15. Calculate the resistive forces acting on a sports car if it is travelling at a steady speed of 25 m/s when the engine is providing 200 kW. gnt V 10X50 - soot Grade 9
Answer:
8000 N
Explanation:
Applying
P = F×V.............. Equation 1
Where P = Power of the sport car, F = resistive force acting on the sport car, V = Speed of the sport car.
make F the subject of the equation
F = P/V............ Equation 2
From the question,
Given: P = 200 kW = 200000 W, V = 25 m/s
Substitute these values into equation 2
F = 200000/25
F = 8000 N
Explain why it takes much more effort to stop a freight train compared with a car?
Answer:
Train wheels and rails are both made of steel, and the steel-steel friction coefficient is around 0.25. As a result, the stopping time and distance will be three to four times that of a car.
The Sun is a type G2 star. Type G stars (from G0 to G9) have a range of temperatures from 5200 to 5900. What is the range of log(T) for G stars
Answer:
Explanation:
I’ll help
state any two effects of gravitational force
what's impulse of
force
Answer:
The impulse experienced by the object equals the change in momentum of the object. In equation form, F.t = m. Δv. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum.
A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the x-axis does the third piece move
Answer:
Explanation:
There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.
momentum of first piece = .320 kg x 2 m/s
= 0.64 kg m/s along x -axis.
momentum of second piece = .355 kg x 1.5 m/s
= 0.5325 kg m/s along y- axis .
Let the velocity of third piece be v and it is making angle of θ with x -axis .
Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ
vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ
For making total momentum in the plane zero
.1 v cosθ = 0.64 kg m/s
.1 v sinθ = 0.5325 kg m/s
Dividing
Tanθ = .5325 / .64 = .83
θ = 40⁰.
The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.
Answer:
8.3 m/s, 2196 degree from + X axis
Explanation:
m = 320 g , u = 2 m/s along X axis
m' = 355 g, u' = 1.5 m/s along Y axis
m'' = 100 g, u'' = v
Let the speed of the third piece is v makes an angle A from the X axis.
use conservation of momentum along X axis
0 = 320 x 2 + 100 x v cos A
v cos A = - 6.4 ..... (1)
Use conservation of momentum along Y axis
0 = 355 x 1.5 + 100 x v sin A
v sinA = - 5.3 ... (2)
Squaring and adding
[tex]v^2 = (-6.4)^2 +(-5.3)^2\\\\v= 8.3 m/s[/tex]
The angle is given by
[tex]tan A = \frac{-5.3}{-6.4}\\\\A = 219.6 degree[/tex] from + X axis
During the Moment of Inertia experiment, a group of students decided to mount a solid sphere on the top of the horizontal disk. The sphere and disk have the same mass, M, and radius, R. They recorded the values for the initial and final positions for the falling mass, and the height above the floor when falling mass is at its lowest position. These values are respectively 90cm, 68.5 cm, and 23.5cm. The radius of the pulley, r, was 1/10 of the radius of the sphere. Furthermore, the mass of the disk, M, was 1.4kg, what was the mass, m, for the falling mass
Answer:
u are the very good person I know if u will do itself u will becam3 a rising star
Question 15
Calculate the velocity of a body if its total energy is three times its rest energy
OA 0.54c
OB. 0.760
OC0.94c
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1. A child slide down an inclined plane of length 10 m at an angle of 45°. If the coefficient friction between the child and the plane is 0.1, evaluate The velocity just before touching the bottom of the plane.
Answer:
The speed at the bottom is 11.2 m/s.
Explanation:
length, s = 10 m
Angle, A = 45 degree
coefficient of friction = 0.1
let the velocity is v.
The acceleration is given by
[tex]a = g sin A - \mu g cos A \\\\a = 9.8 (sin 45 - 0.1 cos 45)\\\\a = 6.24 m/s^2[/tex]
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\v^2 = 0 + 2 \times 6.24 \times 10 \\\\v = 11.2 m/s[/tex]
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:
[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]
where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,
[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]
d = 68.5 x 10⁻⁶ m = 68.5 μm
Compare and contrast how sound waves behave when they
encounter solid (hard) surfaces
encounter soft surfaces
pass through an opening
encounter a barrier
Compare and contrast how light waves are
reflected from concave and convex mirrors
reflected/absorbed in black, red, blue and white paper
refracted in concave and convex lenses
transmitted in opaque, translucent, and transparent materials
Answer:
As sound waves move (or more accurately, when they travel by transferring their energy) they interact with physical objects. Soft surfaces will absorb sound while hard surfaces will reflect it. .Hard surfaces reflect sound and soft surfaces absorb sound.
The convex mirror has a reflecting surface that curves outward, resembling a portion of the exterior of a sphere. Light rays parallel to the optical axis are reflected from the surface in a direction that diverges from the focal point, which is behind the mirror
A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a) the propagation velocity of the wave
Answer:
the propagation velocity of the wave is 274.2 m/s
Explanation:
Given;
length of the string, L = 1.5 m
mass of the string, m = 0.002 kg
Tension of the string, T = 100 N
wavelength, λ = 1.5 m
The propagation velocity of the wave is calculated as;
[tex]v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s[/tex]
Therefore, the propagation velocity of the wave is 274.2 m/s
An astronaut weighs 202 lb. What is his weight in newtons?
An inductive circuit contains resistance of 20 ohm and an inductance of 20 H. If an ac voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be
A 0.0159
A 0.017
A 0.02
A 0.16
Answer:
answer : option (b) 0.016 amp
explanation : resistance of resistor , R = 10 Ω
inductance of inductor , X_LX
L
= 20H
voltage of AC circuit , V = 120volts
frequency, ff =60Hz
so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s
now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=
R
2
+ω
2
L
2
V
= 120/√{10² + (120π)² × 20²}
= 120/√{100 + 14400π² × 400}
after solving this we get, i = 0.016 amp
The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0.850 ATM. A faucet with an opening 1.0 inch diameter is located at the bottom of the container.
Required:
a. What is the net force on the side of the container?
b. How long does it take and how much the water level will drop till water no longer comes out of the faucet?
Answer:
a) F = 2.66 10⁴ N, b) h = 1.55 m
Explanation:
For this fluid exercise we use that the pressure at the tap point is
Exterior
P₂ = P₀ = 1.01 105 Pa
inside
P₁ = P₀ + ρ g h
the liquid is water with a density of ρ=1000 km / m³
P₁ = 0.85 1.01 10⁵ + 1000 9.8 5
P₁ = 85850 + 49000
P₁ = 1.3485 10⁵ Pa
the net force is
ΔP = P₁- P₂
Δp = 1.3485 10⁵ - 1.01 10⁵
ΔP = 3.385 10⁴ Pa
Let's use the definition of pressure
P = Fe / A
F = P A
the area of a circle is
A = pi r² = [i d ^ 2/4
let's reduce the units to the SI system
d = 100 cm (1 m / 100 cm) = 1 m
F = 3.385 104 pi / 4 (1) ²
F = 2.66 10⁴ N
b) the height for which the pressures are in equilibrium is
P₁ = P₂
0.85 P₀ + ρ g h = P₀
h = [tex]\frac{P_o ( 1-0.850)}{\rho \ g}[/tex]
h = [tex]\frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}[/tex]
h = 1.55 m
The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns
A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.
Answer:
C
Explanation:
Similarity
Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến
95oC for 20 minutes.
Answer:
P = 2439.5 W = 2.439 KW
Explanation:
First, we will find the mass of the water:
Mass = (Density)(Volume)
Mass = m = (1 kg/L)(10 L)
m = 10 kg
Now, we will find the energy required to heat the water between given temperature limits:
E = mCΔT
where,
E = energy = ?
C = specific heat capacity of water = 4182 J/kg.°C
ΔT = change in temperature = 95°C - 25°C = 70°C
Therefore,
E = (10 kg)(4182 J/kg.°C)(70°C)
E = 2.927 x 10⁶ J
Now, the power required will be:
[tex]Power = P = \frac{E}{t}[/tex]
where,
t = time = (20 min)(60 s/1 min) = 1200 s
Therefore,
[tex]P = \frac{2.927\ x\ 10^6\ J}{1200\ s}[/tex]
P = 2439.5 W = 2.439 KW
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop
Answer:
The correct solution is "122.2211".
Explanation:
Given:
deceleration,
a = 22 ft/sec²
Initial velocity,
[tex]V_i=50 \ m/h[/tex]
Now,
[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]
[tex]=73.333 \ ft/sec[/tex]
Now,
Final velocity,
[tex]V_f=0[/tex]
Initial velocity,
[tex]V_{initial} = 73.333 \ ft/sec[/tex]
hence,
⇒ [tex]V_f^2=V_i^2+2aD[/tex]
By putting the values, we get
[tex]0=(73.333)^2+2\times( -22) D[/tex]
[tex]44D=(73.333)^2[/tex]
[tex]D=\frac{(73.333)^2}{44}[/tex]
[tex]=122.2211[/tex]
A horizontal rod (oriented in the east-west direction) is moved northward at constant velocity through a magnetic field that points straight down. Make a statement concerning the potential induced across the rod.
a. The east end of the rod is at higher potential than the west end.
b. The bottom surface of the rod is at higher potential than the top surface.
c. The top surface of the rod is at higher potential than the bottom surface.
d. The west end of the rod is at higher potential than the east end.
e. The potential is uniform.
Answer:
a. The east end of the rod is at higher potential than the west end.
Explanation:
The horizontal rod is oriented in the east-west direction. This means that applying right hand rule, the current will flow from the east to the west. Now, if we assume tat it is a closed loop, we know from polarity of voltage that current usually flows positive to negative terminal within a circuit.
This means the east is at a largely positive terminal while the west is at a largely negative terminal.
Thus, we can say that the east end of the rod is at higher potential than the west end of the rod.
A police car travels towards a stationary observer at a speed of 15m/s. the siren on the car emits a sound of frequency 250Hz. Calculate the observer frequency. the speed of sound is 340m/s
Observer Frequency = sound frequency x ( speed of sound / speed of sound - speed of car)
= 250 x (340/( 340-15))
= 261.54 Hz
A boy with a mass of 140 kg and a girl with a mass of 120 kg are on a merry go round. Th merry go round has a radius of 5 meters and its moment of inertia is 986 kg m 2. Beginning from rest the merry go round accelerates with an angular acceleration of 0.040 rad/s2 for 30 seconds then has a constant angular speed.
1. How many revolutions do the kids make before the constant operational speed is reached ?
2. What's the angular speed and magnitude of the tangential of the kids if they are standing at a distance of 1.5m and 2.4 m from the center of the ride.
3. During the ride the kids switch places what is the angular speed and magnitude of the tangential velocities ?
Answer:
we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer
A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with the axis. On the left and right faces, the field has a strength and , respectively. The field along the front and back faces has strengths and . The field at the bottom and top faces has strengths and , respectively. What is the total charge enclosed by the cube
Complete Question
Complete Question is attached below
Answer:
[tex]q=1.558*10^{-9}c[/tex]
Explanation:
From the question we are told that:
Side length s=1.13m
Left field strength [tex]E_l=784.75N/m[/tex]
Right field strength [tex]E_r=776.38 N/m[/tex]
Front field strength [tex]E_f=725.5 N/m[/tex]
Back field strength [tex]E_b=749.54 N/m[/tex]
Top field strength [tex]E_t=944.95 N/m[/tex]
Bottom field strength [tex]E_{bo}=1082.58 N/m[/tex]
Generally, the equation for Charge flux is mathematically given by
[tex]\phi=EAcos\theta[/tex]
Where
Theta for Right,Left,Front and Back are at an angle 90
[tex]cos 90=0[/tex]
Therefore
[tex]\phi =0[/tex] with respect to Right,Left,Front and Back
Generally, the equation for Charge Flux is mathematically also given by
[tex]\phi=\frac{q}{e_o}[/tex]
Where
[tex]Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2[/tex]
Therefore
[tex]Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t[/tex]
[tex]Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)[/tex]
[tex]Q_{net}=176N/C m^2[/tex]
Giving
[tex]q=\phi*e_0[/tex]
[tex]q=176N/C m^2*1.558*10^{-12}c[/tex]
[tex]q=1.558*10^{-9}c[/tex]
To get maximum current in a circuit, the resistance should be in _____
1)series
2)parallel
Answer:
no parallel is the correct answer
Which car has the greatest potential energy? What type of potential energy does the
car have? What characteristics or properties does potential energy depend on? Which
cars have the least potential energy? Explain.
Explanation:
*car A has the greatest PE
*gravitational potential energy
*mass and height
*car D has the least PE
The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy stored in C3 and C4.
Answer:
A) C_{eq} = 15 10⁻⁶ F, B) U₃ = 3 J, U₄ = 0.5 J
Explanation:
In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.
In this case let's start by finding the equivalent capacitance.
A) Let's solve the part where C1 and C3 are. These two capacitors are in serious
[tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex] (you has an mistake in the formula)
[tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]
[tex]\frac{1}{C_{eq1}}[/tex] = 0.1 10⁶
[tex]C_{eq1}[/tex] = 10 10⁻⁶ F
capacitors C₂, C₄ and C₅ are in series
[tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]
[tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]
[tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶
[tex]C_{eq2}[/tex] = 5 10⁻⁶ F
the two equivalent capacitors are in parallel therefore
C_{eq} = C_{eq1} + C_{eq2}
C_{eq} = (10 + 5) 10⁻⁶
C_{eq} = 15 10⁻⁶ F
B) the energy stored in C₃
The charge on the parallel voltage is constant
is the sum of the charge on each branch
Q = C_{eq} V
Q = 15 10⁻⁶ 6
Q = 90 10⁻⁶ C
the charge on each branch is
Q₁ = Ceq1 V
Q₁ = 10 10⁻⁶ 6
Q₁ = 60 10⁻⁶ C
Q₂ = C_{eq2} V
Q₂ = 5 10⁻⁶ 6
Q₂ = 30 10⁻⁶ C
now let's analyze the load on each branch
Branch C₁ and C₃
In series combination the charge is constant Q = Q₁ = Q₃
U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]
U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]
U₃ = 3 J
In Branch C₂, C₄, C₅
since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅
U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]
U₄ = 0.5 J
Catching a wave, a 77-kg surfer starts with a speed of 1.3 m>s, drops through a height of 1.65 m, and ends with a speed of 8.2 m>s. How much nonconservative work was done on the surfer
Answer:
W = 2523.67 J
Explanation:
Given that,
The mass of surfer, m = 77 kg
He starts with a speed of 1.3 m/s
It drops through a height of 1.65 m and ends with a speed of 8.2 m/s.
We know that, the work done is equal to the change in kinetic energy. So,
[tex]W=\dfrac{1}{2}m(v_2^2-v_1^2)\\\\=\dfrac{1}{2}\times 77\times (8.2^2-1.3^2)\\W=2523.67\ J[/tex]
So, the required work done is equal to 2523.67 J.
An oscillating dipole antenna 1.73 m long with a maximum 36.0 mV potential creates a 500 Hz electromagnetic wave. (a) What is the maximum electric field strength created
Answer:
[tex]E_0=0.021v/m[/tex]
Explanation:
From the question we are told that:
Length [tex]l=1.73m[/tex]
Voltage [tex]V=36.0mV[/tex]
Frequency [tex]F=500Hz[/tex]
Generally the equation for maximum Electric Field is mathematically given by
[tex]E_0=\frac{\triangle}{r}[/tex]
[tex]E_0=\frac{36*10^{-3}}{1.73}[/tex]
[tex]E_0=0.021v/m[/tex]
Physics question on picture
Answer:
B. according to Newton's Third Law of Motion, the force of the Moon on the Earth and the force of the Earth on the Moon are equal in magnitude and opposite in direction
describe the movement of the man when the resultant horizontal force is 0 N
can anyone help in both questions please
Answer:
Force A newton Law first law
F = M.A which Force in 0 N as you Questions Above
Force B
Newton Law 3
Action = -Reaction
Hope you can explain this formula as you want to scribe to explaining
What is the electric field strength between two parallel conducting plates separated by 10 cm and having a potential difference between them of 2000 V?
a.
2000 V/m
b.
200 V/m
c.
20 kV/m
d.
200000 V/m
Answer:
• Potential Difference (V) = 2000 V
• Distance b/w the two parallel plates (d) = 10 cm = 10/100 = 1/10 = 0.1 m
• Electric field (E) = ?
[tex]\implies V = E.d[/tex]
[tex]\implies E = \dfrac{V}{d} [/tex]
[tex]\implies E = \dfrac{2000}{0.1} [/tex]
[tex]\implies E = \dfrac{2000}{ {10}^{ - 1} } [/tex]
[tex]\implies E = 2000 \times {10}^{1} [/tex]
[tex]\implies\bf E = 20000 \:V/m[/tex]
[tex]\implies\bf E = 20\:kV/m[/tex]
Hence, option C) the correct answer.
