Answer:
(1) 2.05 x 10^4 N/C
(2) Downward
(3) upward, 9.8 m/s^2
(4) upward, 9.8 m/s^2
Explanation:
q = - 2.1 micro coulomb, m = 0.0044 kg, g = 9.8 m/s^2
(1) The electric force is given by F = - q x E
The magnitude of electric force is balanced by the weight of the charged particle
q x E = m x g
E = mg / q
[tex]E = \frac{0.0044 \times 9.8}{2.1 \times 10^{-6}}[/tex]
E = 2.05 x 10^4 N/C
(2) As the electric force is acting upward and the weight is downward so the elecric field is in downward direction.
(3) The charge is doubled,
then the electric field becomes half.
E = 2.05 x 10^4 / 2 = 1.025 x 10^4 N/C
The direction is same that is in downward direction.
Acceleration = Force / mass
a = mg / m = 9.8 m/s^2 upward
(4) The charge is doubled,
then the electric field becomes half.
E = 2.05 x 10^4 / 2 = 1.025 x 10^4 N/C
The direction is same that is in downward direction.
Acceleration = Force / mass
a = mg / m = 9.8 m/s^2 upward
The magnitude of the electric field and the direction of the electric filed acting on the object is,
(1) The magnitude of the electric field is [tex]2.05\times10^4\rm N/C[/tex](2) The direction of the electric field is downward.(3) The electric charge on the object is doubled while its mass remains the same, then the direction and magnitude is downward and acceleration is 9.8 m/s² respectively.What is electric field?The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
The mass of the object is 0.0044 kg, and the charge on the object is -2.1 μC.
(1) The magnitude of the electric field-As the electric field is the ratio of force (due to gravity in given case) to the charge given on it. Thus the magnitude of the electric field of 0.0044 kg object is,
[tex]E=\dfrac{0.0044\times9.8}{2.1\times10^{-6}}\\E=2.05\times10^{4}\rm N/C[/tex]
(2) The direction of the electric field-The electric force acting on the body is in the upward direction Thus, to balance this, the electric field acts on the body is in the downward direction.
(3) The electric charge on the object is doubled while its mass remains the same, then the direction and magnitude of its acceleration-
When the electric charge on the object is doubled while its mass remains the same, then the strength of the electric field is halved. Therefore,
[tex]E=\dfrac{2.05\times10^4}{2}\\E=1.025\times10^4\rm N/C[/tex]
The electric force acting on the body is in the upward direction Thus, to balance this, the electric field acts on the body is in the downward direction.
Hence, the magnitude of the electric field and the direction of the electric filed acting on the object is,
(1) The magnitude of the electric field is [tex]2.05\times10^4\rm N/C[/tex](2) The direction of the electric field is downward.(3) The electric charge on the object is doubled while its mass remains the same, then the direction and magnitude is downward and acceleration is 9.8 m/s² respectively.Learn more about electric field here;
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A rock is thrown vertically upward from ground level at time t = 0. At t = 1.6 s it passes the top of a tall tower, and 1.0 s later it reaches its maximum height. What is the height of the tower?
The rock has height [tex]y[/tex] at time [tex]t[/tex] according to
[tex]y=v_0t-\dfrac g2t^2[/tex]
where [tex]v_0[/tex] is the velocity with which it was thrown, and g = 9.8 m/s^2 is the acceleration due to gravity.
Complete the square to get
[tex]y=\dfrac{{v_0}^2}{2g}-\dfrac g2\left(t-\dfrac{v_0}g\right)^2[/tex]
which indicates a maximum height of [tex]\dfrac{{v_0}^2}{2g}[/tex] occurs when [tex]t=\dfrac{v_0}g[/tex]. We're told this time is 2.6 s after the rock is thrown:
[tex]2.6\,\mathrm s=\dfrac{v_0}{9.8\frac{\rm m}{\mathrm s^2}}\implies v_0=25.48\dfrac{\rm m}{\rm s}[/tex]
So when t = 1.6 s, the rock reaches the tower's height of
[tex]y=v_0(1.6\,\mathrm s)-\dfrac g2(1.6\,\mathrm s)^2\approx\boxed{28\,\mathrm m}[/tex]
The height of the tower can be calculated by finding the maximum height reached by the rock and then subtracting the initial height of the rock.
Explanation:The height of the tower can be calculated by finding the maximum height reached by the rock and then subtracting the initial height of the rock. Since the rock reaches its maximum height 1.0 s after passing the top of the tower, we can use the formula:
Max height = y1 - y0 - (1/2) * g * t^2
where y1 is the maximum height, y0 is the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes to reach the maximum height.
Since the rock is 8.10 m above its starting point at t = 1.00 s, we can plug in the values:
Max height = 8.10 m - 0 m - (1/2) * (9.8 m/s^2) * (1.0 s)^2
Simplifying this equation, we find that the maximum height reached by the rock is 3.15 m. Therefore, the height of the tower is 8.10 m + 3.15 m = 11.25 m.
A 10 Ω resistor is connected to a 120-V ac power supply. What is the peak current through the resistor?
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]
Suppose you wanted to change the fundamental frequency of an oscillating string. What are three parameters that you could alter and how would these alter the oscillation frequency?
Answer:
Linear density, Length, and tension in the string
Explanation:
The fundamental frequency of an oscillating string is given by:
[tex]f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]
where
L is the length of the string
T is the tension in the string
[tex]\mu[/tex] is the linear density of the string, which can also be rewritten as
[tex]\mu = \frac{m}{L}[/tex]
where m is the mass of the string.
Therefore, we can say that in order to change the fundamental frequency of the string, we can change either its lenght, or its tension or its linear density.
Howuch work does it taje to move a 5kg box up/a 10m tall flight of stairs
Answer:
490 J
Explanation:
The work done to lift an object is equal to its increase in gravitational potential energy, therefore
[tex]W=mgh[/tex]
where
m is the mass of the object
g = 9.8 m/s^2 is the acceleration of gravity
h is the increase in height of the object
In this problem,
m= 5 kg
h = 10 m
Therefore, the work done is
[tex]W=(5 kg)(9.8 m/s^2)(10 m)=490 J[/tex]
A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field?
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 race car goes around a level, circular track with a diameter of 1.00 km at a constant speed of 89 km/h. What is the car's centripetal acceleration in m/s^2?
Answer:
The car's centripetal acceleration is 1.22 m/s².
Explanation:
Given that,
Diameter of circular track [tex]d= 1.00\ km=1000\ m[/tex]
Radius r = 500 m
Constant speed [tex]v = 89\ km/h = 89\times\dfrac{5}{18}=24.722\ m/s[/tex]
The centripetal acceleration is defined as,
[tex]a_{c} = \dfrac{v^2}{r}[/tex]
Where, v = tangential velocity
r = radius
Put the value into the formula
[tex]a_{c}=\dfrac{(24.722)^2}{500}[/tex]
[tex]a_{c}=1.22\ m/s^2[/tex]
Hence, The car's centripetal acceleration is 1.22 m/s².
To find the car's centripetal acceleration, convert the speed to meters per second, then use the formula a = v^2 / r. With a speed of 24.71 m/s and a track radius of 500 m, the acceleration is approximately 1.22069 m/s².
Explanation:The student is asking to determine the centripetal acceleration of a race car moving at a constant speed along a circular track. To calculate this, we can use the formula for centripetal acceleration: a = v^2 / r. First, we need to convert the speed from km/h to m/s, which can be done by multiplying by ⅟ (≅ 0.27778). Thus, 89 km/h equates to 89 × 0.27778 m/s, which is about 24.71 m/s. The diameter of the track is 1.00 km (1000 m), hence the radius r is half of that, 500 m.
Using the formula:
a = (24.71 m/s)² / 500 m
a = 610.3441 m²/s² / 500 m
a = 1.22069 m/s² (approximately)
The centripetal acceleration of the race car is approximately 1.22069 m/s².
For a metal that has an electrical conductivity of 7.1 x 107 (Ω-m)-1, do the following: (a) Calculate the resistance (in Ω) of a wire 2.6 mm in diameter and 6.7 m long. (b) Calculate the current (in A) if the potential drop across the ends of the wire is 0.060 V. (c) Calculate the current density (in A/m^2). (d) Compute the magnitude of the electric field (in V/m) across the ends of the wire.
Answer:
(a) 0.0178 Ω
(b) 3.4 A
(c) 6.4 x 10⁵ A/m²
(d) 9.01 x 10⁻³ V/m
Explanation:
(a)
σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹
d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) π d²
A = (0.25) (3.14) (2.6 x 10⁻³)²
A = 5.3 x 10⁻⁶ m²
L = length of the wire = 6.7 m
Resistance of the wire is given as
[tex]R=\frac{L}{A\sigma }[/tex]
[tex]R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }[/tex]
R = 0.0178 Ω
(b)
V = potential drop across the ends of wire = 0.060 volts
i = current flowing in the wire
Using ohm's law, current flowing is given as
[tex]i = \frac{V}{R}[/tex]
[tex]i = \frac{0.060}{0.0178}[/tex]
i = 3.4 A
(c)
Current density is given as
[tex]J = \frac{i}{A}[/tex]
[tex]J = \frac{3.4}{5.3\times10^{-6}}[/tex]
J = 6.4 x 10⁵ A/m²
(d)
Magnitude of electric field is given as
[tex]E = \frac{J}{\sigma }[/tex]
[tex]E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}[/tex]
E = 9.01 x 10⁻³ V/m
A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at constant speed? Important: Assume that the pushing force is to the right
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.
What is the friction force?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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A monoatomic ideal gas undergoes an isothermal expansion at 300 K, as the volume increased from 0.010 m^3 to 0.040 m^3. The final pressure is 130 kPa. What is the change in the internal (thermal) energy of the gas during this process? (R=8.31 J/mol . K) a. 0.0 kJ b. 3.6 kJ c. 7.2 kJ d.-3.6 kJ e.-7.2 kJ
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
Water flowing through a garden hose of diameter 2.71 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water leaving the end of the hose? Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle? m/s
Answer:
i) [tex]v_1 = 0.40 m/s[/tex]
ii) [tex]v_2 = 3.60 m/s[/tex]
Explanation:
Part A)
As we know that diameter of the hose pipe is 2.71 cm
Now the area of crossection of the pipe will be
[tex]A = \pi (\frac{D}{2})^2[/tex]
[tex]A = \pi (\frac{0.0271}{2})^2 = 5.77 \times 10^{-4} m^2[/tex]
Now the flow rate is defined as the rate of volume
It is given as
[tex]Q = \frac{Volume}{time} = Area \times speed[/tex]
[tex]\frac{20 L\times \frac{10^{-3} m^3}{1L}}{1.45 \times 60 seconds} = 5.77 \times 10^{-4} \times v[/tex]
[tex]v = 0.40 m/s[/tex]
Part b)
As per equation of continuity we know
[tex]A_1 v_1 = A_2 v_2[/tex]
now we have
[tex]\pi (\frac{d_1}{2})^2 v_1 = \pi (\frac{d_2}{2})^2v_2[/tex]
[tex](2.71)^2 (0.40) = (\frac{2.71}{3})^2 v_2[/tex]
[tex]v_2 = 3.60 m/s[/tex]
What does the area under a curve of a graph of velocity versus time yield? A. Average acceleration O B. Average velocity OC.Total distance 0 D. Displacement
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
The average flow rate in the Niagara River is 6.0 × 106 10. kg/s, and the water drops 50 m over Niagara Falls. If all this energy could be harnessed to generate hydroelec- tric power at 90% efficiency, what would be the electric power output?
Answer:
[tex]2.65\cdot 10^9 W[/tex]
Explanation:
The power produced by the water falling down the Falls is
[tex]P=\frac{E}{t}=\frac{mgh}{t}[/tex]
where
E = mgh is the potential energy of the water, with m being the mass, g the gravitational acceleration, h the height
t is the time
In this problem we have
[tex]\frac{m}{t}=6.0\cdot 10^6 kg/s[/tex] us the mass flow rate
h = 50 m is the height
g = 9.8 m/s^2 is the acceleration of gravity
Substituting,
[tex]P=(6.0\cdot 10^6 kg/s)(9.8 m/s^2)(50 m)=2.94\cdot 10^9 W[/tex]
And since the efficiency is only 90%, the power output is
[tex]P_{out} = (0.90) (2.94\cdot 10^9 W)=2.65\cdot 10^9 W[/tex]
The weights of bags filled by a machine are normally distributed with a standard deviation of 0.05 kilograms and a mean that can be set by the operator. At what level should the mean weight be set if it required that only 1% of the bags weigh less than 9.5 kilograms? Round the answer to 2 decimal places.
Answer:
9.62 kg
Explanation:
From a z-score table, P(z<-2.33) ≈ 0.01. So 9.5 should be 2.33 standard deviations below the mean.
z = (x − μ) / σ
-2.33 = (9.5 − μ) / 0.05
-0.1165 = 9.5 − μ
μ = 9.6165
Rounding to 2 decimal places, the mean should be set to 9.62 kg.
An object with initial temperature 130 ∘ F is submerged in large tank of water whose temperature is 50 ∘ F . Find a formula for F ( t ) , the temperature of the object after t minutes, if the cooling constant is k = − 0.2 . Remember Newton's Law of Cooling (the rate of change of temperature with respect to time is equal to k times the difference between the temperature of the object and the surrounding temperature) ! :)
Answer:
T = 50 + 80e^(-0.2t)
Explanation:
Newton's law of cooling says the rate of change of temperature with respect to time is proportional to the temperature difference:
dT/dt = k (T − Tₐ)
Separating the variables and integrating:
dT / (T − Tₐ) = k dt
ln (T − Tₐ) = kt + C
T − Tₐ = Ce^(kt)
T = Tₐ + Ce^(kt)
Given that Tₐ = 50 and k = -0.2:
T = 50 + Ce^(-0.2t)
At t = 0, T = 130.
130 = 50 + Ce^(0)
130 = 50 + C
C = 80
Therefore:
T = 50 + 80e^(-0.2t)
The formula for the object cooling is [tex]T(t) = 50 + 80\cdot e^{-0.2\cdot t}[/tex], where [tex]t[/tex] is in minutes.
The object is cooled by heat mechanism of Convection, Convection is a Heat Transfer mechanism in which is a solid object is cooled due to a fluid in motion and is described by the Newton's Law of Cooling, whose Differential Equation is:
[tex]\frac{dT}{dt} = -r\cdot (T-T_{m})[/tex] (1)
Where:
[tex]T[/tex] - Temperature of the solid, in degrees Fahrenheit.
[tex]r[/tex] - Cooling rate, in [tex]\frac{1}{min}[/tex].
[tex]T_{m}[/tex] - Water temperature, in degrees Fahrenheit.
The solution of this Differential Equation is:
[tex]T(t) = T_{m} + (T_{o}-T_{m})\cdot e^{-r\cdot t}[/tex] (2)
Where [tex]T_{o}[/tex] is the initial temperature of the solid, in degrees Fahrenheit.
If we know that [tex]T_{m} = 50\,^{\circ}F[/tex], [tex]T_{o} = 130\,^{\circ}F[/tex] and [tex]r = 0.2[/tex], then the formula for the object cooling is:
[tex]T(t) = 50 + 80\cdot e^{-0.2\cdot t}[/tex]
The formula for the object cooling is [tex]T(t) = 50 + 80\cdot e^{-0.2\cdot t}[/tex], where [tex]t[/tex] is in minutes.
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Suppose that your mass is 59.1 kg, and you are standing on a scale fastened to the floor of an elevator. The scale measures force and is calibrated in newtons. What does the scale read when the elevator is rising and its speed is decreasing at a rate of 6.87 m/s2? Use 9.80 m/s2 for acceleration due to gravity.
Answer:
173.2 Newtons
Explanation:
Since the scale is a reading of the normal force, the force of gravity and the normal force are in opposite directions, which means the forces can be subtracted from each other. In this situation, we can use Newtons 2nd Law: Net Force =(mass)( acceleration). The net force in this situation would be the Force of Gravity minus the Normal Force. So the equation you will begin with is Fg(which is mg) - Normal Force = ma. We want the normal Force, so our new equation derived for Fnormal will be mg - ma = Fnormal.
The force that a scale in a rising and decelerating elevator reads can be calculated with the equation F= m*(g-a). Given that the mass, acceleration due to gravity, and elevator deceleration are 59.1 kg, 9.80 m/s2, and 6.87 m/s2 respectively, the force would be 172.8 Newtons.
Explanation:The subject of this question is Physics, and it seeks to investigate the reading on a scale in an elevator when it is rising and decelerating. The force that the scale reads can be represented by the equation F= m*(g-a), where 'F' is the force, 'm' is the mass, 'g' is acceleration due to gravity, and 'a' is the rate at which the elevator is decelerating. In your case:
mass (m) = 59.1 kggravitational acceleration (g) = 9.80 m/s2deceleration of the elevator (a) = 6.87 m/s2So, the force the scale reads is F = 59.1 kg * (9.80 m/s2 - 6.87 m/s2). This is equal to 172.8 Newtons.
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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?
The work by tension is ¹¹/₁₀ Mgℓ
[tex]\texttt{ }[/tex]
Further explanationComplete 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]
Learn moreImpacts of Gravity : https://brainly.com/question/5330244Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454The Acceleration Due To Gravity : https://brainly.com/question/4189441Newton's Law of Motion: https://brainly.com/question/10431582Example of Newton's Law: https://brainly.com/question/498822[tex]\texttt{ }[/tex]
Answer detailsGrade: 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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Two capacitors are connected in series and the combination is charged to 120V. There's 90.0V across one capacitor, whose capacitance is 0.28?F What is the capacitance of the second capactior?
Answer:
0.84μF
Explanation:
Charge is same through both the capacitors since they are in series. Total voltage is the sum of the voltages of the individual capacitors.. So voltage across the 2nd capacitor is 120- 90 =30 V.
Charge across first capacitor is Q = C₁V₁ = 90 x0.28 = 25.2μC
Therefore capacitance of 2nd capacitor =
C₂ = Q÷V₂ = 25.2÷30 = 0.84 μF
If Vx = 7.00 units and Vy = -7.60 units, determine the magnitude of V⃗ .
Determine the direction of V⃗ .
|V| = 10.33 units and the direction θ = -47.35° or 312.65°.
Given the x and y components of a vector, we can calculate the magnitude and direction from these components.
Applying the Pythagorean theorem we have that the magnitude of the vector is:
|V| = [tex]\sqrt{Vx^{2}+Vy^{2} }[/tex]
|V| = [tex]\sqrt{(7.00units)^{2}+(-7.60units)^{2}} = \sqrt{106units^{2}} = 10.33units[/tex]
The expression for the direction of a vector comes from the definition of the tangent of an angle:
tan θ = [tex]\frac{Vy}{Vx}[/tex] ------> θ = arc tan [tex]\frac{Vy}{Vx}[/tex]
θ = arc tan [tex]\frac{-7.60units}{7.00units}[/tex]
θ = -47.35° or 312.65°
Final answer:
The magnitude of vector V is found using the Pythagorean theorem, and the direction of V is determined using the arctan function with the known x and y components of V.
Explanation:
To find the magnitude of V, we use the Pythagorean theorem to combine the orthogonal components Vx and Vy. This is given by the equation V = √(Vx² + Vy²). Substituting the given values, we get V = √(7.00² + (-7.60)²) units.
After calculating the magnitude, we determine the direction of V by finding the angle θ relative to the x-axis, using the inverse tangent function (arctan). The formula is θ = tan⁻¹ (Vy/Vx). Since Vy is negative, the vector is in the 4th quadrant, with an angle measured clockwise from the positive x-axis.
A medium-sized pear provides about 102 Calories of energy. HINT (a) Convert 102 Cal to joules. J (b) Suppose that amount of energy is transformed into kinetic energy of a 2.03 kg object initially at rest. Calculate the final speed of the object (in m/s). m/s (c) If that same amount of energy is added to 3.79 kg (about 1 gal) of water at 23.7°C, what is the water's final temperature (in °C)? The specific heat of water is c = 4186 J (kg · °C) . °C
Answer:
(a) 426.8 J
(b) 20.5 m/s
(c) 23.727 °C
Explanation:
(a)
E = Energy of medium-sized pear = 102 cal
we know that , 1 cal = 4.184 J
hence
E = 102 (4.184 J)
E = 426.8 J
(b)
KE = kinetic energy of the object = E = 426.8 J
m = mass of the object = 2.03 kg
v = speed of the object = ?
Kinetic energy of the object is given as
KE = (0.5) m v²
inserting the values
426.8 = (0.5) (2.03) v²
v = 20.5 m/s
(c)
Q = Amount of heat added to water = E = 426.8 J
m = mass of water = 3.79 kg
c = specific heat of water = 4186 J/(Kg °C)
T₀ = initial temperature = 23.7 °C
T = Final temperature = ?
Using the equation
Q = m c (T - T₀)
426.8 = (3.79) (4186) (T - 23.7)
T = 23.727 °C
We first convert Calories to Joules and get 426,768 J. Then, using the kinetic energy equation and the given mass, we find the final speed of the object to be approximately 460 m/s. Using the heat formula to calculate the temperature change, we find that the final temperature of water would be 49.4°C.
Explanation:Let's tackle these parts one by one:
(a) Converting Calories to Joules: The conversion factor is given as 1 Calorie = 4184 Joules (noting that a Calorie in food is actually a kilocalorie). When we multiply 102 Cal by this conversion factor, we get 102 Cal x 4184 J/Cal = 426,768 J.
(b) Kinetic Energy: The formula for kinetic energy is KE = 1/2 mv², where m is mass and v is speed. If all the energy from the pear is transformed to kinetic energy, we can set this equal to 426,768 J, then solve for v. We get: √((2*426768 J) / 2.03 kg) ≈ 460 m/s as the final speed of the object.
(c) Temperature Change in Water: The formula to calculate temperature change is Q = mcΔT, where Q is the heat added (in Joules), m is the mass of the water, c is the specific heat capacity, and ΔT is the temperature change. We know that Q = 426,768 J, m = 3.79 kg, and c = 4186 J/kg°C. We can rearrange the formula to solve for ΔT: ΔT = Q / (m*c). So, ΔT = 426768 J / (3.79 kg * 4186 J/Kg°C) ≈ 25.7°C. So, the final temperature would be the initial temperature plus this change, or 23.7°C + 25.7°C = 49.4°C.
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A person standing 1.20 m from a portable speaker hears its sound at an intensity of 5.50 ✕ 10−3 W/m2. HINT (a) Find the corresponding decibel level. dB (b) Find the sound intensity (in W/m2) at a distance of 36.0 m, assuming the sound propagates as a spherical wave. W/m2 (c) Find the decibel level at a distance of 36.0 m. dB
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.
Explanation:(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].
Explain how adding more copper windings to an electromagnet can increase the magnetic field.
Answer:
This is because copper is a conductor, and adding more copper spreads electricity more, and thus the magnetic field.
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A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance of each bulb does not vary with current. (Note: This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a 25-W, 120-V light bulb dissipates 25 W when connected to a 120-V line.) A. Find the current through the bulbs. B. Find the power dissipated in the 60 W bulb. C. Find the power dissipated in the 200 W bulb. D. One bulb burns out very quickly. Which one? 60-W bulb, 200-W bulb.
A. 0.77 A
Using the relationship:
[tex]P=\frac{V^2}{R}[/tex]
where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.
For the first light bulb, P = 60 W and V = 120 V, so the resistance is
[tex]R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega[/tex]
For the second light bulb, P = 200 W and V = 120 V, so the resistance is
[tex]R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega[/tex]
The two light bulbs are connected in series, so their equivalent resistance is
[tex]R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega[/tex]
The two light bulbs are connected to a voltage of
V = 240 V
So we can find the current through the two bulbs by using Ohm's law:
[tex]I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A[/tex]
B. 142.3 W
The power dissipated in the first bulb is given by:
[tex]P_1=I^2 R_1[/tex]
where
I = 0.77 A is the current
[tex]R_1 = 240 \Omega[/tex] is the resistance of the bulb
Substituting numbers, we get
[tex]P_1 = (0.77 A)^2 (240 \Omega)=142.3 W[/tex]
C. 42.7 W
The power dissipated in the second bulb is given by:
[tex]P_2=I^2 R_2[/tex]
where
I = 0.77 A is the current
[tex]R_2 = 72 \Omega[/tex] is the resistance of the bulb
Substituting numbers, we get
[tex]P_2 = (0.77 A)^2 (72 \Omega)=42.7 W[/tex]
D. The 60-W bulb burns out very quickly
The power dissipated by the resistance of each light bulb is equal to:
[tex]P=\frac{E}{t}[/tex]
where
E is the amount of energy dissipated
t is the time interval
From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.
Final answer:
To solve for the current and power dissipated in each bulb, use the power rating to find their resistances, then apply Ohm's Law to calculate current, and subsequently, determine power dissipated in each. Finally, assess which bulb is likely to burn out first by the dissipated power relative to each bulb's rating.
Explanation:
The question involves finding the current through, and the power dissipated in, a series circuit with two light bulbs with different power ratings. It also inquires which bulb will burn out quickly.
A. Current through the bulbs
Using the power rating (P) and voltage (V) for each bulb, we can find their resistances (R) using the formula P = V^2/R. From there, we can find the total resistance in the series circuit and calculate the current (I) using Ohm's Law, I = V/R.
B. & C. Power dissipated in each bulb
Once the current is known, we can determine the power dissipated (P) in each bulb with the formula P = I^2 * R.
D. Which bulb burns out quickly?
We can infer which bulb burns out based on the power dissipated in each bulb compared to their rated power.
A small object has a mass of 3.0 × 10-3 kg and a charge of -32C. It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of 2.4 × 103 m/s2 in the direction of the +x axis. Determine the electric field, including sign, relative to the +x axis.
The electric field is [tex]-0.225N/C[/tex] towards the negative x-axis.
The electric force (F) acting on a charged object in an electric field (E) is given by the following formula:
[tex]F=qE[/tex]
Here, q is the charge of the object and E is the electric field.
Given:
Acceleration, [tex]a=2.4 \times 10^3 m/s^2[/tex]
Mass, [tex]m=3.0 \times 10^{-3} kg[/tex]
Charge, [tex]q=-32C[/tex]
The force on the charge is computed as:
[tex]qE=ma\\E=\frac{3.0 \times 10^{-3} \times 2.4 \times 10^3}{-32}\\E=-0.225N/C[/tex]
The electric field is towards the negative x-axis.
Therefore, the electric field is [tex]-0.225N/C[/tex] towards the negative x-axis.
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The electric field that causes the object with a mass of 3.0 × 10⁻³kg and a charge of -32C to accelerate at 2.4 × 10³m/s^2 is -7.5 × 10² N/C, and it is in the +x axis direction.
Explanation:The student's question involves determining the electric field that causes an object to experience a specific acceleration. The object has a mass of 3.0 × 10⁻³ kg and a charge of -32 C. To find the electric field E, we can use Newton's second law of motion and the definition of the electric force: F = ma = qE, where F is the force, m is the mass, a is the acceleration, q is the charge, and E is the electric field.
To calculate the electric field, we set F equal to the product of mass and acceleration and then solve for E:
E = F / q
= (m × a) / q
= (3.0 × 10^⁻³ kg × 2.4 × 10³ m/s²) / (-32 C)
= -7.5 × 10²N/C
The negative sign indicates that the electric field's direction is opposite to the charge's motion. Since the charge is negative and it accelerates in the direction of the +x axis, the electric field must be in the +x axis direction.
An electron moves in a straight path with a velocity of +500,000 m/s. It undergoes constant acceleration of +500,000,000,000 meters per second per second. How far will the electron have traveled when it reaches a velocity of +675,000 m/s? A. +0.000000175 m B. +0.00000035 m C. +0.21 m D. +0.41 m
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 child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm is applied to the merry-go-round with a moment of inertia of 50.0 kg m2, what is the angular acceleration in rad/s2?
Answer:
[tex]-0.25 rad/s^2[/tex]
Explanation:
The equivalent of Newton's second law for rotational motions is:
[tex]\tau = I \alpha[/tex]
where
[tex]\tau[/tex] is the net torque applied to the object
I is the moment of inertia
[tex]\alpha[/tex] is the angular acceleration
In this problem we have:
[tex]\tau = -12.5 Nm[/tex] (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)
[tex]I=50.0 kg m^2[/tex] is the moment of inertia
Solving for [tex]\alpha[/tex], we find the angular acceleration:
[tex]\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2[/tex]
A car that is traveling in a straight line at 40 km/h can brake to a stop within 20 m. If the same car is traveling at 120 km/h what would be its stopping distance in this case? Assume the braking force is the same in both cases and ignore air resistance.
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
If the heat necessary to warm 565.0 g of water from a temperature of T1 = 22.0 °C to T2 = 80.0 °C were somehow converted to translational kinetic energy of this amount of water, what would be the speed of this water?
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
A 35-N bucket of water is lifted vertically 3.0m and then returned to its original position. How much work did gravity do on the bucket during this process? (a)180J (b) 90J (c) 45J (d) 0J (e) 900J
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 211-Ω and a 221-Ω resistor are connected in series across an unspecified power supply. If the current through the 211-Ω resistor is 0.22 A, what is the exact current (in A) through the 221-Ω resistor? Do not include units with your answer.
Answer:
0.22 (A)
Explanation:
Two resistors are connected in series when they are connected in the same branch of the circuit. When this occurs, we have the following:
- The current flowing through each resistor is the same (1)
- The total voltage of the circuit is the sum of the voltage drops across each resistor
- The total resistance of the circuit is equal to the sum of the individual resistances:
[tex]R=R_1 + R_2[/tex]
So, given statement (1), in this case the current flowing through each resistor is the same, so the current flowing throught the 221-Ω resistor is also 0.22 A.
A spherical snowball is melting in such a way that its radius is decreasing at a rate of 0.1 cm/min. At what rate is the volume of the snowball decreasing when the radius is 8 cm. (Note the answer is a positive number).
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.
Explanation: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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