An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has specific heat capacity 950 J/kg K melting point 680*C, latent heat of fusion 450 kJ/kg, the efficiency of furnace is 85% and cost of energy is 20 cents/kWhr Calculate: (a) Power required (b) The cost of operating the furnace for 30 hours

Answer : Units of momentum are :

1. Kg m/s

2. N-s

Explanation:

The momentum of an object is given by the product of its mass and velocity with which it is moving. Mathematically, it is given by :

P = mv

Where

m is in kilogram

v is in m/s

Option (1) : N-m = It is not a unit of momentum. It includes the product of force and distance.

Option (2) : Kg s/m = It is again not a unit of momentum.

Option (3) : Kg m/s =

Since, p = mv

p = Kg × m/s

It can be the unit of momentum.

(4) Option (4) : N-s = The change in momentum is equal to the impulse applied on an object. It is given by the product of force and short duration of time. It can be the unit of momentum.

(5) Option (5) : Kg/m²/s² = It is not the unit of momentum.

Hence, the correct options are (c) and (d).

The quantities that are units of momentum among the options provided are C) kg m/s and D) N-s. The other options correspond to different physical quantities.

The concept in question pertains to the momentum of an object, which, in physics, is a vector quantity defined as the product of an object's mass and its velocity. The standard international (SI) unit of momentum is kilogram meter per second (kg m/s).

Examining each giver option: A) Newton meter (N m) is a unit of work, not momentum. B) Kilogram second/meter (kg s/m) does not align with the definition of momentum. C) Kilogram meter/second (kg m/s), this is the correct SI unit for momentum. D) Newton-second (N-s) is also a correct unit for momentum as Newton is equivalent to kg m/s2. E) Kilogram meter2/second2 (kg m2/s2) is the unit for kinetic energy, not momentum.

So, C) kg m/s and D) N-s are the units of momentum among the given choices.

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Answer:

Amount of radioactive particles left after 60 years = 100 particles.

Explanation:

Amount of radioactive particles before 60 years = 400

Amount of radioactive particles present today = 200

That is radio active particles reduced to half. That is 60 years is half life of this radio active material.

After 60 years this 200 radio active particles will reduce to half.

Amount of radioactive particles left after 60 years = 0.5 x 200 = 100 particles.

Final answer:

The sample will have 100 radioactive particles remaining 60 years from today, based on the half-life of the material being 60 years.

Explanation:

The question concerns the concept of radioactive decay and specifically the half-life of a radioactive sample. In this case, the sample's quantity of radioactive particles was observed to decrease from 400 to 200 over a span of 60 years. Thus, the half-life of the material is 60 years, which is the time it takes for half of the radioactive atoms (parent nuclei) to decay into their decay products (daughter elements).

Given that the sample has 200 particles today, we can predict that in another 60 years, the number of radioactive particles will again be halved. Therefore, after 60 years from today, we expect there to be 100 radioactive particles remaining in the sample.

Answer:

The magnitude and direction of the acceleration of the particle is [tex]a= 0.3296\ \hat{k}\ m/s^2[/tex]

Explanation:

Given that,

Mass [tex]m = 1.81\times10^{-3}\ kg[/tex]

Velocity [tex]v = (3.00\times10^{4}\ m/s)j[/tex]

Charge [tex]q = 1.22\times10^{-8}\ C[/tex]

Magnetic field [tex] B= (1.63\hat{i}+0.980\hat{j})\ T[/tex]

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

[tex]F = ma=q(v\times B)[/tex]

[tex]a =\dfrac{q(v\times B)}{m}[/tex]

We need to calculate the value of [tex]v\times B[/tex]

[tex]v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})[/tex]

[tex]v\times B=4.89\times10^{4}[/tex]

Now, put the all values into the acceleration 's formula

[tex]a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}[/tex]

[tex]a= -0.3296\ \hat{k}\ m/s^2[/tex]

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is [tex]a= 0.3296\ \hat{k}\ m/s^2[/tex]

The magnitude and direction of the particle’s acceleration produced by a uniform magnetic field  [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^ is [tex]\bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]

A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^?

A charged particle is a particle with an electric charge. Whereas electric charge is the matter physical property that causes to experience a force when placed in an electromagnetic field. Uniform magnetic field is the condition when magnetic field lines are parallel then magnetic force experienced by an object is same at all points in that field

From Newton's second law, the force is given by:

[tex]F=ma[/tex]

Magnetic force is

[tex]F= qv \times B[/tex]

[tex]ma = qv \times B[/tex]

[tex]a = \frac{qv \times B}{m}[/tex]

Subsituting with the givens above we get

[tex]a = \frac{(1.22 \times 10^{-8} C) (3 \times 10^{4} m/s) (1.63 T ) (\hat{j} \times \hat{i})}{1.81 \times 10^{-3} kg} = -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]

Therefore the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field  [tex]B =(1.63T)i[/tex]^[tex]+(0.980T)j[/tex]^ is [tex]\bold{{a}}= -(0.330 m/s^2) \bold{\hat{{k}}}[/tex]

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Answer:

Height reached will be 28.35 m

Explanation:

Here we can use the work energy theorem to find the maximum height

As we know by work energy theorem

Work done by gravity + work done by friction = change in kinetic energy

[tex]-mgh - F_f h = 0 - \frac{1}{2}mv_i^2[/tex]

now we will have

[tex]-1.60(9.8)(h) - 0.900(h) = - 470[/tex]

[tex]-16.58 h = -470[/tex]

[tex]h = 28.35 m[/tex]

so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction

Answer:

The speed of light in the medium is [tex]\dfrac{2c}{3}[/tex]

(c) correct option.

Explanation:

Given that,

Speed of light in vacuum = c

Refraction index = 1.5

We need to calculate the value of speed of light in the medium

The refractive index is equal to the speed of light in vacuum divide by the speed of light in medium.

Using formula of refractive index

[tex]\mu = \dfrac{c}{v}[/tex]

[tex]v=\dfrac{c}{\mu}[/tex]

Where, c = speed of light in vacuum

v = speed of light in medium

Put the value into the formula

[tex]v=\dfrac{c}{1.5}[/tex]

[tex]v=\dfrac{2c}{3}[/tex]

Hence, The speed of light in the medium is [tex]\dfrac{2c}{3}[/tex]

The work by tension is ¹¹/₁₀ Mgℓ

[tex]\texttt{ }[/tex]

Complete Question:

An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g/10 . If it has been raised a distance ℓ from rest, how much work has been done by the tension in the string?

[tex]\texttt{ }[/tex]

Given:

Mass of the object = M

Acceleration of the object = g/10

Distance = ℓ

Asked:

Work by Tension = W = ?

Solution:

Let's find the magnitude of tension as follows:

[tex]\Sigma F = ma[/tex]

[tex]T - Mg = Ma[/tex]

[tex]T = Mg + Ma[/tex]

[tex]T = M(g + a)[/tex]

[tex]T = M(g + \frac{1}{10}g)[/tex]

[tex]T = M(\frac{11}{10}g)[/tex]

[tex]T = \frac{11}{10}Mg[/tex]

[tex]\texttt{ }[/tex]

[tex]W = T \times L[/tex]

[tex]W = \frac{11}{10}Mg \times L[/tex]

[tex]W = \frac{11}{10}MgL[/tex]

[tex]\texttt{ }[/tex]

The work by tension is ¹¹/₁₀ Mgℓ

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Answer:

The work done by tension in string is [tex]\dfrac{11}{10}MgL}[/tex].

Explanation:

Given data:

Mass of object is, M.

Acceleration of object is, a = g/10.

Distance covered vertically is, L.

The work done by tension in the string is given as,

[tex]W = T \times L[/tex]  .......................................................... (1)

Here, T is the tension force on string.

Apply the equilibrium of forces on string as,

[tex]T- Mg=Ma[/tex]

Here, g is gravitational acceleration.

[tex]T- Mg=M(\dfrac{g}{10} )\\T=M(\dfrac{g}{10} )+Mg\\T=\dfrac{11}{10}Mg[/tex]

Substituting value in equation (1) as,

[tex]W = \dfrac{11}{10}Mg \times L\\W = \dfrac{11}{10}MgL}[/tex]

Thus, the work done by tension in string is [tex]\dfrac{11}{10}MgL}[/tex].

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Answer:

Acceleration, [tex]a=\dfrac{1}{8}(-i+9j)\ m/s^2[/tex]

Explanation:

Initial velocity of a particle in vector form, u = (-5i - 2j) m/s

Final velocity of particle in vector form, v = (-6i + 7j) m/s

Time taken, t = 8 seconds

We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{(-6i+7j)\ m/s-(-5i-2j)\ m/s}{8\ s}[/tex]    

[tex]a=\dfrac{(-i+9j)}{8\ s}\ m/s^2[/tex]    

or

[tex]a=\dfrac{1}{8}(-i+9j)\ m/s^2[/tex]

Hence, the value of acceleration vector is solved.

Answer: 80.384 cubic cm /min

Explanation:

Let V denote the volume and r denotes the radius of the spherical snowball .

Given : [tex]\dfrac{dr}{dt}=-0.1\text{cm/min}[/tex]

We know that the volume of a sphere is given by :-

[tex]V=\dfrac{4}{3}\pi r^3[/tex]

Differentiating on the both sides w.r.t. t (time) ,w e get

[tex]\dfrac{dV}{dt}=\dfrac{4}{3}\pi(3r^2)\dfrac{dr}{dt}\\\\\Rightarrow\ \dfrac{dV}{dt}=4\pi r^2 (-0.1)=-0.4\pi r^2[/tex]

When r= 8 cm

[tex]\dfrac{dV}{dt}=-0.4(3.14)(8)^2=-80.384[/tex]

Hence, the volume of the snowball decreasing at the rate of 80.384 cubic cm /min.

The volume of the snowball is decreasing at a rate of 8.03 cubic cm per minute when the radius of the snowball is 8 cm.

The rate at which the volume of the spherical snowball is decreasing significantly depends on the rate of decrease in the sphere's radius. The volume formula of a sphere is V = 4/3πR³. With differentiation, volume change in the sphere over time, or dV/dt, can be represented as dV/dt = 4πR² * dR/dt. Plugging in the given values, dR/dt = -0.1 cm/min and R = 8cm, we find that dV/dt = -8.03 cm³/min. This indicates that the volume of the snowball is decreasing at a rate of 8.03 cm³/min. Remember, the answer is given as a positive number, i.e., without the negative sign, which represents a decrease.

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The decibel level at 1.20 m is approximately 93 dB. The sound intensity at 36.0 m is approximately 1.58 × 10^-13 W/m². The decibel level at 36.0 m is approximately 83 dB.

(a) To find the decibel level, we can use the formula B(dB) = 10 log10(I/Io), where I is the sound intensity and Io is the reference intensity. Plugging in the given values, we get B = 10 log10(5.50 ✕ 10^-3/10^-12). After evaluating this expression, we find that the decibel level is approximately 93 dB.

(b) To find the sound intensity at a distance of 36.0 m, we can use the inverse square law for sound, which states that the intensity is inversely proportional to the square of the distance. Using the formula I2 = I1(d1^2/d2^2), where I1 is the initial intensity, I2 is the final intensity, d1 is the initial distance, and d2 is the final distance, we find that the sound intensity at 36.0 m is approximately 1.58 × 10^-13 W/m².

(c) To find the decibel level at a distance of 36.0 m, we can use the same formula as in part (a), but with the intensity at 36.0 m as the new value for I. Plugging in the values, we get B = 10 log10(1.58 × 10^-13/10^-12). After evaluating this expression, we find that the decibel level at a distance of 36.0 m is approximately 83 dB.

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We calculated the decibel level at [tex]1.20 m[/tex] to be approximately [tex]97.4 dB[/tex] . The sound intensity at [tex]36.0 m[/tex]   is about [tex]6.11 \times 10^{-6} W/m^2[/tex] , and its corresponding decibel level is approximately [tex]67.9 dB[/tex].

(a) Finding the decibel level

The decibel (dB) level can be found using the formula:

[tex]\[\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)\][/tex]

where [tex]I[/tex]  is the sound intensity, and I0 is the reference intensity ([tex]10^{-12} W/m^2[/tex]). Given, I[tex]= 5.50 \times 10^{-3} W/m^2[/tex],

[tex]\[\beta = 10 \log_{10} \left( \frac{5.50 \times 10^{-3}}{10^{-12}} \right)\]\beta = 10 \log_{10} \left( 5.50 \times 10^{9} \right)\]\beta \approx 10 \times 9.74 = 97.4 \, \text{dB}\][/tex]

So, the corresponding decibel level is approximately [tex]97.4 dB[/tex].

(b) Finding the sound intensity at a distance of [tex]36.0 m[/tex]

Sound intensity decreases with the square of the distance from the source. We can use the inverse square law:

[tex]\[I_2 = I_1 \left( \frac{r_1^2}{r_2^2} \right)\][/tex]

Given, [tex]\[I_1 = 5.50 \times 10^{-3} \, \text{W/m}^2, \quad r_1 = 1.20 \, \text{m}, \quad r_2 = 36.0 \, \text{m}\][/tex],

[tex]\[I_2 = 5.50 \times 10^{-3} \left( \frac{1.20^2}{36.0^2} \right)\]I_2 = 5.50 \times 10^{-3} \left( \frac{1.44}{1296} \right)\]I_2 \approx 6.11 \times 10^{-6} \, \text{W/m}^2\][/tex]

The sound intensity at 36.0 m is approximately [tex]6.11 \times 10^{-6} W/m^2[/tex].

(c) Finding the decibel level at a distance of [tex]36.0 m[/tex]

Using the decibel formula again:

[tex]\[\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)\][/tex]

where

[tex]I = 6.11 \times 10^{-6} \, \text{W/m}^2\]\beta = 10 \log_{10} \left( \frac{6.11 \times 10^{-6}}{10^{-12}} \right)\]\beta = 10 \log_{10} \left( 6.11 \times 10^{6} \right)\]\beta \approx 10 \times 6.79 = 67.9 \, \text{dB}\][/tex]

The corresponding decibel level at [tex]36.0 m[/tex] is approximately [tex]67.9 dB[/tex].

Explanation:

It is given that,

Charge acquired on rubbing a balloon on your hair, q = -1.93 C

If you move the balloon 25 cm away from your hair, r = 25 cm = 0.25 m

Electric field acting between the balloon and your hair is given by :

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

k = electrostatic constant

[tex]E=\dfrac{9\times 10^9\ Nm^2/C^2\times 1.93\ C}{(0.25\ m)^2}[/tex]

[tex]E=2.779\times 10^{11}\ N/C[/tex]

or

[tex]E=2.78\times 10^{11}\ N/C[/tex]

Hence, this is the required solution.

Answer:

A) 0.0 kJ

Explanation:

Change in the internal energy of the gas is a state function

which means it will not depends on the process but it will depends on the initial and final state

Also we know that internal energy is a function of temperature only

so here the process is given as isothermal process in which temperature will remain constant always

here we know that

[tex]\Delta U = \frac{3}{2}nR\Delta T[/tex]

now for isothermal process since temperature change is zero

so change in internal energy must be ZERO

Explanation:

It is given that,

Frequency of diagnostic ultrasound, f = 3.82 MHz = 3820 Hz

The speed of the sound in air, v = 343 m/s

(a) We need to find the wavelength in air of such a sound wave. Let it is given by λ₁

i.e. [tex]\lambda=\dfrac{v}{\nu}[/tex]

[tex]\lambda_1=\dfrac{343\ m/s}{3820\ Hz}[/tex]

[tex]\lambda_1=0.089\ m[/tex]

(b) If the speed of sound in tissue is 1650 m/s .

[tex]\lambda_2=\dfrac{v}{\nu}[/tex]

[tex]\lambda_2=\dfrac{1650\ m/s}{3820\ Hz}[/tex]

[tex]\lambda_2=0.43\ m[/tex]

Hence, this is the required solution.

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

[tex]h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m[/tex]

or in kilometers,

[tex]h_2 = 47 km[/tex]

the first altitude in kilometers is

[tex]h_1 = 155 km[/tex]

so the difference between the two altitudes is

[tex]\Delta h = 155 km - 47 km = 108 km[/tex]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

           v = u + at

           0 = 19.09 - 9.81 t

            t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

     Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

     So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

              S = ut + 0.5 at²

              H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

    Maximum height of the ball = 18.57 m

Answer:

Peak current = 16.9 A

Explanation:

Given that

RMS voltage = 120 Volts

[tex]V_{rms} = 120 V[/tex]

AC is connected across resistance

[tex]R = 10 ohm[/tex]

now by ohm's law

[tex]V = i R[/tex]

[tex]120 = i (10)[/tex]

[tex]i_{rms} = \frac{120}{10} = 12 A[/tex]

now peak value of current will be given as

[tex]i_{peak} = \sqrt{2} i_{rms}[/tex]

[tex]i_{peak} = \sqrt2 (12) = 16.9 A[/tex]

Answer:

Distance travelled, d = 0.21 m

Explanation:

It is given that,

Initial velocity of electron, u = 500,000 m/s

Acceleration of the electron, a = 500,000,000,000 m/s²

Final velocity of the electron, v = 675,000 m/s

We need to find the distance travelled by the electron. Let distance travelled is s. Using third equation of motion as :

[tex]v^2-u^2=2as[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{(675000\ m/s)^2-(500000\ m/s^2)^2}{2\times 500000000000\ m\s^2}[/tex]

s = 0.205 m

or

s = 0.21 m

So, the electron will travel a distance of 0.21 meters. Hence, this is the required solution.

(a) 168.2 ft/s

The vertical position of the ball is given by

[tex]s = 144t - 16t^2[/tex]

where t is the time.

By differentiating this expression, we find the velocity:

[tex]v = 144-32 t[/tex]

The maximum height is reached when the velocity is zero, so:

[tex]0 = 144 - 32 t[/tex]

From which we find

[tex]t = \frac{144}{32}=4.5 s[/tex]

And substituting this value into the equation for s, we find the maximum height:

[tex]s = 144(4.5 s)-16(4.5 s)^2=324 ft[/tex]

(b) 16 ft/s

We want to find the velocity of the ball when the position of the ball is

s = +320 ft

Substituting into the equation for the position,

[tex]320 = 144t-16t^2[/tex]

[tex]16t^2 -144t +320 = 0[/tex]

Solving for t, we find two solutions:

t = 4 s

t = 5 s

The first one corresponds to the instant in which the ball is still on its way up: Substituting into the equation for the velocity, we find the velocity of the ball at that time

[tex]v = 144 - 32 t=144- 32(4 s)=16 ft/s[/tex]

(c) -16 ft/s

Now we want to find the velocity of the ball when the position of the ball is

s = +320 ft

but on its way down. In the previous part, we found

t = 4 s

t = 5 s

So the second time corresponds to the instant in which the ball is at s = 320 ft but on the way down.

Substituting t = 5 s into the equation for the velocity, we find:

[tex]v = 144 - 32 t=144- 32(5 s)=-16 ft/s[/tex]

And the negative sign means the direction is downward.

The answers for the ball thrown vertically upward with a velocity of 144 ft/s and with a height after t seconds of s = 144t - 16t² are:

a) The maximum height reached by the ball is 324 ft.

b) The velocity of the ball when it is 320 ft above the ground on its way up is 16 ft/s.

c) The velocity of the ball when it is 320 ft above the ground on its way down is -16 ft/s.

a) The maximum height reached by the ball can be calculated with the given equation:

[tex] s = 144t - 16t^{2} [/tex]   (1)

Where:

s: is the height

t: is the time

We can find the time with the following equation:

[tex] v_{f} = v_{i} - gt [/tex]   (2)

Where:

[tex] v_{f} [/tex]: is the final velocity = 0 (at the maximum height)

[tex] v_{i} [/tex]: is the initial velocity = 144 ft/s

g: is the acceleration due to gravity = 32 ft/s²    

Solving equation (2) for t and entering into equation (1), we can find the maximum height:

[tex]s = 144t - 16t^{2} = 144(\frac{v_{i}}{g}) - 16(\frac{v_{i}}{g})^{2} = 144(\frac{144 ft/s}{32 ft/s^{2}}) - 16(\frac{144 ft/s}{32 ft/s{2}})^{2} = 324 ft[/tex]  

Hence, the maximum height is 324 ft.

b) To find the velocity of the ball when it is 320 ft above, we can use the following equation:                       

[tex] v_{f}^{2} = v_{i}^{2} - 2gs [/tex]

[tex]v_{f}^{2} = (144 ft/s)^{2} - 2*32 ft/s^{2}*320 ft[/tex]    

The above equation has two solutions:

[tex]v_{f_{1}} = 16 ft/s[/tex]

[tex]v_{f_{2}} = -16 ft/s[/tex]

Since the question is for the velocity of the ball on its way up and considering the way up as the positive direction, the answer is the positive value [tex]v_{f_{1}} = 16 ft/s[/tex].                      

c) The velocity of the ball when it is 320 ft above the ground on its way down is -16 ft/s (we take the negative value calculated above, [tex] v_{f_{2}}[/tex]). We consider the way down as the negative direction.

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Answer:

(a) 446.88 N

(b) 241.08 N

Explanation:

m = 60 kg, μs = 0.760, μk = 0.410

(a) To just start the crate moving: the coefficient of friction is static.

F = μs m g

F = 0.760 x 60 x 9.8

F = 446.88 N

(b) To slide the crate with constant speed: the coefficient of friction is kinetic.

F = μk m g

F = 0.410 x 60 x 9.8

F = 241.08 N

Static friction is the friction that occurs in the body when the body is just about to move. The static friction force will be 446.88 N.While the kinetic friction force will be 241.08 N.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically it is defined as the product of the coefficient of friction and normal reaction.

m is the mass of crate  = 60 kg

μs is the coefficient of static friction = 0.760

μk  is the coefficient of kinetic friction= 0.410

(a) Static friction is the friction that occurs in the body when the body is just about to move.

[tex]\rm F_s = \mu_s m g\\\\\rm F_s = 0.760\times60\times9.8\\\\\rm F_s = 446.88 N[/tex]

Hence static friction force will be 446.8 N.

(b) When the body is moving in a straight and inclined plane the value of friction force acting on the body is known as kinetic friction.

[tex]\rm F_k = \mu_km g\\\\\rm F_k = 0.410\times60\times9.8\\\\\rm F_k= 241.08\; N[/tex]

Hence kinetic friction force will be 241.08 N.

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Answer:

180 m

Explanation:

Case 1.

U = 40 km/h = 11.1 m/s, V = 0, s = 20 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0 = 11.1 × 11.1 - 2 × a × 20

a = 3.08 m/s^2

Case 2.

U = 220 km/h = 33.3 m/s, V = 0

a = 3.08 m/s^2

Let the stopping distance be x.

Again use third equation of motion

0 = 33.3 × 33.3 - 2 × 3.08 × x

X = 180 m

Answer:

12.3 m/s

Explanation:

m = mass of the box sliding on frictionless flat surface = 12.5 kg

x = compression of the spring = 14.1 cm = 0.141 m

k = spring constant = 94.5 kN/m = 94500 N/m

v = initial speed of the box

Using conservation of energy

Kinetic energy of the box = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

Inserting the values

(12.5) v² = (94500) (0.141)²

v = 12.3 m/s

Answer:

71.8 N

Explanation:

T = Tension force in the strap

W = net work done = 752 J

f = force of friction = 8 N

d = displacement = 15 m

θ = angle between tension force and horizontal displacement = 36 deg

work done by frictional force is given as

W' = - f d

Work done by the tension force is given as

W'' = T d Cos36

Net work done is given as

W = W' + W''

W = T d Cos36 - f d

752 = T (15) Cos36 - (8) (15)

T = 71.8 N

Answer:

option (D)

Explanation:

If we plot a graph between the velocity of the object and the time taken, the slope of graph gives the value of acceleration of the object and the area under the graph gives the product of velocity and time taken that means it is displacement

Answer:

D

Explanation:

Calculate v = (v + u) / 2. ...

Average velocity (v) of an object is equal to its final velocity (v) plus initial velocity (u), divided by two.

The average velocity calculator solves for the average velocity using the same method as finding the average of any two numbers. ...

Given v and u, calculate v. ...

Given v and v calculate u.

Answer is D

Answer:

This is because copper is a conductor, and adding more copper spreads electricity more, and thus the magnetic field.

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Answer:

Angle the wire make with respect to the magnetic field is 30°.

Explanation:

It is given that,

Length of wire, L = 0.6 m

Current flowing in the wire, I = 2 A

Magnetic field strength, B = 0.3 T

It is placed in the magnetic field. It will experience a force of, F = 0.18 N. We need to find the angle the wire make with respect to the magnetic field. The force acting on the wire is given by :

[tex]F=I(L\times B)[/tex]

[tex]F=ILB\ sin\theta[/tex]

[tex]\theta=sin^{-1}(\dfrac{F}{ILB})[/tex]

[tex]\theta=sin^{-1}(\dfrac{0.18\ N}{2\ A\times 0.6\ m\times 0.3\ T})[/tex]

[tex]\theta=30^{\circ}[/tex]

So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.

By substituting the given values and solving for sin(θ), we determined that the wire makes a 30° angle with the magnetic field. Therefore, the wire experiences the magnetic force at this angle.

To determine the angle between a wire carrying a current and a uniform magnetic field, we can use the formula for the magnetic force on a current-carrying wire:

F = I * L * B * sin(θ)

Where:
F is the magnetic force (0.18 N)
I is the current (2.0 A)
L is the length of the wire (0.60 m)
B is the magnetic field strength (0.30 T)
θ is the angle between the wire and the magnetic field

Rearrange the formula to solve for sin(θ):

sin(θ) = F / (I * L * B)

Substitute the given values:

sin(θ) = 0.18 N / (2.0 A * 0.60 m * 0.30 T)

Simplify the expression:

sin(θ) = 0.18 / 0.36 = 0.50

Take the sin (inverse sine) to find θ:

θ = sin⁻¹(0.50) = 30°

Therefore, the wire makes an angle of 30° with respect to the magnetic field.

(a) 7861 N

Along the vertical direction, the plane is moving at constant velocity: this means that the net vertical acceleration is zero, so the vertical component of the 8420 N upward force is balanced by the weight (pointing downward).

The vertical component of the upward force is given by:

[tex]F_y = F sin \theta[/tex]

where

F = 8420 N is the magnitude of the force

[tex]\theta=69.0^{\circ}[/tex] is the angle above the horizontal

Substituting,

[tex]F_y = (8420 N)(sin 69.0^{\circ}) =7861 N[/tex]

This means that the weight of the plane is also 7861 N.

(b) 3.87 m/s^2

From the weight of the plane, we can calculate its mass:

[tex]m=\frac{W}{g}=\frac{7861 N}{9.8 m/s^2}=802 kg[/tex]

Where g = 9.8 m/s^2 is the acceleration due to gravity.

Along the horizontal direction, the 8420 N is not balanced by any other backward force: so, there is a net acceleration along this direction.

The horizontal component of the force is given by

[tex]F_x = F cos \theta = (8420 N)(cos 69.0^{\circ})=3107 N[/tex]

According to Newton's second law, the net force along the horizontal direction is equal to the product between the plane's mass and the horizontal acceleration:

[tex]F_x = m a_x[/tex]

so if we solve for a_x, we find:

[tex]a_x = \frac{F_x}{m}=\frac{3107 N}{802 kg}=3.87 m/s^2[/tex]

Answer:

696.83 m/s

Explanation:

m = mass of water = 565 g = 0.565 kg

c = specific heat of water = 4186 J/(kg⁰C)

ΔT = Change in temperature = T₂ - T₁ = 80 - 22 = 58 ⁰C

v = speed gained by water

Using conservation of energy

Kinetic energy gained by water = heat required to warm water

(0.5) m v² = m c ΔT

(0.5) v² = c ΔT

(0.5) v² = (4186) (58)

v = 696.83 m/s

The magnitude of the electric field is -1.39 x 10^6 N/C and it is directed inward.

To determine the magnitude and direction of the electric field along the axis of the rod at a point 34.0 cm from its center, we can use the formula for the electric field due to a uniformly charged rod. The formula is given by:

E = (k * Q * L) / (x^2 * sqrt(L^2 + x^2))

where E is the electric field, k is the Coulomb's constant, Q is the total charge on the rod, L is the length of the rod, and x is the distance from the center of the rod to the point where we want to find the electric field.

Substituting the given values:

E = (9.0 x 10^9 Nm^2/C^2 * (-21.0 x 10^-6 C) * 0.10 m) / (0.34 m)^2 * sqrt((0.10 m)^2 + (0.34 m)^2) = -1.39 x 10^6 N/C

The negative sign indicates that the electric field is directed inward.

Answer:

Work done, W = 0 J

Explanation:

It is given that,

Weight of the bucket, W = F = 35 N

It is lifted vertically 3 m and then returned to its original position. We need to find the work gravity do on the bucket during this process.

Work done when the bucket is lifted vertically, W₁ = -mgh

Work done when the bucket returned to its original position, W₂ = +mgh

Net work done, W = W₁ + W₂

W = 0 J

So, the work done on the bucket is zero. Hence, this is the required solution.

Final answer:

The total work done by gravity on a bucket of water that is lifted and then returned to its original position is zero joules (0 J), because gravity does equal amounts of positive and negative work on the bucket during the lifting and lowering phases, respectively.

Explanation:

The question examines the concept of work done by a force, which in physics is defined as the product of the force applied to an object and the distance over which that force is applied, provided the force is applied in the direction of motion. When a bucket of water is lifted vertically and then returned to its original position, gravity does work on the bucket on the way down, but because the bucket returns to its starting position, the total work done by gravity over the entire journey is zero joules (d). This is because gravity does positive work as the bucket is lowered and an equal amount of negative work as the bucket is lifted, resulting in a net work of zero.

Concretely, when the bucket is lifted, work is done against gravity and when it is lowered, gravity does work on the bucket. However, since the starting and ending points are the same, the net work done by gravity over the entire process is zero. It's important to notice that this is true regardless of the path taken; as long as the initial and final positions are the same, the work done by a conservative force such as gravity will be zero.

(a) [tex]y(t)=250 - 4.9 t^2[/tex]

For an object in free-fall, the vertical position at time t is given by:

[tex]y(t) = h + ut - \frac{1}{2}gt^2[/tex]

where

h is the initial vertical position

u is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

In this problem,

h = 250 m

u = 0 (the stone starts from rest)

So, the vertical position of the stone is given by

[tex]y(t) = 250 - \frac{1}{2}(9.8) t^2 = 250 - 4.9 t^2[/tex]

(b) 7.14 s

The time it takes for the stone to reach the ground is the time t at which the vertical position of the stone becomes zero:

y(t) = 0

Which means

[tex]y(t) = h - \frac{1}{2}gt^2=0[/tex]

So for the stone in the problem, we have

[tex]250 - 4.9 t^2 = 0[/tex]

Solving for t, we find:

[tex]t=\sqrt{\frac{250}{4.9}}=7.14 s[/tex]

(c) -70.0 m/s (downward)

The velocity of an object in free fall is given by the equation

[tex]v(t) = u - gt[/tex]

where

u is the initial velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have

u = 0

So if we substitute t = 7.14 s, we find the velocity of the stone at the time it reaches the ground:

[tex]v=0-(9.8 m/s^2)(7.14 s)=-70.0 m/s[/tex]

The negative sign means the direction of the velocity is downward.

(d) 6.94 s

In this situation, the stone is thrown downward with an initial speed of 2 m/s, so its initial velocity is

u = -2 m/s

So the equation of the vertical position of the stone in this case is

[tex]y(t) = h + ut - \frac{1}{2}gt^2=250 - 2t - 4.9 t^2[/tex]

By solving the equation, we find the time t at which the stone reaches the ground.

We find two solutions:

t = -7.35 s

t = 6.94 s

The first solution is negative, so it has no physical meaning, therefore we discard it. So, the time it takes for the stone to reach the ground is:

t = 6.94 s

An object in free fall, like a stone dropped from a tower, has its motion governed by the acceleration due to gravity. Using the physics of motion, we can find the height of the stone at any given time, the time it takes to reach the ground, the velocity it strikes the ground, and the time taken if it is initially thrown downward.

To solve these types of questions, we need to use the physics of motion. In the case of an object in free fall like a stone dropped from a tower, the only force acting on it is gravity, which pulls it downwards.

(a) The formula h(t) = 250 - 1/2gt^2 represents the height of the stone above ground level at any time t. Here, g is the acceleration due to gravity (9.8 m/s^2).

(b) The stone will reach the ground when h(t) = 0. Solving the equation 250 - 1/2*9.8*t^2 = 0 gives t ≈ 7.18 seconds (rounded to two decimal places).

(c) The velocity v with which the stone strikes the ground can be found using v = gt. Substituting g = 9.8 m/s^2 and t = 7.18 s gives v ≈ 70.4 m/s (rounded to one decimal place).

(d) If the stone is thrown downward with initial velocity of 2 m/s, the equation for time becomes 250 -2t- 1/2*9.8*t^2 = 0. Solving this gives t ≈ 7.04 seconds (rounded to two decimal places).

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