Answer:
Magnitude of the average induced emf = 36.4 mv
Explanation:
Given:
inductance of the coil (L) = 7 mH
Initial current (Ci) = 0.200 A
Final current (Cf) = 1.50 A
Interval = 0.250 s
Find:
Magnitude of the average induced emf
Computation:
E = L(di/dt)
E = 7 [(1.5-0.20) / 0.250]
E = 7[5.2]
Magnitude of the average induced emf = 36.4 mv
What is the wavelength (in 10-15 m) of a proton traveling at 13.2% of the speed of light?
Answer:
The wavelength is [tex]\lambda = 10.01 *10^{-15} \ m[/tex]
Explanation:
From the question we are told that
The speed is [tex]v = 0.132 c[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
Generally the wavelength is mathematically represented as
[tex]\lambda = \frac{h}{m* v }[/tex]
where m is the mass of the proton with the value [tex]m = 1.6726 ^{-27} \ kg[/tex]
h is the Planck's constant with value [tex]h = 6.626 *10^{-34} \ J\cdot s[/tex]
=> [tex]\lambda = \frac{6.626 *10^{-34}}{1.6726 *10^{-34}* 0.132*3.0*10^8 }[/tex]
=> [tex]\lambda = 10.01 *10^{-15} \ m[/tex]
A large, cylindrical water tank with diameter 2.40 mm is on a platform 2.00 mm above the ground. The vertical tank is open to the air and the depth of the water in the tank is 2.00 mm. There is a hole with diameter 0.600 cmcm in the side of the tank just above the bottom of the tank. The hole is plugged with a cork. You remove the cork and collect in a bucket the water that flows out the hole.
A) When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank?
B) How long does it take you to collect 1.00 gal of water in the bucket?
Answer:
Explanation:
A ) radius of tank r = 1.2 m
depth of water in the tank = 2 m
1 gal of water = 1 / 264.17 m³
= 3.785 x 10⁻³ m³
Let h be the change in height of water in the tank .
volume of water flowing out
= π r² x h = 3.785 x 10⁻³
3.14 x 1.2² x h = 3.785 x 10⁻³
h = 83.71 x 10⁻⁵ m
= .84 mm .
B )
change in height is negligible .
velocity of efflux of water from the hole at the bottom
v = √ 2 gh
h is height of water level which is 2 m
v = √ (2 x 9.8 x 2 )
= 6.26 m / s
radius of hole = .3 x 10⁻² m
area of cross section
= π r²
= 3.14 x ( .3 x 10⁻² )²
= 28.26 x 10⁻⁶ m²
volume of water flowing through the hole per unit area
= area of cross section x velocity of efflux
= 28.26 x 10⁻⁶ x 6.26
If t be the time required ,
28.26 x 10⁻⁶ x 6.26 x t = 3.785 x 10⁻³
t = 21.4 s
Which planet orbits in a different plane than all of the others?
Although they're all 'close', none of the planets orbits in the same plane as any other planet. They're all in slightly different planes.
The farthest out compared to all the others is Pluto, with an orbit inclined about 17 degrees compared to the ecliptic plane (Earth's orbit). But Pluto is officially not a planet, so I don't think it's a good answer.
The next greatest inclination compared to Earth's orbit is Mercury. That one is about 7 degrees.
The other six planets are all in different orbital planes inclined less than 7 degrees compared to Earth's orbit.
If a car is travelling 120 miles southbound for 3 hours, what is the velocity?
Answer:
40 miles / hour south
Explanation:
120 miles/3 hours = 40 miles / hour
Answer:
v = 40 miles / hour
Explanation:
using velocity formula v = d / t
where d = distance and t = time
120 miles
v = --------------
3 hours
v = 40 miles / hour
In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separation between adjacent bright fringes on a screen 5 m from the slits is: CONVERT FIRST
Answer:
Δx = 2.5 x 10⁻³ m = 2.5 mm
Explanation:
The distance between two consecutive fringes, also known as fringe spacing, in Young's Double Slit Experiment, is given as follows:
Δx = λL/d
where,
Δx = distance between consecutive fringes = ?
λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m
L = Distance between slits and screen = 5 m
d = slit separation = 1 mm = 1 x 10⁻³ m
Therefore,
Δx = (5 x 10⁻⁷ m)(5 m)/(1 x 10⁻³ m)
Δx = 2.5 x 10⁻³ m = 2.5 mm
The separation between adjacent bright fringes on a screen 5 m from the slits is: 2.5 mm
We are given;
Wavelength of light; λ = 500 nm = 500 × 10⁻⁹ m
Distance of slit separation; d = 1mm = 0.001 m
Distance between slit and the screen; D = 5 m
Now, formula for fringe width is;
β = λD/d
Plugging in the relevant values gives;
β = (500 × 10⁻⁹ × 5)/0.001
β = 2.5 × 10⁻³ m
Converting to mm gives;
β = 2.5 mm
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A large, cylindrical water tank with diameter 3.00 m is on a platform 2.00 m above the ground. The vertical tank is open to the air and the depth of the water in the tank is 2.00 m. There is a hole with diameter 0.420 cm in the side of the tank just above the bottom of the tank. The hole is plugged with a cork. You remove the cork and collect in a bucket the water that flows out the hole.
A. When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank? Express your answer in millimeters.
B. How long does it take you to collect 1.00 gal of water in the bucket? Express your answer in seconds.
Answer:
1999.46 mm
45.59 s
Explanation:
given that
cylindrical water tank with diameter, D = 3 m
Height of the tank above the ground, h = 2 m
Depth of the water in the tank, d = 2 m
Diameter of hole, d = 0.420 cm
We start by calculating the volume of water in the tank, which is given as
Volume = πr²h
V = (πD²)/4 * h
V = (3.142 * 3²)/4 * 2
V = 28.278/4 * 2
V = 7.07 * 2
V = 14.14 m³
If 1.0 gal of water is equal to 0.0038m³, then
1 gal is 0.0038 = A * h
the area of the tank is 7.07 m²
therefore, 0.0038 = 7.07 * h
h₁ =0.00054 m = 0.54 mm is the height of water that flow out
the change in height of water in the tank = h - h₁ = 2 - 0.00054 = 1.99946 m
b)
Like we stated earlier, 1.0 gal of water is 0.0038m³
to solve this we use the formula
Q = Cd * A * √2gH
where Cd is a discharge coefficient, and is given by 0.9 for water
A is the area of the small hole
A = (πD²)/4
A = (π * 0.0042²)/4
A = 5.54*10^-5 / 4
A = 1.39*10^-5 m²
H= height of the hole from the tank water level = 2m - 0.0042 = 1.9958 m
g = 9.8 m/s²
Q = 0.9 * 1.39*10^-5 m² * √2 * 9.8 * 1.9958
Q = 1.251*10^-5 * 6.25
Q = 7.82*10^-5 m³/s
Q = V/t
t = V/Q = 0.0038m³ / 7.82*10^-5 m³/s
t = 45.59 s
an electromagnetic wave has an electric field with peak value 120. What is the averge energy delievered to a surface
Answer:
The average energy delivered to a surface is 19.116 W/m².
Explanation:
Given;
maximum electric field, E₀ = 120 v/m
The average energy delivered by the wave to a surface is given by
[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2}[/tex]
where;
c is the speed of light, = 3 x 10⁸ m/s
ε₀ is the permittivity of free space = 8.85 x 10⁻¹² c²/Nm²
[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2} \\\\I_{avg} = \frac{(3*10^8)(8.85*10^{-12})( 120)^2}{2}\\\\ I_{avg} =19.116 \ W/m^2[/tex]
Therefore, the average energy delivered to a surface is 19.116 W/m².
Water flows through a valve with inlet and outlet velocities of 3 m/s. If the loss coefficient of the valve is 2.0, and the specific weight of water is 9800 N/m3, the pressure drop across the valve is most nearly:
Answer:
9,000 kg/ms^2
Explanation:
The computation of the pressure fall across the valve is shown below:
It is to be computed by using the following formula
[tex]\Delta P = \frac{1}{2}\times K\times P\times v^2[/tex]
where,
[tex]\Delta P[/tex] = Fall in pressure
k = Coefficent loss
P = Loss of density
V = velocity of water
But before reach to the final solution first we have to determine the loss of density which is
[tex]P = \frac{r}{g}\\\\ = \frac{9,800 N/m^{3}}{9.81 m/s^{2}}\\\\ = 999kg/m^{3}\\\\ = 1000kg/m^{3}[/tex]
Now put all other values to the given formula
So,
[tex]= 2 \times \frac{1}{2} \times 1000 \times 3^2 \\\\ = 9,000 kg/ms^2[/tex]
Which of the following statements is true regarding electromagnetic waves traveling through a vacuum?
a. All waves have the same wavelength.
b. All waves have the same frequency.
c. All waves have the same speed.
d. The speed of the waves depends on their wavelength.
e. The speed of the waves depends on their frequency.
Answer:
C. All waves have the same speed.
Explanation:
Wave equation is given as;
V = fλ
where;
V is the speed of the wave
f is the frequency of the wave
λ is the wavelength
The speed of the wave depends on both wavelength and frequency
The speed of the electromagnetic waves in a vacuum is 3 x 10⁸ m/s, this also the speed of light which is constant for all electromagnetic waves.
Therefore, the correct option is "C"
C. All waves have the same speed.
How many significant figures are in 246.32
Answer:5
Explanation:
Decimal :2
Significant notation :2.4632× 10^2
an electric field of magnitude 200 N/C in the positive x- direction. calculate the acceleration in (m/s^2) of a charged particle of mass 1g and charge 1mC that is released from rest in this field?
1/200 that should get your
answer
A ball bearing of radius of 1.5 mm made of iron of density
7.85 g cm is allowed to fall through a long column of
glycerine of density 1.25 g cm. It is found to attain a
terminal velocity of 2.25 cm s-'. The viscosity of glycerine is
Answer:
[tex] \boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise} [/tex]
Given:
Radius of ball bearing (r) = 1.5 mm = 0.15 cm
Density of iron (ρ) = 7.85 g/cm³
Density of glycerine (σ) = 1.25 g/cm³
Terminal velocity (v) = 2.25 cm/s
Acceleration due to gravity (g) = 980.6 cm/s²
To Find:
Viscosity of glycerine ([tex] \sf \eta [/tex])
Explanation:
[tex] \boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}[/tex]
[tex] \sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v} [/tex]
Substituting values of r, ρ, σ, v & g in the equation:
[tex] \sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \times 64.7196[/tex]
[tex]\sf \implies \eta = 2 \times 7.191[/tex]
[tex]\sf \implies \eta = 14.382 \: poise[/tex]
Please answer all and will give Brainly and get points.
Answer:
16 is D, 17 is A, 18 is C, 19 is B ( would approve of brainliest)
Explanation:
16 is D, 17 is A, 18 is C, 19 is B
In the First option, distance is constant so D shows the correct graph,
In the second option, distance is increasing with time so, the velocity graph A is correct,
In the third option distance is constantly increasing with time, so C is the correct option.
In the last option distance is decreasing with time, so option B is the correct.
What is Distance time graph?
A distance-time graph is defined as how far an object has traveled in a given amount of time which is a simple line graph that shows the plot of distance versus time on a graph. Distance is plotted on the Y-axis while time is plotted on the X-axis.
The graphs which is shown in the question is distance-time graphs for various types of body motion.
When the body is steady or stationary,When thebody is moving non-uniformly with increasing speed,When the body is moving at a uniform speed, andWhen the body is moving non-uniformly with decreasing speed.Thus, the correct options for 16, 17, 18 and 19 are D, A, C and B respectively.
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A truck filled with gravel parks on a roadside scale that measures 8.00m by 6.80m. If the gravel and truck has a mass of 7,400kg. What pressure does the scale put on the spring below?
Explanation:
Pressure = force / area
P = (7400 kg × 10 m/s²) / (8.00 m × 6.80 m)
P = 1360 Pa
In certain metal, the stopping potential is found to be 3.70 V. When 235 nm light is incident on the metal, electrons are emitted. What is the maximum kinetic energy given to the electrons in eV and J?
Answer:
3.7 eV
5.92*10^-19 J
Explanation:
Given that.
Potential difference of the metal, V = 3.7 V
Wavelength of the light, n = 235 nm
maximum kinetic energy given to the electrons is giving them the formula
K(max) = e.V(s), where
KE(max) is the maximum kinetic energy needed
V = potential difference of the metal
KE(max) = e * 3.7
KE(max) = 3.7eV
converting our answer to Joules, we have
3.7eV = 3.7eV * 1.6*10^-19 J/eV
3.7eV = 5.92*10^-19 J
Therefore, the maximum kinetic energy in both eV and Joules is 3.7eV and 5.92*10^-19 Joules respectively
Answer:
Explanation:
d dnnd
A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i v, start subscript, i, end subscript. Later, its speed increases to 4v_i4v i 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases
Answer and Explanation: Centripetal Acceleration is the change in velocity caused by a circular motion. It is calculated as:
[tex]a_{c}=\frac{v^{2}}{r}[/tex]
v is linear speed
r is radius of the curve the object in traveling along
For its first lap:
[tex]a_{c}_{1}=\frac{v_{i}^{2}}{R}[/tex]
After a while:
[tex]a_{c}_{2}=\frac{(4v_{i})^{2}}{R}[/tex]
[tex]a_{c}_{2}=\frac{16v_{i}^{2}}{R}[/tex]
Comparing accelerations:
[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]
[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=\frac{16.v_{i}^{2}}{R}.\frac{R}{v_{i}^{2}}[/tex]
[tex]\frac{a_{c}_{2}}{a_{c}_{1}}=16[/tex]
[tex]a_{c}_{2}=16a_{c}_{1}[/tex]
With linear speed 4 times faster, centripetal acceleration is 16 times greater.
Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to find c). You have a spring with spring constant k 5 N/m. You take the spring, you attach it to the mass and fix it to a wall. Then you pull on the spring and let the mass go. You find that the mass oscillates with frequency 1 Hz. What is the friction
Answer:
b = 0.6487 kg / s
Explanation:
In an oscillatory motion, friction is proportional to speed,
fr = - b v
where b is the coefficient of friction
when solving the equation the angular velocity has the form
w² = k / m - (b / 2m)²
In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction
let's call
w₀² = k / m
w² = w₀² - b² / 4m²
b² = (w₀² -w²) 4 m²
Let's find the angular velocities
w₀² = 5 / 0.1
w₀² = 50
w = 2π f
w = 2π 1
w = 6.2832 rad / s
we subtitute
b² = (50 - 6.2832²) 4 0.1²
b = √ 0.42086
b = 0.6487 kg / s
The coefficient friction of the mass during the measurement is 0.648 kg/s.
The given parameters;
mass, m = 0.1 kgspring constant, k = 5 N/mfrequency of the mass, F = 1 HzDuring oscillatory motion, friction is directly proportional to speed.
[tex]F_k = -vb[/tex]
where;
b is the coefficient of frictionThe angular velocity is given as;
[tex]\omega ^2 = \frac{k}{m} - \frac{b^2}{4m^2} \\\\\omega ^2 = \omega _0^2 - \frac{b^2}{4m^2}\ \ ---\ (1)[/tex]
From the equation above, we will have the following;
[tex]\omega_0^2 = \frac{k}{m} \\\\\omega_0^2 = \frac{5}{0.1} \\\\\omega_0^2 = 50[/tex]
Also, the instantaneous angular speed is calculated as;
[tex]\omega = 2\pi f\\\\\omega = 2\pi \times 1\\\\\omega = 2\pi\\\\\omega = 6.284 \ rad/s[/tex]
From equation (1), the coefficient of friction is calculated as follows;
[tex]\omega ^2 = \omega ^2_0 - \frac{b^2}{4m^2} \\\\ \frac{b^2}{4m^2} = \omega ^2_0 - \omega ^2 \\\\b^2 = 4m^2( \omega ^2_0 - \omega ^2)\\\\b= \sqrt{ 4m^2( \omega ^2_0 - \omega ^2)}\\\\b = \sqrt{ 4\times 0.1^2\times ( 50 - 6.284^2)}\\\\b = 0.648 \ \ kg/s[/tex]
Thus, the coefficient friction of the mass during the measurement is 0.648 kg/s.
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What amount of heat is required to increase the temperature of 75.0 grams of gold from 150°C to 250°C? The specific heat of gold is 0.13 J/g°C.A. 750 joulesB. 980 joulesC. 1300 joulesD. 1500 joulesE. 2500 joules
Answer:
B. 980 joulesExplanation:
Given the following data
initial temperature T1= 150 °C
final temperature T2= 250 °C
specific heat of gold c= 0.13 J/g°C
mass of gold m= 75.0 grams
we can use the expression stated below to solve for the quantity of heat
[tex]Q= mc(T2-T1)---------1[/tex]
Substituting our known data into the expression we can solve for the value of Q
[tex]Q= 75*0.13(250-150)---------1\\\\Q= 75*0.13(100)\\\\Q= 975 Joules[/tex]
The quantity of heat need to raise the temperature from 150°C to 250°C is 975 J
Answer:
B. 980 joules
Explanation:
A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass is at the 50 cm mark.
a) If a mass m1 = 80 g is suspended at the 30 cm mark, at which cm mark would a mass m2 = 110 g need to be suspended for the system to be in equilibrium?
b) If a mass m1=80g is suspended at the 25cm mark,and a mass m2 =110g is suspended at the 60 cm mark, from what cm mark would a mass m3 = 45 g need to be suspended for the system to be in equilibrium?
Answer:
a) 800N × 20 cm = 1100N × x cm
16000= 1100x
x= 14.5
therefore it must be placed on the (50 + 14.5)cm mark
= 64.5 cm mark
b) 800N × 25 cm = (1100N × 10 cm)+(450N × x cm)
20000 = 11000 + 450x
450x = 9000
x = 20 cm
therefore it must be placed on the (50 + 20)cm mark
= 70 cm mark
a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.
b) The distance at which mass m₃(45 g) should be suspended is 70 cm.
What is meant by principle of moments?According to the Principle of Moments, when a body is balanced or is at equilibrium, the total clockwise and anticlockwise moments about a given point are equal.
a) m₁ = 80 g
m₂ = 110 g
r₁ = 30 cm
According to the Principle of Moments,
m₁r₁ = m₂r₂
Therefore, the distance,
r₂ = m₁r₁/m₂
r₂ = 80 x 20/110
r₂ = 14.54 cm
So, the distance at which mass m₂ should be suspended is,
r' = 50 + 14.54
r' = 64.54 cm
b) m₁ = 80 g
m₂ = 110 g
m₃ = 45 g
r₁ = 25 cm
r₂ = 60 cm
According to the Principle of Moments,
m₁r₁ = m₂r₂ + m₃r₃
80 x 25 = (110 x 10) + (45 x r₃)
45 x r₃ = 2000 - 1100
r₃ = 900/45
r₃ = 20 cm
So, the distance at which mass m₃ should be suspended is,
r' = 50 + 20
r' = 70 cm.
Hence,
a) The distance at which the mass m₂(110 g) should be suspended is 64.54 cm.
b) The distance at which mass m₃(45 g) should be suspended is 70 cm.
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A thin rod of mass M and length l hangs from a pivot at its upper end. A ball of clay of mass m and of horizontal velocity v strikes the lower end at right angles andremains stuck (total inelastic collision).
Required:
How high will the rod swing after this collision?
Answer:
P = 2923.89 W
Explanation:
Power is
P = F v
so This exercise will solve them in parts using the conservation of momentum and then using the conservation of energy
To use the conservation of the momentum we must define a system, formed by the bodies, so that the forces during the collision have internal forces and the moment is conserved
initial instant, before the crash
p₀ = m v
final instant. Right after the crash
[tex]p_{f}[/tex] = (M + m) v₂
p₀ =p_{f}
m v = (M + m) v₂
v₂ = m / (m + M) v
this is the speed with which two come out, now we can apply the conservation of energy to the system formed by the two bodies together
Starting point. Lower
Em = K = ½ (M + m) v²
Final point. Highest point
Em = U = (M + m) g h
Eo₀ = [tex]Em_{f}[/tex]
½ (M + m) v2 = (M + m) g h
h = 1/2 v2 / g
h = ½ [m / (m + M) v] 2 / g
h = 1/2 (m / m + M) 2 / g we must calculate the force, let's use Newton's second law, let's set a coordinate system with a parallel axis flat and the other axis (y) perpendicular to the plane
X Axis
Fe - Wₓ = 0
F = Wₓ
Y Axis
N - [tex]W_{y}[/tex] = 0
let's use trigonometry for the components of the weight
sin 6 = Wₓ / W
cos 6 = [tex]W_{y}[/tex] / W
Wx = W sin 6
W_{y}= W cos 6
F = mg cos 6
F = 75 9.8 cos 6
F = 730.97 N
let's calculate the power
P = F v
P = 730.97 4.0
P = 2923.89 W
Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is 2.7 m. Consider a point on the screen 1.3 cm from the central maximum. Calculate (a) θ for that point, (b) α, and (c) the ratio of the intensity at that point to the intensity at the central maximum.
Answer:
0.276
0.9
0.756
Explanation:
Given that
Wavelength of the light, λ = 588 nm
Distance from the slit to the screen, L = 2.7 m
Width of the slit, a = 0.0351 mm
a point on the screen, y = 1.3 cm = 0.013 m
Sinθ = y/L
Sinθ = 0.013/2.7
sinθ = 0.0081
θ = sin^-1 0.00481
θ = 0.276°
α = (π.a.sinθ)/λ
α = (3.142 * 3.51*10^-5 * sin 0.276) / 588*10^-9
α = 5.3*10^-7 / 588*10^-9
α = 0.9 rad
I/i(m) = ((sinα)/α)²
I/I(m) = ((sin 0.9) / 0.9)²
I/I(m) = (0.783/0.9)²
I/I(m) = 0.87²
I/I(m) = 0.756
Note, our calculator has to be set in Rad instead of degree for part C, to get the answer
Land, labor, and capital are examples of...
Answer:
The factors of production are resources that are the building blocks of the economy; they are what people use to produce goods and services. Economists divide the factors of production into four categories: land, labor, capital, and entrepreneurship
When stable air is forced to rise, any clouds that are produced are generally thin and flat lying.
a) true
b) false
Answer:
a) true
Explanation:
One of the important factors behind the formation of clouds is the stability of the atmosphere. Air gets condensed with the increase in the height, while it becomes warm with a decrease in its height. A stable air is the type of air that can sink. The air which has low temperature has more density than the air it is surrounded by. When clouds are formed with stable air, the clouds formed are thin and horizontal.
Solve for A if F=MA and F=100 and M=25?
Answer:
[tex] \boxed{ \boxed{ \bold{A = 4}}}[/tex]Explanation:
Given,
F = 100 , M = 25
Now, let's find the value of A
[tex] \sf{F = MA}[/tex]
plug the values
⇒[tex] \sf{100 = 25 A}[/tex]
Swap the sides of the equation
⇒[tex] \sf{25A = 100}[/tex]
Divide both sides of the equation by 25
⇒[tex] \sf{ \frac{25A}{25} = \frac{100}{25} }[/tex]
Calculate
⇒[tex] \sf{A = 4}[/tex]
Hope I helped!
Best regards!!
Given:-
Force,F = 100 NMass,m = 25 kgTo find out:-
Calculate the acceleration, a ?
Formula applied:-
F = m × a
Solution:-
We know,
[tex] \sf{F = m × a}[/tex]
Substituting the values of mass and acceleration,we get
[tex] \sf\implies \: 100 = 25 \times a[/tex]
[tex] \sf \implies a = \cancel \dfrac{100}{25} [/tex]
[tex] \sf \implies a = 4 \: ms {}^{ - 1} [/tex]
I don’t understand can someone break it down for me
Answer:
a = (v² – v₀²)/ 2(s – s₀)
Explanation:
v² = v₀² + 2a (s – s₀)
We can make 'a' the subject of the above expression as follow:
v² = v₀² + 2a (s – s₀)
Subtract v₀² from both side
v² – v₀² = v₀² + 2a (s – s₀) – v₀²
v² – v₀² = v₀² – v₀² + 2a (s – s₀)
v² – v₀² = 2a (s – s₀)
Divide both side by (s – s₀)
(v² – v₀²)/ (s – s₀) = 2a
Divide both side by 2
(v² – v₀²)/ (s – s₀) ÷ 2 = a
(v² – v₀²)/ (s – s₀) × 1/2 = a
(v² – v₀²)/ 2(s – s₀) = a
a = (v² – v₀²)/ 2(s – s₀)
If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged?
Answer:
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Explanation:
The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.
The first minimum occurs for m = 1, so the diffraction equation of a slit remains
a sin θ = λ
in general, the diffraction patterns occur at very small angles, so
sin θ = θ
θ = λ / a
in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.
θ = 1.22 λ /a
In this exercise we are told that the opening changes
a’ = 2 a
we substitute
θ ‘= 1.22 λ / 2a
θ' = (1.22 λ / a) 1/2
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
(II) A baseball pitcher throws a baseball with a speed of 43 m????s. Estimate the average acceleration of the ball during the throwing motion. In throwing the baseball, the pitcher accelerates it through a displacement of about 3.5 m, from behind the body to the point where it is released
Answer:
a = 264.14 m/s²
Explanation:
From the question;
Initial velocity; u will be 0 m/s since the ball will start from rest.
Final velocity; v = 43 m/s
distance covered by the motion; s = 3.5m
To get the acceleration, we will make use of Newton's third equation of motion which is;
v² = u² + 2as
Making a the subject, we have;
a = (v² - u²)/2s
Plugging in the relevant values to give;
a = (43² - 0)/(2 × 3.5)
a = 264.14 m/s²
The average acceleration of the ball during the throwing motion is 265.14m/s².
In order to get the acceleration, the Newton's third law of motion will be used. This will be:
v² = u² + 2as
We'll make a to be the subject of the formula and this will be:
a = (v² - u²) / 2s
We'll plug in the value into the equation and this will be:
a = (43² - 0) / (2 × 3.5)
a = 1849 / 7
= 264.14 m/s²
Therefore, the acceleration is 265.14m/s.
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https://brainly.com/question/17071337
A 36.5 mA current is carried by a uniformly wound air-core solenoid with 430 turns, a 18.5 mm diameter, and 11.5 cm length. (a) Compute the magnetic field inside the solenoid. (b) Compute the magnetic flux through each turn. Tm2 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current?i) magnetic field inside the solenoid ii) magnetic flux through each turn inductance of the solenoid
Answer:
Explanation:
a )
magnetic field inside the solenoid B = μ₀ n I where n is no of turns per unit length , I is current and μ₀ = 4 π x 10⁻⁷ .
Putting the values in the equation
B = 4 π x 10⁻⁷ x (430 / .115 ) x 36.5 x 10⁻³
= 1.7 x 10⁻⁴ T .
b ) magnetic flux through each turn
= B x A where A is cross sectional area of solenoid .
= 1.7 x 10⁻⁴ x π x 9.25² x 10⁻⁶
= 456.73 x 10⁻¹⁰ Tm² .
c ) Inductance of solenoid
L = flux associated with all turns / current
= 456.73 x 10⁻¹⁰ x 430 / (36.5 x 10⁻³)
= 5381 x 10⁻⁷
= 538 x 10⁻⁶ H
= 538 μH .
d )
magnetic field inside the solenoid depends upon current
magnetic flux through each turn depends upon current
inductance of the solenoid does not depend upon current because current is divided from total flux with solenoid.
Name the Sl base units that are ilnportant in chem-istry. Give the Sl units for expressing the following:
(a) length.
(b) volume.
(c) mass,
(d) time,
(e) energy,
(f) temperature
Answer:
Explanation:
SI unit of
length=meter
volume =dm^3
mass =kilogram
time=second
energy= joule
temperature =kelvin
A turntable of radius R1 is turned by a circular rubberroller of radius R2 in contact with it at their outeredges. What is the ratio of their angular velocities,ω1 / ω2 ?
Answer:
The ratio is [tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]
Explanation:
From the question we are told that
The first radius is [tex]R_1[/tex]
The second radius is [tex]R_2[/tex]
Generally the angular speed of the turntable is mathematically represented as
[tex]w_1 = \frac{ v_k }{R_1 }[/tex]
Generally the angular speed of the rubber roller is mathematically represented as
[tex]w_2 = \frac{ v_k }{R_2 }[/tex]
Where [tex]v_k[/tex] is the velocity of both turntable and rubber roller
So
[tex]\frac{w_1}{w_2} = \frac{\frac{v_k}{R_1} }{\frac{v_k}{R_2} }[/tex]
[tex]\frac{w_1}{w_2} = \frac{R_2}{R_1}[/tex]