortest length

Explanation:

From,

1cm = [tex]10^{-2}m[/tex]

1mm = [tex]10^{-3}m[/tex]

1nanometer = [tex]10^{-9[/tex]

1micrometer = [tex]10^{-6[/tex]

Therefore,

A) [tex]10^0[/tex] meters = 1meter

B) [tex]10^2[/tex] cm = [tex]10^2 * 10^{-2} = 1meter[/tex]

C) [tex]10^4[/tex] mm = [tex]10^4 * 10^{-3} = 10meter[/tex]

D) [tex]10^5[/tex] micrometer = [tex]10^5 * 10^{-6} = 0.1meter[/tex]

E) [tex]10[/tex] nanometer = [tex]10 * 10^-9 = 1*10^{-8}[/tex]

Therefore 10nanometers is the shortest length

For more information on length conversions, visit

https://brainly.com/subject/physics

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