Answer:
- 1 m/s, 20 m
Explanation:
u = 9 m/s, a = - 2 m/s^2, t = 5 sec
Let s be the displacement and v be the velocity after 5 seconds
Use first equation of motion.
v = u + a t
v = 9 - 2 x 5 = 9 - 10 = - 1 m/s
Use second equation of motion
s = u t + 1/2 a t^2
s = 9 x 5 - 1/2 x 2 x 5 x 5
s = 45 - 25 = 20 m
Final answer:
The runner's displacement in the first 5 seconds after crossing the finish line is 22.5 meters. Her velocity 5 seconds after crossing the finish line is 1 m/s in the positive direction. These calculations are based on the formulas for displacement and velocity under constant acceleration.
Explanation:
As a runner crosses the finish line, she decelerates from a velocity of 9 m/s at a rate of 2 m/s2. To find her displacement during the first 5 seconds after crossing the finish line, we use the formula for displacement under constant acceleration, δx = v0t + ½at2, where v0 = 9 m/s, a = -2 m/s2 (deceleration means acceleration is in the opposite direction to velocity), and t = 5 s. Substituting these values gives us a displacement of 22.5 meters. To find her velocity 5 seconds after crossing the finish line, we use the formula v = v0 + at, which yields a final velocity of 1 m/s in the positive direction.
This result makes sense as the runner is slowing down but not yet stopped 5 seconds after crossing the finish line. The calculated displacement of 22.5 meters is the total distance covered during these 5 seconds of deceleration.
Calculate the speed of an electron that has fallen through a potential difference of
(a) 125 volts and
(b) 125 megavolts.
Explanation:
We need to find the speed of an electron that has fallen through a potential difference of 125 volts. It can be calculated using the De-broglie hypothesis as :
(a) V = 125 volts
[tex]\dfrac{1}{2}mv^2=qV[/tex]
Where
v = speed of electron
V is potential difference
[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 125\ V}{9.1\times 10^{-31}}}[/tex]
v = 6629935.44 m/s
[tex]v=6.62\times 10^6\ m/s[/tex]
(b) V = 125 megavolts
[tex]V=1.25\times 10^8\ V[/tex]
[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.25\times 10^8\ V}{9.1\times 10^{-31}}}[/tex]
[tex]v=6.62\times 10^9\ m/s[/tex]
Hence, this is the required solution.
Taking the speed of light in vacuum to be 3.000 x 10^8 m/s, find the speed of light in: a. air b. diamond c. crown glass d. water Data: nair =1.0003; ndiamond = 2.420; nwater = 1.340 ncrown glass = 1.500
Explanation:
The speed of light in vacuum is, c = 3 × 10⁸ m/s
We have to find the speed of light :
(a) In air :
[tex]n_1_{air}=1.0003[/tex]
The equation of refractive index is given as :
[tex]n_1=\dfrac{c}{v_1}[/tex]
[tex]v_1=\dfrac{c}{n_1}[/tex]
[tex]v_1=\dfrac{3\times 10^8\ m/s}{1.0003}[/tex]
[tex]v_1=299910026.9\ m/s[/tex]
[tex]v_1=2.99\times 10^8\ m/s[/tex]
(b) In diamond :
[tex]n_2_{diamond}=2.42[/tex]
The equation of refractive index is given as :
[tex]n_2=\dfrac{c}{v_2}[/tex]
[tex]v_2=\dfrac{c}{n_2}[/tex]
[tex]v_2=\dfrac{3\times 10^8\ m/s}{2.42}[/tex]
[tex]v_2=123966942.1\ m/s[/tex]
[tex]v_2=1.23\times 10^8\ m/s[/tex]
(c) In crown glass :
[tex]n_3_{glass}=1.5[/tex]
The equation of refractive index is given as :
[tex]n_3=\dfrac{c}{v_3}[/tex]
[tex]v_3=\dfrac{c}{n_3}[/tex]
[tex]v_3=\dfrac{3\times 10^8\ m/s}{1.5}[/tex]
[tex]v_3=200000000\ m/s[/tex]
[tex]v_3=2\times 10^8\ m/s[/tex]
(4) In water :
[tex]n_4_{glass}=1.34[/tex]
The equation of refractive index is given as :
[tex]n_4=\dfrac{c}{v_4}[/tex]
[tex]v_4=\dfrac{c}{n_4}[/tex]
[tex]v_4=\dfrac{3\times 10^8\ m/s}{1.34}[/tex]
[tex]v_4=223880597.01\ m/s[/tex]
[tex]v_4=2.23\times 10^8\ m/s[/tex]
Hence, this is the required solution.
The total solar irradiance (TSI) at Earth orbit is 1400 watts/m2 . Assuming this value can be represented as a single Poynting flux. Find the corresponding flux at the solar visible-light surface. Explain your methods.
(Distance between Sun & Earth is 150 million km)
Answer:
3.958 × 10²⁶ watt
Explanation:
Given:
Distance between earth and sun, [tex]R_e[/tex] = 150 ×10⁸ m
Total solar irradiance at earth orbit = 1400 watt/m²
Now,
Area irradiated ([tex]A_e[/tex]) will be = [tex]4\pi R_e^2[/tex]
⇒ [tex]A_e[/tex] = [tex]4\pi \times (150\times 10^{8})^2[/tex]
⇒ [tex]A_e[/tex] = [tex]2.827\times 10^{23}m^2[/tex]
Therefore, the flux = Total solar irradiance at earth orbit × [tex]A_e[/tex]
the flux = [tex]1400watt/m^2\times 2.827\times 10^{23}m^2[/tex]
⇒the flux = 3.958 × 10²⁶ watt
13. A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V and the secondary is connected to a floodlight that draws 5A, what is the power output? Please show ALL of your work.
Answer:
The output power is 2 kW
Explanation:
It is given that,
Number of turns in primary coil, [tex]N_p=250[/tex]
Number of turns in secondary coil, [tex]N_s=500[/tex]
Voltage of primary coil, [tex]V_p=200\ V[/tex]
Current drawn from secondary coil, [tex]I_s=5\ A[/tex]
We need to find the power output. It is equal to the product of voltage and current. Firstly, we will find the voltage of secondary coil as :
[tex]\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}[/tex]
[tex]\dfrac{250}{500}=\dfrac{200}{V_s}[/tex]
[tex]V_s=400\ V[/tex]
So, the power output is :
[tex]P_s=V_s\times I_s[/tex]
[tex]P_s=400\ V\times 5\ A[/tex]
[tex]P_s=2000\ watts[/tex]
or
[tex]P_s=2\ kW[/tex]
So, the output power is 2 kW. Hence, this is the required solution.
A batter hits a 0.140-kg baseball that was approaching him at 50.0 m/s and, as a result, the ball leaves the bat at 35.0 m/s in the direction of the pitcher. What is the magnitude of the impulse delivered to the baseball?
Answer:
Impulse, J = 2.1 kg-m/s
Explanation:
Given that,
Mass of baseball, m = 0.14 kg
It was approaching him at 50.0 m/s and, as a result, the ball leaves the bat at 35.0 m/s in the direction of the pitcher. We need to find the magnitude of Impulse delivered to the baseball.
The change in momentum is equal to the Impulse imparted to the ball i.e.
[tex]J=m(v-u)[/tex]
[tex]J=0.14(-35-50)[/tex]
J = -2.1 kg-m/s
So, the Impulse delivered to the baseball is 2.1 kg-m/s
Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of Gaussian surface B. The flux of electric field through Gaussian surface A is A) nine times larger than the flux of electric field through Gaussian surface B. B) three times smaller than the flux of electric field through Gaussian surface B. C) unrelated to the flux of electric field through Gaussian surface B. D) equal to the flux of electric field through Gaussian surface B. E) three times larger than the flux of electric field through Gaussian surface B
The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Therefore, the flux through surface A is three times larger than the flux through surface B.
The flux of the electric field through a Gaussian surface depends on the charge enclosed by the surface and the area of the surface. In this scenario, Gaussian surfaces A and B enclose the same positive charge +Q, but the area of surface A is three times larger than that of surface B. Since the flux of electric field is proportional to the area of the surface, the flux through surface A is three times larger than the flux through surface B. Therefore, the correct answer is E) three times larger than the flux of electric field through Gaussian surface B.
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Final answer:
The correct answer is D) equal to the flux of the electric field through Gaussian surface B. The electric flux depends on the charge enclosed by the Gaussian surface, not on its size or shape, according to Gauss's law.
Explanation:
The question asks whether Gaussian surfaces A and B, enclosing the same positive charge +Q, with surface A having three times the area of surface B, have different electric fluxes. According to Gauss's law, the electric flux (Φ) through a closed surface is proportional to the charge enclosed (Φ = Q/ε0). Since both surfaces enclose the same charge, the flux through each surface must be the same, regardless of their respective areas.
Thus, the correct answer is D) equal to the flux of the electric field through Gaussian surface B because the flux depends only on the amount of enclosed charge, not on the size or shape of the Gaussian surface.
When tuning a guitar, by comparing the frequency of a string that is struck against a standard sound source (of known frequency), what does the one adjusting the tension in the string listen to?
wo kids are on a seesaw. The one on the left has a mass of 75 kg and is sitting 1.5 m from the pivot point. The one on the right has a mass of 60 kg and is sitting 1.8 m from the pivot point. What is the net torque on the system?
Answer:
4.5 Nm (Anticlockwise)
Explanation:
Let the 75 kg kid is sitting at the left end and the 60 kg kid is sitting on the right end.
Anticlockwise Torque = 75 x 1.5 = 112.5 Nm
clockwise Torque = 60 x 1.8 = 108 Nm
Net torque = Anticlockwise torque - clockwise torque
Net Torque = 112.5 - 108 = 4.5 Nm (Anticlockwise)
Rearrange each of the following combinations of units to obtain a units that involves Joule ) Show kg m s intermediate steps, ( ).
Answer:
Explanation:
Joule is SI unit of work
Work = force x distance
Work = mass x acceleration x distance
Unit of mass is kg
Unit of acceleration is m/s^2
Unit of distance is m
So, unit of work = kg x m x m /s^2
So, Joule = kg m^2 / s^2
Calculate the wavelength of 100-MHz microwaves in muscle and in fat.
Answer:
Wavelength of microwaves is 3 m.
Explanation:
In this question, we need to find the wavelength of 100 MHz microwaves in muscle and in fat.
Frequency of the microwaves, [tex]\nu=100\ MHz=100\times 10^6\ Hz=10^8\ Hz[/tex]
The relation between the frequency and the wavelength is given by :
[tex]c=\nu\times \lambda[/tex]
[tex]\lambda=\dfrac{c}{\nu}[/tex]
Where
c is the sped of light
[tex]\lambda=\dfrac{3\times 10^8\ m/s}{10^8\ Hz}[/tex]
[tex]\lambda=3\ m[/tex]
So, the wavelength of microwaves is 3 m. Hence, this is the required solution.
lectric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an electric potential difference of 12 V is placed across its leads. What is the resistance of the heater wire?
Answer:
The resistance of the heater wire is of R= 0.68 Ω.
Explanation:
1 cal/s = 4.184 W
P= 50 cal/s = 209.2 W
V= 12V
P= V* I
I= P/V
I= 17.43 A
P= I² * R
R= P / I²
R= 0.68 Ω
How far would you go if you traveled at a speed of 30 miles per hour for 3 hours?
Answer:
90 miles
Explanation:
speed = 30 miles per hour, time = 3 hour
The formula for the speed is given by
speed = distance / time
Distance = speed x time
distance = 30 x 3 = 90 miles
Answer:
90 Miles
Explanation:
This is more simple than you thought...
The equation is 30 miles per hour,in 3 hours you would travel (30*3)=90 miles.
Formula used Distance = Speed x Time
Speed = 30 mph
Time = 3 hours
So, Distance = Speed * Time = 30 * 3 = 90 miles.
So all in all, the answer is 90 miles
In a car lift, compressed air exerts a force on a piston with a radius of 2.62 cm. This pressure is transmitted to a second piston with a radius of 10.8 cm. a) How large a force must the compressed air exert to lift a 1.44 × 104 N car
Answer:
847.45 N
Explanation:
F₁=force of exerted by smaller piston
F₂=force of exerted by larger piston=1.44×10⁴ N
A₁=Area of smaller piston= 2.62 cm =0.0265 m
A₂=Area of larger piston= 10.8 cm =0.108 m
Pressure exerted by both the pistons will be equal
[tex]P_1=P_2\\\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=\frac{F_2}{A_2} A_1\\\Rightarrow F_1=\frac{14400}{\pi\times 0.108^2}\pi\times 0.0262^2\\\Rightarrow F_1=847.45\ N[/tex]
Hence, force exerted to lift a 14400 N car is 847.45 N
After an afternoon party, a small cooler full of ice is dumped onto the hot ground and melts. If the cooler contained 5.50 kg of ice and the temperature of the ground was 43.0 °C, calculate the energy that is required to melt all the ice at 0 °C. The heat of fusion for water is 80.0 cal/g.
Answer:
[tex]Q = 4.40 \times 10^5 Cal[/tex]
Explanation:
Here we know that initial temperature of ice is given as
[tex]T = 0^o C[/tex]
now the latent heat of ice is given as
[tex]L = 80 Cal/g[/tex]
now we also know that the mass of ice is
[tex]m = 5.50 kg[/tex]
so here we know that heat required to change the phase of the ice is given as
[tex]Q = mL[/tex]
[tex]Q = (5.50 \times 10^3)(80)[/tex]
[tex]Q = 4.40 \times 10^5 Cal[/tex]
The total energy required to melt all the 5.50 kg (or 5500g) of ice is 440,000 calories. This is calculated using the given heat of fusion (Lf) value of 80 cal/g for water and the heat exchange formula, Q=mLf.
Explanation:To calculate the amount of energy required to melt all the ice, we need to use the formula for heat exchange: Q = mLf. Where Q is the heat required, m is the mass, and Lf is the heat of fusion. Given that Lf is 80 cal/g for water, the mass m is 5.50 kg (or 5500g), and you are trying to find Q, you can simply replace the known quantities into the equation:
Q = (5500g) * (80cal/g)
So, the total energy required to melt all the ice is 440,000 calories. This heat is absorbed by the ice, providing the energy required to break the intermolecular bonds in the ice and facilitate the phase transition from solid ice to liquid water. The energy required for a phase change like this is significant, explaining why ice can take a while to melt even on a hot summer day.
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A.
Calculate the specific weight, density and specific gravity of one litre of a liquid, which weighs 7N.
Select one:
1. 7000 N/m3, 713.5 kg/m3, 0.7135
2. 700 N/m3, 71.35 kg/m3, 0.07135
3. 70 N/m3, 7.135 kg/m3, 0.007135
4. None of the above.
B.
The multiplying factor for converting one stoke into m2/s is
Select one:
1. 102
2. 104
3. 10-2
4. 10-4
Answer:
A) Option 1 is the correct answer.
B) Option 4 is the correct answer.
Explanation:
A) Weight of liquid = 7 N
Volume of liquid = 1 L = 0.001 m³
Specific weight = [tex]\frac{7}{0.001}=7000N/m^3[/tex]
Density = [tex]\frac{7000}{9.81}=713.5kg/m^3[/tex]
Specific gravity = [tex]\frac{713.5}{1000}=0.7135[/tex]
Option 1 is the correct answer.
B) The Stokes(St) is the cgs physical unit for kinematic viscosity, named after George Gabriel Stokes.
We have
1 St = 10⁻⁴ m²/s
Option 4 is the correct answer.
After evaluating all the options we have:
A. From all of the options of specific weight, density, and specific gravity of 1 liter of liquid, the correct is option 1: 7000 N/m³, 713.5 kg/m³, 0.7135.
B. The correct option of the multiplying factor for converting one stoke into m²/s is 4: 10⁻⁴.
A. Let's calculate the specific weight, density, and specific gravity of 1 L of the liquid that weighs 7 N.
The specific weight is given by:[tex] \gamma = dg [/tex] (1)
Where:
γ: is the specific weight
d: is the density
g: is the gravity = 9.81 m/s²
We need to find the density which is:
[tex] d = \frac{m}{V} [/tex] (2)
Where:
m: is the mass
V: is the volume = 1 L = 0.001 m³
The mass can be found knowing that the liquid weighs (W) 7 N, so:
[tex] W = mg [/tex]
[tex] m = \frac{W}{g} [/tex] (3)
By entering equations (3) and (2) into (1) we have:
[tex] \gamma = dg = \frac{mg}{V} = \frac{W}{V} = \frac{7 N}{0.001 m^{3}} = 7000 N/m^{3} [/tex]
Hence, the specific weight is 7000 N/m³.
The density can be found as follows:[tex] d = \frac{m}{V} = \frac{W}{gV} = \frac{7 N}{9.81 m/s^{2}*0.001 m^{3}} = 713.5 kg/m^{3} [/tex]
Then, the density is 713.5 kg/m³.
The specific gravity (SG) of a liquid can be calculated with the following equation:[tex] SG = \frac{d}{d_{H_{2}O}} = \frac{713.5 kg/m^{3}}{1000 kg/m^{3}} = 0.7135 [/tex]
Hence, the specific gravity is 0.7135.
Therefore, the correct option is 1: 7000 N/m³, 713.5 kg/m³, 0.7135.
B. A Stokes is a measurement unit of kinematic viscosity.
One m²/s is equal to 10⁴ stokes, so to convert 1 stokes to m²/s we need to multiply for 10⁻⁴.
Hence, the correct option is 4: 10⁻⁴.
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The diameter of a 12-gauge copper wire is about 0.790 mm. If the drift velocity of the electrons is 3.25 mm/s what is the electron current in the wire? The number of electron carriers in 1.0 cm3 of copper is 8.5 × 1022.
Answer:
21.6 A
Explanation:
n = number density of free electrons in copper = 8.5 x 10²² cm⁻³ = 8.5 x 10²⁸ m⁻³
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
d = diameter of copper wire = 0.790 mm = 0.790 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.790 x 10⁻³)²
A = 4.89 x 10⁻⁷ m²
v = drift speed = 3.25 mm/s = 3.25 x 10⁻³ m /s
the electric current is given as
i = n e A v
i = (8.5 x 10²⁸) (1.6 x 10⁻¹⁹) (4.89 x 10⁻⁷ ) (3.25 x 10⁻³)
i = 21.6 A
Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. What is the ratio of the power drawn from the battery for the two cases (i.e. P series: P parallel)? Note: The battery supplies voltage V in both cases. A) 1:N B) 1:N2 C) N:1 D) N2:1
Answer:
option (b)
Explanation:
Let the resistance of each resistor is R.
In series combination,
The effective resistance is Rs.
rs = r + R + R + .... + n times = NR
Let V be the source of potential difference.
Power in series
Ps = v^2 / Rs = V^2 / NR ..... (1)
In parallel combination
the effective resistance is Rp
1 / Rp = 1 / R + 1 / R + .... + N times
1 / Rp = N / R
Rp = R / N
Power is parallel
Rp = v^2 / Rp = N V^2 / R ..... (2)
Divide equation (1) by equation (2) we get
Ps / Pp = 1 / N^2
At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values of L and C. larger than R equal to R Which of the following does the quality factor of the circuit depend on? (Select all that apply.)
Answer:
at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all
Explanation:
we know that for series RLC circuit impedance is given by
[tex]Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2[/tex]
but we know that at resonance [tex]X_L=X_C[/tex]
putting [tex]X_L=X_C[/tex] in impedance formula , impedance will become
Z=R so at resonance impedance of series RLC is equal to resistance only
now quality factor of series resonance is given by
[tex]Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}[/tex] so from given expression it is clear that quality factor depends on R L and C
A current of 0.2 A flows through a 3 m long wire that is perpendicular to a 0.3 T magnetic field. What is the magnitude of the force on the wire in units of newtons?
Answer:
Magnetic force, F = 0.18 N
Explanation:
It is given that,
Current flowing in the wire, I = 0.2 A
Length of the wire, L = 3 m
Magnetic field, B = 0.3 T
It is placed perpendicular to the magnetic field. We need to find the magnitude of force on the wire. It is given by :
[tex]F=ILB\ sin\theta[/tex]
[tex]F=0.2\ A\times 3\ m\times 0.3\ T\ sin(90)[/tex]
F = 0.18 N
So, the magnitude of force on the wire is 0.18 N. Hence, this is the required solution.
The 400 kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F= (3200) N, where t is in seconds. If the car has an initial velocity V1= 2m/s at s 0 and t= 0, determine the distance it moves the plane when (a) t 1 s and (b) f-5 s.
Answer:
(a) 110 m/s
(b) 42 m/s
Explanation:
mass, m = 400 kg, F = 3200 N, V1 = 2 m/s,
acceleration, a = Force / mass = 3200 / 400 = 8 m/s^2
(a) Use first equation of motion
v = V1 + a t
v = 2 + 8 x 1 = 10 m/s
(b) Again using first equation of motion
v = V1 + a t
v = 2 + 8 x 5 = 42 m/s
Thus, the velocity of plane after 1 second is 10 m/s and after 5 second the velocity is 42 m/s.
The 400 kg mine car moves a distance of 6 m after one second and 110 m after five seconds given an initial velocity of 2 m/s and a force of 3200 N.
Explanation:The problem describes a physics scenario where a 400 kg mine car is pulled up an incline by a cable and motor. The force on the cable is given as F = 3200N and the initial velocity, V1, is given as 2 m/s. We can calculate the distance the car moves on the plane at different times using the physics equations of motion.
Let's use the equation of motion: s = ut + 1/2 at², where 's' is the distance moved, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time.
Given that the net force is equal to mass times acceleration (F = ma), we can calculate the acceleration, 'a', as F/m. So, a = 3200N/400kg = 8 m/s².
(a) t = 1s: The distance moved is s = 2m/s * 1s + 1/2 * 8 m/s² * (1s)² = 2m + 4m = 6m.(b) t = 5s: The distance moved is s = 2m/s * 5s + 1/2 * 8 m/s² * (5s)² = 10m + 100m = 110m.Learn more about Physics of Motion here:https://brainly.com/question/13966796
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Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 250 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 30 centimeters to 60 centimeters? N-cm
Answer:
112.5 J
Explanation:
F = 250 N, x = 30 cm = 0.3 m
Let the spring constant be K.
By using the Hooke's law
F = k x
250 = k x 0.3
k = 833.3 N / m
xi = 30 cm = 0.3 m, xf 60 cm = 0.6 m
Work done = 1/2 k (xf^2 - xi^2)
Work done = 0.5 x 833.33 x (0.6^2 - 0.3^2)
Work done = 112.5 J
A 3.9 kg block is pushed along a horizontal floor by a force ModifyingAbove Upper F With right-arrow of magnitude 27 N at a downward angle θ = 40°. The coefficient of kinetic friction between the block and the floor is 0.22. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block’s acceleration.
Answer:
a) 12.23 N
b) 2.2 m/s²
Explanation:
m = mass of the block = 3.9 kg
F = applied force = 27 N
θ = angle of the applied force with the horizontal = 40°
μ = Coefficient of kinetic friction = 0.22
[tex]F_{n}[/tex] = normal force
[tex]F_{g}[/tex] = weight of the block = mg
Along the vertical direction, force equation is given as
[tex]F_{n}[/tex] = F Sinθ + [tex]F_{g}[/tex]
[tex]F_{n}[/tex] = F Sinθ + mg
Kinetic frictional force is given as
f = μ [tex]F_{n}[/tex]
f = μ (F Sinθ + mg)
f = (0.22) (27 Sin40 + (3.9)(9.8))
f = 12.23 N
b)
Force equation along the horizontal direction is given as
F Cosθ - f = ma
27 Cos40 - 12.23 = 3.9 a
a = 2.2 m/s²
Capacitors, C1 = 1.0 F and C2 = 1.0 F, are connected in parallel to a 6.0 volt battery (ΔV = 6.0V). If the battery is disconnected and the capacitors are connected to a 33 ohm resistor, how long should it take for the voltage to cross the capacitors to drop to 2.2 volts (36.8% of the original 6.0 volts)?
Answer:
66.2 sec
Explanation:
C₁ = 1.0 F
C₂ = 1.0 F
ΔV = Potential difference across the capacitor = 6.0 V
C = parallel combination of capacitors
Parallel combination of capacitors is given as
C = C₁ + C₂
C = 1.0 + 1.0
C = 2.0 F
R = resistance = 33 Ω
Time constant is given as
T = RC
T = 33 x 2
T = 66 sec
V₀ = initial potential difference across the combination = 6.0 Volts
V = final potential difference = 2.2 volts
Using the equation
[tex]V = V_{o} e^{\frac{-t}{T}}[/tex]
[tex]2.2 = 6 e^{\frac{-t}{66}}[/tex]
t = 66.2 sec
The output of a generator is 440 V at 20 A. It is to be transmitted on a line with resistance of 0.60 Ω. To what voltage must the generator output be stepped up with a transformer if the power loss in transmission is not to exceed 0.010% of the original power?
Answer:
The voltage of the generator is 7.27 kV.
Explanation:
Given that,
Output of generator = 440 V
Current = 20 A
Resistance = 0.60 Ω
Power loss =0.010%
We need to calculate the total power of the generator
Using formula of power
[tex]P=VI[/tex]
Where, V = voltage
I = current
Put the value into the formula
[tex]P=440\times20[/tex]
[tex]P=8800\ W[/tex]
Th power lost on the transmission lines
[tex]P_{L}=0.010\% P[/tex]
[tex]P_{L} = 0.010\%\times8800[/tex]
[tex]P_{L}=0.88\ W[/tex]
The current passing through the transmission line
[tex]I'=\sqrt{\dfrac{P_{L}}{R}}[/tex]
[tex]I'=\sqrt{\dfrac{0.88}{0.60}}[/tex]
[tex]I'=1.211\ A[/tex]
We need to calculate the voltage of the generator
Using formula of voltage
[tex]V_{g}=\dfrac{P}{I'}[/tex]
Put the value into the formula
[tex]V_{g}=\dfrac{8800}{1.211}[/tex]
[tex]V_{g}=7.27\times10^{3}\ V[/tex]
[tex]V_{g}=7.27\ kV[/tex]
Hence, The voltage of the generator is 7.27 kV.
The voltage of the generator output needs to be stepped up with a transformer to limit power losses in transmission.
Explanation:To limit power losses in transmission, the voltage of the generator output needs to be stepped up with a transformer. The formula to calculate power loss is P_loss = I^2 x R. Given that the power loss should not exceed 0.010% of the original power, we can calculate the maximum allowable power loss. By rearranging the formula, we can find the voltage required for the generator output with a transformer.
First, we calculate the original power using P = V x I. Then, we calculate the maximum allowable power loss as a percentage of the original power. Next, we substitute the given values into the formula and solve for the new voltage using the rearranged formula.
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What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body?
Answer:
Weight, w = mg
Mass is an intrinsic property.
Explanation:
Mass is the measure of amount of matter in an object. It is an intrinsic that is unchanging property of a body. Mass cannot be destroyed nor be created.
Weight is the product of mass and acceleration due to gravity.
It changes with value of acceleration due to gravity. That is weight in Earth not equal to weight in moon since the value of acceleration due to gravity is different. So, weight is not an intrinsic property. It is a changing property for a body.
Weight, w = mg, where m is the mass and g is the acceleration due to gravity value.
Mass is an intrinsic and constant property of a body representing its matter content, while weight, the force of gravity on a body, varies with the gravitational environment. Despite common misuse in everyday language, they are distinct concepts in physics.
Explanation:Mass and weight are two different but closely related concepts. Mass is an intrinsic property of a body, representing the amount of matter it contains, and it remains constant regardless of the body's location, be it on Earth, the moon, or in orbit.
On the other hand, weight is the force exerted on a body due to gravity. It's a product of the body's mass and the acceleration due to gravity, represented by the formula 'Weight = Mass x Gravity'. Unlike mass, weight changes with the gravitational environment. For example, a person's weight on the moon is only one-sixth of their weight on Earth due to the moon's lower gravity, while their mass remains the same.
In everyday language, mass and weight are often used interchangeably, but in the field of physics, it is important to distinguish between them. For example, when we refer to our 'weight' in kilograms, we're technically referring to our mass. The proper unit of weight, consistent with its definition as a force, is the newton in the International System of Units (SI).
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A 0.42 kg football is thrown with a velocity of 17 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.017 S. What is the force exerted on the receiver? Answer in units of N
Answer:
420 N
Explanation:
m = 0.42 Kg, u = 17 m/s, v = 0 m/s, t = 0.017 s
By first law of Newtons' laws of motion, the rate of change in momentum is force, F = m (v - u) / t
F = 0.42 x ( 0 - 17) / 0.017
F = - 420 N
Negative sign shows hat the force is resistive that means the ball finally comes to rest.
Two objects that may be considered point masses are initially separated by a distance d. The separation distance is then decreased to d/4. How does the gravitational force between these two objects change as a result of the decrease?
Answer:
Increased by 16 times
Explanation:
F = Gravitational force between two bodies
G = Gravitational constant = 6.67408 × 10⁻¹¹ m³/kg s²
m₁ = Mass of one body
m₂ = Mass of other body
d = distance between the two bodies
[tex]F=\frac{Gm_1m_2}{d^2}\\ F=\frac{1}{d^2}\quad \text {(as G and masses are constant)}[/tex]
[tex]F_{new}=\frac{1}{\left (\frac{d}{4}\right )^2}\\\Rightarrow F_{new}=\frac{1}{\frac{d^2}{16}}\\\Rightarrow F_{new}={16}\times \frac{1}{d^2}\\\Rightarrow F_{new}=16\times F[/tex]
∴Force will increase 16 times
The return-air ventilation duct in a home has a cross-sectional area of 900 cm^2. The air in a room that has dimensions 7.0 m ×× 11.0 m ×× 2.4 m is to be completely circulated in a 40-min cycle. What is the speed of the air in the duct? (Express your answer to two significant figures.) (m/s)
Answer:
0.86 m/s
Explanation:
A = cross-sectional area of the duct = 900 cm² = 900 x 10⁻⁴ m²
v = speed of air in the duct
t = time period of circulation = 40 min = 40 x 60 sec = 2400 sec
V = Volume of the air in the room = volume of room = 7 x 11 x 2.4 = 184.8 m³
Volume of air in the room is given as
V = A v t
inserting the values
184.8 = (900 x 10⁻⁴) (2400) v
v = 0.86 m/s
An electron (mass m=9.11×10−31kg) is accelerated from the rest in the uniform field E⃗ (E=1.45×104N/C) between two thin parallel charged plates. The separation of the plates is 1.90 cm . The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, with what speed does it leave the hole?
Explanation:
It is given that,
Mass of an electron, [tex]m=9.11\times 10^{-31}\ kg[/tex]
Electric field, [tex]E=1.45\times 10^4\ N/C[/tex]
Separation between the plates, d = 1.9 cm = 0.019 m
The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. We need to find the speed of the electron as it leave the hole.
The force due to accelerating electron is balanced by the electrostatic force i.e
qE = ma
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\ C\times 1.45\times 10^4\ N/C}{9.11\times 10^{-31}\ kg}[/tex]
[tex]a=2.54\times 10^{15}\ m/s^2[/tex]
Let v is the speed as it leave the hole. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2ad[/tex]
[tex]v=\sqrt{2ad}[/tex]
[tex]v=\sqrt{2\times 2.54\times 10^{15}\times 0.019\ m}[/tex]
v = 9824459.27 m/s
or
[tex]v=9.82\times 10^6\ m/s[/tex]
So, the speed of the electron as it leave the hole is [tex]9.82\times 10^6\ m/s[/tex]. Hence, this is the required solution.
An object at the surface of Earth (at a distance R from the center of Earth) weighs 166 N. What is its weight (in N) at a distance 4R from the center of Earth? Round your answer to the nearest tenth.
Answer:
The weight at a distance 4R from the center of earth is 10.37 N.
Explanation:
Given that,
Weight = 166 N
Distance = 4R
Let m be the mass of the object.
We know that,
Mass of earth [tex]M_{e}=5.98\times10^{24}\ kg[/tex]
Gravitational constant[tex]G = 6.67\times10^{-11}\ N-m^2/kg^2[/tex]
Radius of earth [tex]R = 6.38\times10^{6}\ m[/tex]
We need to calculate the weight at a distance 4 R from the center of earth
Using formula of gravitational force
[tex]W = \dfrac{GmM_{e}}{R^2}[/tex]
Put the value in to the formula
[tex]166=\dfrac{6.67\times10^{-11}\times m\times5.98\times10^{24}}{(6.38\times10^{6})^2}[/tex]
[tex]m=\dfrac{166\times(6.38\times10^{6})^2}{6.67\times10^{-11}\times5.98\times10^{24}}[/tex]
[tex]m=16.94 kg[/tex]
Now, Again using formula of gravitational
[tex]W=\dfrac{6.67\times10^{-11}\times 16.94\times5.98\times10^{24}}{(4\times6.38\times10^{6})^2}[/tex]
[tex]W=10.37 N[/tex]
Hence, The weight at a distance 4R from the center of earth is 10.37 N.