A hydraulic lift is used to raise an automobile of mass 1510 kg. The radius of the shaft of the lift is 8.0 cm and that of the piston is 1.0 cm. How much force must be applied to the piston to raise the automobile?\

Answer: [tex]2(10)^{42}kg [/tex]

Explanation:

Approaching the orbit of the Large Magellanic Cloud around the Milky Way to a circular orbit, we can use the equation of velocity in the case of uniform circular motion:

[tex]V=\sqrt{G\frac{M}{r}}[/tex] (1)  

Where:

[tex]V=300km/s=3(10)^{5}m/s[/tex] is the velocity of the Large Magellanic Cloud's orbit, which is assumed as constant.

[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]  

[tex]M[/tex] is the mass of the Milky Way

[tex]r=160000ly=1.51376(10)^{21}m[/tex] is the radius of the orbit, which is the distance from the center of the Milky Way to the Large Magellanic Cloud.  

Now, if we want to know the estimated mass of the Milky Way, we have to find [tex]M[/tex] from (1):

[tex]M=\frac{V^{2} r}{G}[/tex] (2)  

Substituting the known values:

[tex]M=\frac{(3(10)^{5}m/s)^{2}(1.51376(10)^{21}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex] (3)  

[tex]M=\frac{1.362384(10)^{32}\frac{m^{3}}{s^{2}}}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex]  

Finally:

[tex]M=2.0416(10)^{42}kg\approx 2(10)^{42}kg[/tex] >>>This is the estimated mass of the Milky Way

To estimate the Milky Way's mass within 160,000 light-years from the center, we can use the orbital velocity law. The estimated mass of the Milky Way within 160,000 light-years from the center is approximately 4.12 × 10^41 kg.

To estimate the Milky Way's mass within 160,000 light-years from the center, we can use the orbital velocity law. The formula for this law is v = sqrt(GM/r), where v is the velocity, G is the gravitational constant, M is the Milky Way's mass, and r is the distance from the center. Rearranging the equation to solve for M, we get M = v^2 * r / G.

Using the given values of v = 300 km/s and r = 160,000 light-years, we need to convert them to appropriate units. 1 light-year is approximately 9.461 × 10^15 meters. After performing the necessary conversions and calculations, the estimated mass of the Milky Way within 160,000 light-years from the center is approximately 4.12 × 10^41 kg.

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Answer:

615.15 K/m

33 K

Explanation:

A = 0.897 m^2, Heat per second = 27700 W, ΔT/Δx = ?, Δx = 0.0537 m

K = 50.2 W/mK

heat per second = K x A x ΔT/Δx

27700 = 50.2 x 0.897 x ΔT/Δx

ΔT/Δx = 615.15 K/m

Now ΔT = ΔT/Δx x Δx

ΔT = 615.15 x 0.0537 = 33 K

Answer:

The answer is E) 20kPa  

Explanation:

Here we use the definition of Pressure: Pressure is a physical quantity that measures the projection of force in the perpendicular direction per unit area. In this case, wehave a 200N Force over a 100cm^2 Area

In order to have the pressure in international units we need to convert those cm^2 to m^2 this way:

We know 100cm is equal to 1m

[tex](100cm^2)*((1m^2)/(100cm)^2)[/tex]

In this case, the area will be equal to 0,01m^2

With this information we use the Pressure equation, which based on the definition will be:

[tex]P=F/A[/tex]

Solving:

[tex]P=(200N)/(0,01m^2)[/tex]

[tex]P=20000Pa[/tex]

Then P is the same as 20kPa  

Explanation:

There are two types of collision. First one is elastic and other one is inelastic collision.

The linear momentum of two objects before and after the collision remains the same in case of elastic collision. Also, the kinetic energy in elastic collision is conserved. But in case of inelastic collision, the momentum of two objects before and after the collision remains constant but the kinetic energy is in conversed. It changes form one form of energy to another.

Hence, the correct option is (E) "None of the above".

C) It is transformed into momentum such that momentum is conserved. (This statement is incorrect. Conservation of momentum applies to all collisions, not just the change in kinetic energy)

In an ideal elastic collision, the following happens:

Total Kinetic Energy is Conserved: The total kinetic energy of the system (colliding objects) before the collision is equal to the total kinetic energy after the collision. No energy is lost or gained in terms of kinetic energy.

Here's why the other options are incorrect:

A) Heat and Deformation: While real collisions might involve some energy loss due to heat and deformation, an ideal elastic collision is considered perfectly elastic, meaning no energy is lost in these ways.

B) Potential Energy: The collision doesn't necessarily involve a change in potential energy of the objects.

C) Momentum: Momentum is always conserved in any collision, elastic or inelastic.

D) All of the Above: Only the conservation of total kinetic energy applies to an ideal elastic collision.

E) None of the Above: In an ideal elastic collision, kinetic energy is conserved.

Therefore, the correct answer is:

C) It is transformed into momentum such that momentum is conserved. (This statement is incorrect. Conservation of momentum applies to all collisions, not just the change in kinetic energy)

Answer:

83.3 rad/s^2

Explanation:

r = 3.6 cm = 0.036 m

a = 3 m/s^2

the relation  between the linear acceleration and angular acceleration is given by

a = r α

where, α is angular acceleration, a be the linear acceleration.

α = a / r

α = 3 / 0.036 = 83.3 rad/s^2

Answer:

a) 2.85 kW

b) $ 432

c) $ 76.95

Explanation:

Average price of electricity = 1 $/40 MJ

Q = 20 kW

Heat energy production = 20.0 KJ/s

Coefficient of performance,  K = 7

also

K=(QH)/Win

Now,

Coefficient of Performance, K = (QH)/Win = (QH)/P(in) = 20/P(in) = 7

where

P(in) is the input power

Thus,

P(in) = 20/7 = 2.85 kW

b) Cost = Energy consumed × charges

Cost = ($1/40000kWh) × (16kW × 300 × 3600s)

cost = $ 432

c) cost = (1$/40000kWh) × (2.85 kW × 200 × 3600s) = $76.95

Final answer:

To heat a home for one month, an electric heater requires $432, while a heat pump with a coefficient of performance of 7 costs significantly less at $61.71, showcasing the efficiency and cost-effectiveness of using a heat pump for home heating.

Explanation:

To calculate the cost of heating for one month using a 16.0 kW electric heater and a heat pump with a coefficient of performance of 7.00, we start with understanding how a heat pump works. It essentially moves heat from the outdoors indoors, making it more efficient than directly converting electricity into heat. The first step is to calculate the total energy used by the electric heater and then compare it to the energy utilized efficiently by the heat pump.

Cost Calculation for Electric Heater:

Power usage of heater = 16.0 kW

Total hours run per month = 300 hours

Total energy used per month = 16.0 kW × 300 hours = 4800 kWh

Since 1 kWh equals 3.6 MJ, the total energy in MJ = 4800 × 3.6 = 17280 MJ

Average price for electricity = 40 MJ per dollar

Cost = 17280 MJ / (40 MJ/dollar) = $432

Cost Calculation for Heat Pump:

The coefficient of performance (COP) of the heat pump = 7.00

Efficiency implies that for every unit of energy (work) used, 7 units of heat are transferred inside.

To provide the same heat as the electric heater, the energy consumed by the heat pump = 4800 kWh / 7 = 685.71 kWh

Total energy in MJ = 685.71 × 3.6 = 2468.56 MJ

Cost = 2468.56 MJ / (40 MJ/dollar) = $61.71

Thus, using a heat pump is significantly more cost-efficient for home heating compared to using a direct electric heater.

Answer:

Power, P = 112500 watts

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Initial velocity of the car, u = 0

Final velocity of the car, v = 30 m/s

Time taken, t = 6 s

The minimum average power delivered by the engine is given by :

P = W/t

Where

W = work done

t = time taken

Work done = change in kinetic energy

So, [tex]P=\dfrac{\Delta K}{t}=\dfrac{\dfrac{1}{2}m(v^2-u^2)}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}\times 1500\ kg\times (30\ m/s)^2}{6\ s}[/tex]

P = 112500 watts

So, the minimum average power delivered by the engine is 112500 watts. Hence, this is the required solution.

Answer:

The Volume is V= 644.89 L

Explanation:

m= 2 kg

T= 120 º C = 393.15 K

P= 3.5 bar = 3.45 atm

R= 0.08205746 atm L / K mol

M= 0.029 Kg / mol

n= m/M

n= 68.96 moles

p*V= m*R*T

V= m*R*T/p

V=644.89 L

Answer:

[tex]f = 8.89 cm[/tex]

Explanation:

As we know that Far sighted person has near point shifted to 80 cm distance

so he is able to see the object 80 cm

now the distance of lens from eye is 2 cm

and the person want to see the objects at distance 10 cm

so here the image distance from lens is 80 cm and the object distance from lens is 8 cm

now from lens formula we have

[tex]\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}[/tex]

[tex]-\frac{1}{80} + \frac{1}{8} = \frac{1}{f}[/tex]

[tex]f = 8.89 cm[/tex]

Answer:

The angle is 1.33°.

Explanation:

Given that,

Distance [tex]d=84\times10^{-6}\ m[/tex]

Wavelength = 650 nm

Number of fringe = 3

We need to calculate the angle

Using formula of angle for brightness

[tex]d\sin\theta=m\lambda[/tex]

[tex]\theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]

Where, d = distance

[tex]\lambda[/tex] =wavelength

m = number of fringe

Put the value into the formula

[tex]\theta=\sin^{-1}\dfrac{3\times650\times10^{-9}}{84\times10^{-6}}[/tex]

[tex]\theta=1.33^{\circ}[/tex]

Hence, The angle is 1.33°.

Answer:

125.09 Hz

Explanation:

Given : A person stands 5.20 from one speaker and 4.10 m away from the other speaker.

Distance between the speakers is 5.20 - 4.10 =1.10 m

We know that

For destructive interferences

        n . λ / 2

where n =1,3,5,7....

Therefore difference between the speakers is

5.10 - 4.20 = 1 X 0.5 λ

λ = 2.2

Given velocity of sound is V = 344 m/s

Therefore frequency, f = [tex]\frac{v}{\lambda }[/tex]

                                      = [tex]\frac{344}{2.2 }[/tex]

                                      = 156.36 Hz

Now, the third lowest frequency is given by

              λ = (5.20-4.10) X 5 X 0.5

                 = 2.75 m

Therefore frequency, f = [tex]\frac{v}{\lambda}[/tex]

                                      = [tex]\frac{344}{2.75}[/tex]

                                      = 125.09 Hz

Therefore third lowest frequency is 125.09 Hz

Answer:

Charge on each plate = 2.31 x 10⁻⁸ C

Explanation:

We have the equations

           [tex]E=\frac{V}{d}\texttt{ and }V=\frac{Q}{C}[/tex]

Combining both equations

           [tex]E=\frac{\left (\frac{Q}{C}\right )}{d}=\frac{Q}{Cd}[/tex]

We also have the equation for capacitance

           [tex]C=\frac{\epsilon A}{d}[/tex]

That is

          [tex]E=\frac{Q}{\frac{\epsilon A}{d}\times d}=\frac{Q}{\epsilon A}\\\\Q=\epsilon AE[/tex]

Substituting

           [tex]Q=8.85\times 10^{-12}\times 31.5\times 10^{-4}\times 8.30\times 10^5=2.31\times 10^{-8}C[/tex]

Charge on each plate = 2.31 x 10⁻⁸ C

Answer:

a) Average velocity of the car for the time interval t=0 to t = 1.95 s is 2.62 m/s.

b) Average velocity of the car for the time interval t=0 to t = 3.96 s is 4.92 m/s.

c) Average velocity of the car for the time interval t=1.95 to t = 3.96 s is 7.15 m/s.

Explanation:

We have  x(t)= α t²− β t³

That is x(t)= 1.44 t²− 5 x 10⁻² t³

Average velocity is ratio of distance traveled to time.

a)Average velocity of the car for the time interval t=0 to t = 1.95 s

  x (0) = 1.44 x 0²− 5 x 10⁻² x 0³ = 0 m

  x (1.95) = 1.44 x 1.95²− 5 x 10⁻² x 1.95³ = 5.10 m

  Time difference = 1.95 - 0 = 1.95 s

  Average velocity [tex]=\frac{5.10}{1.95}=2.62m/s[/tex]

b)Average velocity of the car for the time interval t=0 to t = 3.96 s

  x (0) = 1.44 x 0²− 5 x 10⁻² x 0³ = 0 m

  x (3.96) = 1.44 x 3.96²− 5 x 10⁻² x 3.96³ = 19.48 m

  Time difference = 3.96 - 0 = 3.96 s

  Average velocity [tex]=\frac{19.48}{3.96}=4.92m/s[/tex]

c)Average velocity of the car for the time interval t=0 to t = 3.96 s

  x (1.95) = 1.44 x 1.95²− 5 x 10⁻² x 1.95³ = 5.10 m

  x (3.96) = 1.44 x 3.96²− 5 x 10⁻² x 3.96³ = 19.48 m

  Time difference = 3.96 - 1.95 = 2.01 s

  Average velocity [tex]=\frac{19.48-5.10}{2.01}=7.15m/s[/tex]

Final answer:

The total magnetic flux through the coil before rotation is calculated using the formula Φ = B × A × cos(θ), with B being the Earth's magnetic field strength, A the area of the coil, and θ the angle between the field and the normal to the coil, resulting in a flux of 7.2 × 10⁻⁸ Tm².

Explanation:

The total magnetic flux Φ through the coil before it is rotated is given by the formula Φ = B × A × cos(θ), where B is the magnetic field strength, A is the area enclosed by the coil, and θ is the angle between the magnetic field and the normal to the plane of the coil. Before the rotation, the plane of the coil is perpendicular to Earth's magnetic field, making θ = 0° (or cos(θ) = 1), so the cos(θ) term does not affect the calculation.

To calculate the magnetic flux, convert the area from cm² to m² by multiplying by 10⁻⁴, and use the given values:

Thus, the total magnetic flux through the coil before it is rotated is 7.2 × 10⁻⁸ Tm².

Answer:

Charge, q = 3.48 μC

Explanation:

It is given that,

Electric field, [tex]E=1.25\times 10^5\ N/C[/tex]

Distance form positive charge, r = 0.5 meters

We need to find the charge. It is given by the formula as follows :

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

[tex]q=\dfrac{Er^2}{k}[/tex]

k = electrostatic constant

[tex]q=\dfrac{1.25\times 10^5\ N/C\times (0.5\ m)^2}{9\times 10^9}[/tex]

q = 0.00000347 C

or

[tex]q=3.48\times 10^{-6}\ C[/tex]

[tex]q=3.48\ \mu C[/tex]

So, the charge is 3.48 μC. Hence, this is the required solution.

Answer:

v = 335.7 m/s

Explanation:

As we know that speed of sound in air is given by the formula

[tex]v = \sqrt{\frac{\gamma RT}{M}}[/tex]

now we have

[tex]\gamma = 1.4[/tex] For air

M = 29 g/mol = 0.029 kg/mol

T = 8 degree Celcius = 273 + 8 = 281 K

R = 8.31 J/mol K

now from above formula we have

[tex]v = \sqrt{\frac{(1.4)(8.31)(281)}{0.029}}[/tex]

[tex]v = 335.7 m/s[/tex]

Answer:

Increase in temperature =  269.54 °C

Explanation:

We have equation for thermal expansion

          ΔL = LαΔT

Change in length, ΔL = 0.08 m

Length, L = 56 m

Coefficient of thermal expansion, α = 5.3 x 10⁻⁶ °C⁻1

Change in temperature, ΔT = T - 253

Substituting

          0.08 = 56 x 5.3 x 10⁻⁶ x (T - 253)

         (T - 253) = 269.54

           T = 522.54 °C

Increase in temperature =  269.54 °C      

Answer:

[tex]x=0\\x=4\\x=-2[/tex]

Explanation:

The zeros of a function are all the values of x for which f (x) = 0.

Therefore to find the zeros of the function I must equal f(x) to zero and solve for x.

[tex]f(x) = x(x - 4)(x + 2)=0[/tex]

[tex]x(x - 4)(x + 2)=0[/tex]

We have the multiplication of 3 factors x, (x-4) and (x + 2)

Then the function will be equal to zero when one of the factors is equal to zero, that is:

[tex]x = 0\\(x-4) = 0,\ x = 4\\(x + 2) = 0,\ x = -2[/tex]

Note that [tex]f(x) = x (x - 4) (x + 2)[/tex] is a cubic function of positive principal coefficient, the graph starts from [tex]-\infty[/tex] and cuts to the x-axis at [tex]x = -2[/tex], then decreases and cuts by second once to the x-axis at [tex]x = 0[/tex], it finally cuts the x-axis for the third time at [tex]x = 4[/tex] and then tends to [tex]\infty[/tex]

The zeros of the function f(x) = x(x - 4)(x + 2) are 0, 4, and -2. The end behavior of the function indicates that as x approaches infinity, f(x) approaches infinity, and as x approaches negative infinity, f(x) approaches negative infinity.

To determine the zeros of the function f(x) = x(x - 4)(x + 2), we need to find the values of x that make f(x) equal to zero. The function will be zero when any of the factors are zero. Therefore, the zeros of f(x) are 0, 4, and -2.

The end behavior of the polynomial function is determined by the leading term. Since our function has a leading term x3 with a positive coefficient, as x approaches positive infinity, f(x) also approaches positive infinity, and as x approaches negative infinity, f(x) approaches negative infinity. This is characteristic of an odd-degree polynomial with a positive leading coefficient.

Answer:

57 %

Explanation:

input power = 16.4 kW = 16.4 x 10^3 W = 16400 W

Water pumped per second = 67 L/s

Mass of water pumped per second, m = Volume of water pumped epr second x density of water

m = 67 x 10^-3 x 1000 = 67 kg/s

height raised, h = 14 m

Output Power = m x g x h / t = 67 x 10 x 14 = 9380 W

efficiency = output power / input power = 9380 / 16400 = 0.57

% efficiency = 57 %

thus, the efficiency of the pump is 57 %.

Answer:

Part a)

F = 0.056 N

Part b)

a = 28.13 m/s/s

Explanation:

Part a)

Electrostatic force between two charged balls is given as

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

now we will have

[tex]q_1 = q_2 = 50 nC[/tex]

r = 2 cm

now we will have

[tex]F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.02^2}[/tex]

so here we have

[tex]F = 0.056 N[/tex]

Part b)

Now due to above force the acceleration of one ball which is released is given as

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{0.056}{2 \times 10^{-3}}[/tex]

[tex]a = 28.13 m/s^2[/tex]

Answer:

159241.048 cm³/s

Explanation:

r = Radius = 3×height = 3h

h = height = 16 cm

Height of the pile increases at a rate = [tex]\frac{dh}{dt}=22\ cm/s[/tex]

[tex]\text{Volume of cone}=\frac{1}{3}\pi r^2h\\\Rightarrow V=\frac{1}{3}\pi (3h)^2h\\\Rightarrow V=3\pi h^3[/tex]

Differentiating with respect to time

[tex]\frac{dv}{dt}=9\pi h^2\frac{dh}{dt}\\\Rightarrow \frac{dv}{dt}=9\pi 16^2\times 22\\\Rightarrow \frac{dv}{dt}=159241.048\ cm^3/s[/tex]

∴ Rate is the sand leaving the bin at that​ instant is 159241.048 cm³/s

Answer:

(a)  1.3 x 10^6 Hz

(b) 76.73 cm

Explanation:

(a)

the formula for the frequency is given by

f = B q / 2 π m

where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.

B = 46.7 micro tesla = 46.7 x 10^-6 T

q = 1.6 x 10^-19 C

m = 9.1 x 10^-31 kg

f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz

(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J

K = 1/2 mv^2

182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 6.3 x 10^6 m/s

r = m v / B q

Where, r be the radius of circular path

r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)

r = 0.7673 m = 76.73 cm  

Explanation:

It is given that,

Mass of the artillery shell, m = 1100 kg

Acceleration of he shell, [tex]a=2.4\times 10^4\ m/s^2[/tex]

(a) The magnitude of force exerted on the shell can be calculated using Newton's second law of motion as :

F = ma

[tex]F=1100\ kg\times 2.4\times 10^4\ m/s^2[/tex]

F = 26400000 N

or

[tex]F=2.6\times 10^7\ N[/tex]

(b) The magnitude of the force exerted on the ship by the artillery shell will be same but the direction is opposite as per Newton's third law of motion. i.e [tex]2.6\times 10^7\ N[/tex]

The magnitude of the force exerted on the ship by the artillery shell is 2.64 × 10^7 N, and it acts in the opposite direction of the shell's motion.

(a) Net External Force on the Artillery Shell:

We can use Newton's second law of motion (ΣF = ma) to solve this problem. Here, ΣF represents the net external force acting on the object (artillery shell), m is the mass of the shell (1100.0 kg), and a is the acceleration of the shell (2.40 × 10^4 m/s²).

Force (F) = mass (m) × acceleration (a)

F = 1100.0 kg × 2.40 × 10^4 m/s²

F = 2.64 × 10^7 N

Therefore, the net external force exerted on the artillery shell is 2.64 × 10^7 N.

(b) Force on the Ship and Explanation:

According to Newton's third law of motion, for every action, there is an equal and opposite reaction. In this case, the force exerted on the artillery shell to accelerate it (calculated in part (a)) is equal in magnitude but opposite in direction to the force exerted by the shell on the ship.

Therefore, the magnitude of the force exerted on the ship by the artillery shell is 2.64 × 10^7 N, and it acts in the opposite direction of the shell's motion.

Answer:

a) 3.5 Ω

b) 11.5 Ω

Explanation:

In series:

R₁ + R₂ = 15.0

In parallel:

1/R₁ + 1/R₂ = 1/2.67

Multiply both sides by R₁ R₂:

R₂ + R₁ = R₁ R₂ / 2.67

Solve the system of equations with substitution:

15.0 = R₁ (15.0 − R₁) / 2.67

40.1 = 15.0 R₁ − R₁²

R₁² − 15.0 R₁ + 40.1 = 0

R₁ = [ 15 ± √(225 − 4(1)(40.1)) ] / 2

R₁ = 11.5 Ω

R₂ = 15.0 − R₁

R₂ = 3.5 Ω

Answer:

32.225  and angle is 48.7 degree.

Explanation:

u = 15, θ = 35 degree

v = 18, θ = 60

First represent the u and v in vector form.

u = 15 (Cos 35 i + Sin 35 j ) = 12.287 i + 8.6 j

v = 18 ( Cos 60 i + Sin 60 j ) = 9 i + 15.6 j

The sum of the two vectors is given by

u + v = 12.287 i + 8.6 j + 9 i + 15.6 j = 21.28 i + 24.2 j

Magnitude of u + v = [tex]\sqrt{21.28^{2}+24.2^{2}}[/tex] = 32.225

Let Ф be the angle

tan Ф = 24.2 / 21.28

Ф = 48.7 degree

The magnitude and direction of the resultant vector, for given vectors u and v, can be calculated by first finding the horizontal and vertical components of each vector. The magnitude of the resultant vector is then found using the Pythagorean theorem and the direction using the inverse tangent function.

To find the resulting vector of u + v where u and v are vectors and the given magnitudes and directions, we begin by solving for the horizontal and vertical components of each vector (u and v), using the formulas ux = u cos θ and uy = u sin θ for vector u and similarly for vector v. Then, add the respective horizontal and vertical components to find the horizontal and vertical components of the resultant vector.

The magnitude of the resulting vector (u + v) can be found using the Pythagorean theorem, which is: abs(u + v) = sqrt((ux+vx)² + (uy+vy)²).

The direction angle of the resulting vector can be determined using the formula: θ = atan((uy+vy) / (ux+vx)), where atan is the inverse tangent function. Round the magnitude to the nearest tenth and the direction angle to the nearest whole degree as per the question's instructions.

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Answer:

Option D is the correct answer.

Explanation:

Refer the figure given.

By Pascal's principle we have

            [tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]

F2 = 2 x 10⁴ N

[tex]A_1=\frac{\pi\times (12\times 10^{-3})^2}{4}=1.13\times 10^{-4}m^2\\\\A_2=\frac{\pi\times (36\times 10^{-3})^2}{4}=1.02\times 10^{-3}m^2[/tex]

Substituting

      [tex]\frac{F_1}{1.13\times 10^{-4}}=\frac{2\times 10^4}{1.02\times 10^{-3}}\\\\F_1=2.22\times 10^3N[/tex]

Option D is the correct answer.

Answer:

0.587 m/s

Explanation:

m = mass of the object = 2.7 kg

k = spring constant = 280 N/m

[tex]w[/tex] = angular frequency

Angular frequency is given as

[tex]w =\sqrt{ \frac{k}{m}}[/tex]

[tex]w =\sqrt{ \frac{280}{2.7}}[/tex]

[tex]w[/tex] = 10.2 rad/s

x = position relation to equilibrium position = 0.020 m

A = amplitude

[tex]v[/tex]  = speed at position "x" = 0.55 m/s

speed is given as

[tex]v = w\sqrt{A^{2} - x^{2}}[/tex]

[tex]0.55 = (10.2)\sqrt{A^{2} - 0.02^{2}}[/tex]

A = 0.0575 m

[tex]v_{max}[/tex] = maximum speed of the object

maximum speed of the object is given as

[tex]v_{max}=A w[/tex]

[tex]v_{max}=(0.0575) (10.2)[/tex]

[tex]v_{max}[/tex] = 0.587 m/s

The maximum speed attained by the object,

[tex]\rm v_m_a_x=0.587\;m/sec[/tex]

Given :

Mass of the object, m = 2.7 Kg

Spring constant, K = 280 N/m

Speed = 0.55 m/sec

Solution :

We know that the angular velocity is given by,

[tex]\rm \omega = \sqrt{\dfrac{K}{m}}[/tex]

[tex]\rm \omega = \sqrt{\dfrac{280}{2.7}}[/tex]

[tex]\rm \omega = 10.2\;rad/sec[/tex]

Now, speed is given as

[tex]\rm v = \omega\sqrt{A^2-x^2}[/tex]

[tex]0.55=(10.2)\sqrt{A^2-0.02^2}[/tex]

[tex]\rm A = 0.0575\;m[/tex]

Now, maximum speed of the object is,

[tex]\rm v_m_a_x=A\omega[/tex]

[tex]\rm v_m_a_x=0.0575\times10.2=0.587\;m/sec[/tex]

For more information, refer the link given below

https://brainly.com/question/12446100?referrer=searchResults

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

[tex]\mu mg = \frac{mv^2}{R}[/tex]

here we have

[tex]\mu g = \frac{v^2}{R}[/tex]

now we know that

[tex]v = \sqrt{\mu Rg}[/tex]

here we have

[tex]\mu = 0.600[/tex]

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

[tex]v = \sqrt{(0.600)(35)(9.81)}[/tex]

[tex]v = 14.35 m/s[/tex]

Answer:

10^-7 C

Explanation:

m = 1 g = 10^-3 kg, E = 200,000 N/C, a = 20 m/s^2, u = 0

Let q be the charge on bead

Force = m a = q E

a = q E / m

q = m a / E = (10^-3 x 20) / 200000 = 10^-7 C