Answer: 10.58 C has flowed during the lightning bolt
Explanation:
Given that;
Time of flow t = 1.2 × 10⁻³
perpendicular distance r = 21 m
Magnetic field B = 8.4 x 10⁻⁵ T
Now lets consider the expression for magnetic field;
B = u₀I / 2πr
the current flow is;
I = ( B × 2πr ) / u₀
so we substitute
I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷
= 0.01107792 / 0.000001256
= 8820 A
Hence the charge flows during lightning bolt will be;
q = It
so we substitute
q = 8820 × 1.2 × 10⁻³
q = 10.58 C
therefore 10.58 C has flowed during the lightning bolt
A 1200 kg car is accelerating at 4.5 m/s2. What is the force on the car?
Answer:
5400 N
Explanation:
f=ma
f= 1200*4.5
f=5400N
What is the energy contained in a 1.30 m3 volume near the Earth's surface due to radiant energy from the Sun
Answer:
The energy contained is 5.856 x 10⁻⁶ J
Explanation:
Average energy density of electromagnetic radiation per unit volume is given by the equation;
[tex]U_{avg} = \frac{1}{2} \epsilon _o E_o[/tex]²
where;
[tex]\epsilon _o[/tex] is permittivity of free space
[tex]E_o[/tex] is maximum electric field strength, this can be calculated from the intensity of sun reaching the Earth's surface.
[tex]E_o = \sqrt{\frac{2I}{\epsilon_o C} }[/tex]
The intensity of sun reaching the Earth is 1350 W/m²
[tex]E_o = \sqrt{\frac{2*1350}{8.885*10^{-12}*3*10^8 } } \\\\E_o = 1008.96 \ V/m\\[/tex]
Average energy density of electromagnetic radiation per unit volume;
[tex]U_{avg} = \frac{1}{2} \epsilon_o E_o^2\\\\U_{avg} = \frac{1}{2} (8.85*10^{-12})(1008.96)^2\\\\U_{avg} = 4.505 *10^{-6} \ J/m^3[/tex]
The energy contained in a 1.30 m³ volume is given by;
E = (4.505 x 10⁻⁶)(1.3)
E = 5.856 x 10⁻⁶ J
Therefore, the energy contained is 5.856 x 10⁻⁶ J
You are hired to lift 30 kg crates a vertically 0.90 m from the ground onto a truck. How many crates would you have to load onto the truck in 1 minute for your average power output in lifting the crates to be 100 W
Answer:
22 crates
Explanation:
Power = Force ×Distance/time taken
Power = m×a×d/t
Power = 30×9.81×0.9/60 (1min was converted to second)
Power = 264.87/60
Power = 4.4145Watts
If my average power output us 100aw, then;
Number of crate to load = average power/4.4145
Number of crates to load = 100/4.4146
Number of crates to load = 22.6
Hence I will have to load about 22 crates onto the truck in 1 minute for my average power output in lifting the crates to be 100 W.
Converting compound units
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?
Answer:
Explanation:
Given the density of silicon as 2.33g/cm³
We are to convert this to kg/cm³
We will be using the following conversion factors
1000g = 1kg
2.33g = x
Cross multiply
1000x = 2.33
x = 2.33/1000
x = 0.00233kg
Also we need to convert 1cm³ to 1m³
1cm = 0.01m
1cm³ = 0.01×0.01×0.01
1cm³ = 0.000001m³
Substituting into the density value of silicon
2.33g/cm³ = 0.00233kg/0.000001m³
= 2330kg/m³
A model rocket is shot directly upward, rises to its maximum height and then returns to its launch position in 10.0 s. Assuming free fall conditions, what was the rocket's initial upward velocity?
a. 98.0 m/s
b. 123 m/s
c. 24.5 m/s
d. 49.0 m/s
d. 49.0 m/s
this is your answer. ....OK. ..
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A 75kg man climbs the stairs to the fifth floor of a building. A total hieght of 16m. His potential energy has increased by
Answer:
11760 joules
Explanation:
Given
Mass (m) = 75kg
Height (h) = 16m
Required
Determine the increment in potential energy (PE)
This is calculated as thus:
PE = mgh
Where g = 9.8m/s²
Substitute values for m, g and h.
P.E = 75 * 9.8 * 16
P.E = 11760 joules
The potential energy of the man in the fifth floor of the building has increased by 11760J.
Given the data in the question;
Mass of the man; [tex]m = 75kg[/tex]Height; [tex]h = 16m[/tex]Potential energy; [tex]P_E =\ ?[/tex]
Potential energy is the energy possessed by a particle due to its position relative to other particles. It is expressed as:
[tex]P_E = mgh[/tex]
Where m is mass of the particle, h is its height above ground level and g is acceleration due to gravity( [tex]g = 9.8m/s^2[/tex] ).
We substitute our values into the equation
[tex]P_E = 75kg\ *\ 9.8m/s^2\ *\ 16m\\\\P_E = 11760kg.m/s^2\\\\ P_E = 11760J[/tex]
Therefore, the potential energy of the man in the fifth floor of the building has increased by 11760J.
Learn more: https://brainly.com/question/16850959
What energy transformation takes place when you stretch a bungee cord?
Answer:
potential energy
Explanation:
A 75 Kg skateboarder is riding downhill, exerting 25 N. What is their acceleration?
Answer:
[tex]a=0.33\frac{m}{s^2}[/tex]
Explanation:
Hello.
In this case, since the force is defined in terms of the mass and acceleration as follows:
[tex]F=ma[/tex]
Given the force and the mass, we can compute the acceleration as shown below:
[tex]a=\frac{F}{m}=\frac{25N}{75kg}=\frac{25kg\frac{m}{s^2} }{75kg}\\ \\a=0.33\frac{m}{s^2}[/tex]
Best regards.
Bird A, with a mass of 2.2 kg, is stationary while Bird B, with a mass of 1.7 kg, is moving due north from Bird A at 3 m/s. What is the velocity of the center of mass for this system of two birds
Answer:
1.3 m/s
Explanation:
It is given that,
Mass of bird A, [tex]m_A=2.2\ kg[/tex]
Mass of bird B, [tex]m_B=1.7\ kg[/tex]
Initial speed of bird A is 0 as it was at rest
Initial speed of bird B is 3 m/s
We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,
[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]
So, the center of mass for this system is 1.3 m/s.
Two charged objects are separated by distance, d. The first charge has a larger magnitude (size) than the second charge. Which one exerts the most force?
A .05 kg rubber ball is dropped and hits the floor with an initial velocity of 10 m/s. It rebounds away from the floor with a final speed of 7 m/s after being in contact with the floor for .01 seconds. Find the magnitude of the force exerted by the floor on the rubber ball.
Answer:the answer is 3
Explanation:
Chris races his Audi north down a road for 1000 meters in 20 seconds, what is his velocity?
Answer:
I think it would be 50 I am not really sure
Explanation:
I think you would have to divid 1000 by 20 Again I'm not sure
In which medium does the light move faster, water or diamond?
color code of electrical resistors
Answer:
Tolerance: [tex]\pm 10\%[/tex]
Explanation:
Resistor Color Codes
Resistor Color Coding uses colored bands to quickly identify the resistive value or resistors and its percentage of tolerance.
Since the question does not provide a specific color table, we'll use the table attached below.
The colors of the resistor shown in the question are:
First band: orange
Second band: blue
Third band: brown
Fourth band: silver
The colors relate to the following numbers respectively:
3, 6, 10Ω, [tex]\pm 1\%[/tex]
The first two colors form the number 36
The third color is the multiplier: 36*10Ω = 360Ω
And the fourth color is the tolerance or the possible variation of the resistance [tex]\pm 1\%[/tex]
Resistance: 360Ω
Tolerance: [tex]\pm 10\%[/tex]
3. How are force, work, and power related?
Answer:
Work is the energy needed to apply a force to move an object a particular distance, where force is parallel to the displacement. Power is the rate at which that work is done.
Answer: Work is the energy needed to apply a force to move an object a particular distance, where force is parallel to the displacement. Power is the rate at which that work is done.
Explanation:
A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis
ball moving 30 m/s to the left. After the collision, the tennis ball is moving 34
m/s to the right. What is the velocity of the basketball after the collision?
Assume an elastic collision occurred.
O A. 11.4 m/s to the left
O B. 11.4 m/s to the right
O C. 1.4 m/s to the right
O D. 1.4 m/s to the left
Answer:
1.4 m/s to the left
Explanation:
just took it c:
A soccer ball is kicked with a velocity of 8 m/s at an angle of 23°. What is the
ball's acceleration in the vertical direction as it flies through the air?
A. -7.4 m/s2
B. O m/s2
C. 3.1 m/s2
D. -9.8 m/s2
Answer: -9.8 m/s2
Explanation:
What is the moment of inertia I of an object that rotates at 13.0 rev/min13.0 rev/min about an axis and has a rotational kinetic energy of 16.0 J?
Answer:
The moment of inertia of the object is 17.276 kilogram-square meters.
Explanation:
According to the statement, we find that object has rotation and no translation. From Rotation Physics we get that rotational kinetic energy ([tex]K_{R}[/tex]), measured in joules, is represented by the following formula:
[tex]K_{R} = \frac{1}{2}\cdot I_{G}\cdot \omega^{2}[/tex] (Eq. 1)
Where:
[tex]I_{G}[/tex] - Moment of inertia with respect to center of mass, measured in kilogram-square meters.
[tex]\omega[/tex] - Angular speed, measured in radians per second.
Now we clear the moment of inertia:
[tex]I_{G} = \frac{2\cdot K_{R}}{\omega^{2}}[/tex]
If we know that [tex]K_{R} = 16\,J[/tex] and [tex]\omega \approx 1.361\,\frac{rad}{s}[/tex], then the moment of inertia of the object is:
[tex]I_{G} = \frac{2\cdot (16\,J)}{\left(1.361\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]I_{G} =17.276\,kg\cdot m^{2}[/tex]
The moment of inertia of the object is 17.276 kilogram-square meters.
The moment of inertia of the object will be "17.276 kg/m²".
Moment of inertiaRotational Kinetic energy, [tex]K_R[/tex] = 16 J
Angular speed, ω = 1.361 rad/s
By using the Rotation Physics, the relation will be:
→ [tex]K_R[/tex] = [tex]\frac{1}{2}[/tex] × [tex]I_G[/tex] × ω²
the,
The moment of inertia be:
→ [tex]I_G[/tex] = [tex]\frac{2\times K_R}{\omega^2}[/tex]
By substituting the values, we get
= [tex]\frac{2\times 16}{(1.361)^2}[/tex]
= [tex]\frac{32}{(1.361)^2}[/tex]
= 17.276 kg.m²
Thus the above answer is correct.
Find out more information about Kinetic energy here:
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At time t = 0 the point at x = 0 has velocity v0 and displacement y0. The phase constant φ is given by tanφ =:
This question is incomplete, the complete question is;
The displacement of a string carrying a traveling sinusoidal wave is given by y(x,t)=ymsin(kx - ωt -φ) .
At time t = 0, the point at x = 0 has velocity v₀ and displacement y₀.
The phase constant φ is given by tanφ =:
A) ωv₀ /y₀
B) ωv₀ y₀
C) v₀ /ωy₀
D) y₀ /ωv₀
E) ωy₀ /v₀
Answer:
E) ωy₀ /v₀
Explanation:
Given that;
displacement of a wave is; y(x,t) = ym sin (kx - ωt - φ)
we differentiate the given equation with respect to time
d/dt (y(x,t)) = d/dt(ym sin(kx - ωt - φ) )
v(0,0)) = -ym ωcos (k(0) - ω(0) - φ) )
v₀ = -ym ωcos (-φ) ......... lets leave thisas equ 1
At t = 0, x = 0
the displacement of the wave is
y(0,0) = ym sin (k(0) - ω(0) - φ)
y₀ = ym sin(-φ) ..............let this be equ 2
y₀/v₀ = (ym sin(-φ)) / (-ym ωcos (-φ)) = ( -ym sin(φ)) / (-ym ωcos (φ))
(tanφ)/ω = y₀/v₀
tanφ = y₀ω/v₀
therefore the required value is y₀ω/v₀
option (E).
You measure the radius of a sphere as (6.45 ± 0.30) cm, and you measure its mass as (1.79 ± 0.08) kg. What is the density and uncertainty in the density of the sphere, in kilograms per cubic meter?
Answer:
[tex](1630.13\pm 300.10)\ kg/m^3[/tex]
Explanation:
Given that,
The radius of a sphere is (6.45 ± 0.30) cm
Mass of the sphere is (1.79 ± 0.08) kg
Density = mass/volume
For sphere,
[tex]d=\dfrac{m}{V}\\\\d=\dfrac{m}{\dfrac{4}{3}\pi r^3}\\\\d=\dfrac{1.79\ kg}{\dfrac{4}{3}\pi (6.4\times 10^{-2}\ m)^3}\\\\d=1630.13\ kg/m^3[/tex]
We can find the uncertainty in volume as follows :
[tex]\dfrac{\delta V}{V}=3\dfrac{\delta r}{r}\\\\=3\times \dfrac{0.3\times 10^{-2}}{6.45\times 10^{-2}}\\\\=0.1395[/tex]
Uncertainty in mass,
[tex]\dfrac{\delta m}{m}=\dfrac{0.08}{1.79}\\\\=0.0446[/tex]
Now, the uncertainty in density of sphere is given by :
[tex]\dfrac{\delta d}{d}=\dfrac{\delta m}{m}+\dfrac{\delta V}V}\\\\=0.0446+0.1395\\\\\dfrac{\delta d}{d}=0.1841\\\\\delta d=0.1841\times d\\\\\delta d=0.1841\times 1630.13\\\\\delta d = 300.10\ kg/m^3[/tex]
Hence, the density pf the sphere is [tex](1630.13\pm 300.10)\ kg/m^3[/tex]
You pull a wagon with a force of 20 N. The wagon has a mass of 10 kg. What is the wagon's acceleration?
Answer:
The answer is 2 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
f is the force
m is the mass
From the question
f = 20 N
m = 10 kg
We have
[tex]a = \frac{20}{10} \\ [/tex]
We have the final answer as
2 m/s²Hope this helps you
Answer:.
Explanation:.
Please provide explanation!!!
Thank you.
Answer:
(a) 102 cm/s
(b) 0.490 cm²
Explanation:
(a) Use Bernoulli equation.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0
½ ρ v₁² + ρgh₁ = ½ ρ v₂²
½ v₁² + gh₁ = ½ v₂²
½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²
v = 102 cm/s
(b) The flow rate is constant.
v₁ A₁ = v₂ A₂
(25.0 cm/s) (2.00 cm²) = (102 cm/s) A
A = 0.490 cm²
gold has a density of 19.32g/cm3. if you have a 25 cm3 sample of gold what is the mass of the sample
Answer:
ggggggggggggggggggggggggggggg
Explanation:
Answer:
The volume of the sample of gold is
16.51 [tex]cm^{3}[/tex]
Explanation:
The formula for density is:
D= [tex]\frac{M}{V}[/tex].
where:
D is density,
M is mass, and
V is volume.
Rearrange the density formula to isolate volume.
V= [tex]\frac{M}{D}[/tex]
V= [tex]\frac{318.97g Au}{19.32g cm^{3}}[/tex]
V= 318.97∅ × [tex]\frac{1 cm^{3} Au}{19.32g cm^{3} }[/tex]← Multiply by the multiplicative inverse of the density.
V= 16.51 cm³ Au.
A car is traveling south at 8.77 m/s. It then begins a uniform acceleration until it reaches a velocity of 47.8 m/s over a period of 3.84s. What is the car's acceleration?
Please help !
Answer:
The acceleration of the car is 10.16m/s²
Explanation:
Given parameters:
Initial velocity = 8.77m/s
Final velocity = 47.8m/s
Time duration = 3.84s
Unknown:
Acceleration of the car = ?
Solution:
To find the acceleration, we must bear in mind that this physical quantity is the change in velocity with time;
Acceleration = [tex]\frac{V - U}{T}[/tex]
V is the final velocity
U is the initial velocity
T is the time taken
Input the parameters and solve for acceleration;
Acceleration = [tex]\frac{47.8 - 8.77}{3.84}[/tex] = 10.16m/s²
The acceleration of the car is 10.16m/s²
Find the angle between the two unitless vectors: F1 = 8.92 i + 17.37 j F2 = 12.44 i + 7.11 j Answer in degrees, and to the fourth decimal place.
Answer:
θ = 33.0705°
Explanation:
The angle between the two vectors is given by the formula;
Cos θ = (F1 • F2)/(|F1| × |F2|)
We are given;
F1 = 8.92i + 17.37j
F2 = 12.44i + 7.11j
Thus;
Cos θ = [(8.92i + 17.37j) • (12.44i + 7.11j)]/[√(8.92² + 17.37²) × √(12.44² + 7.11²)]
Cos θ = (110.9648 + 123.5007)/(19.5265 × 14.3285)
Cos θ = 0.8380
θ = cos^(-1) 0.8380
θ = 33.0705°
21. A toy car starts from rest and begins to
accelerate at 11.0 m/s2. What is the toy car's
final velocity after 6.0 seconds?
Answer:
v = 66 m/s
Explanation:
Given that,
The initial velocity of a car, u = 0
Acceleration of the car, a = 11 m/s²
We need to find the final velocity of the toy after 6 seconds.
Let v is the final velocity. It can be calculated using first equation of motion. It is given by :
v = u +at
v = 0 + 11 m/s² × 6 s
v = 66 m/s
So, the final velocity of the car is 66 m/s.
Please provide an explanation.
Thank you!!
Answer:
(a) 22 kN
(b) 36 kN, 29 kN
(c) left will decrease, right will increase
(d) 43 kN
Explanation:
(a) When the truck is off the bridge, there are 3 forces on the bridge.
Reaction force F₁ pushing up at the first support,
reaction force F₂ pushing up at the second support,
and weight force Mg pulling down at the middle of the bridge.
Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)
∑τ = Iα
(Mg) (0.3 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L)
F₁ = ½ Mg
F₁ = ½ (44.0 kN)
F₁ = 22.0 kN
(b) This time, we have the added force of the truck's weight.
Using the same logic as part (a), we sum the torques about the second support:
∑τ = Iα
(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)
F₁ = ½ Mg + ⅔ mg
F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)
F₁ = 36.0 kN
Now sum the torques about the first support:
∑τ = Iα
-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)
F₂ = ½ Mg + ⅓ mg
F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)
F₂ = 29.0 kN
Alternatively, sum the forces in the y direction.
∑F = ma
F₁ + F₂ − Mg − mg = 0
F₂ = Mg + mg − F₁
F₂ = 44.0 kN + 21.0 kN − 36.0 kN
F₂ = 29.0 kN
(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)
As x increases, F₁ decreases and F₂ increases.
(d) Using our equation from part (c), when x = 0.6 L, F₂ is:
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)
F₂ = ½ Mg + mg
F₂ = ½ (44.0 kN) + 21.0 kN
F₂ = 43.0 kN
Answer:
a. Left support = Right support = 22 kNb. Left support = 36 kN Right support = 29 kNc. Left support force will decrease Right support force will increase.d. Right support = 43 kNExplanation:
given:
weight of bridge = 44 kN
weight of truck = 21 kN
a) truck is off the bridge
since the bridge is symmetrical, left support is equal to right support.
Left support = Right support = 44/2
Left support = Right support = 22 kN
b) truck is positioned as shown.
to get the reaction at left support, take moment from right support = 0
∑M at Right support = 0
Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0
Left support = 44 (0.3) + 21 (0.4)
0.6
Left support = 36 kN
Right support = weight of bridge + weight of truck - Left support
Right support = 44 + 21 - 36
Right support = 29 kN
c)
as the truck continues to drive to the right, Left support will decrease
as the truck get closer to the right support, Right support will increase.
d) truck is directly under the right support, find reaction at Right support?
∑M at Left support = 0
Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0
Right support = 44 (0.3) + 21 (0.6)
0.6
Right support = 43 kN
A chef places an open sack of flour on a kitchen scale. The scale reading of
40 N indicates that the scale is exerting an upward force of 40 N on the sack. The magnitude of this force equals the magnitude of the force of Earth’s gravitational attraction on the sack. The chef then exerts an upward force of
10 N on the bag and the scale reading falls to 30 N.Draw a free-body diagram of the latter situation.
Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:
40N
Explanation:
trust
in the figure shown if angle i increases slightly angle r will
Answer:
we need the image to do so.
Explanation:
sorry
An airplane, starting at rest, takes off on a 600. m long runway accelerating at a rate of 12 m/s/s. How many seconds does it take to reach the end of the runway?
Answer:
10 seconds
Explanation:
As it starts from rest, then u=0
and by III rd equation of motion: