Answer:
75
Explanation:
i am not sure but if 200N boy is being held back then the force that's holding him back must be equal to or greater than his weight. if 125N is already exerted then the tension will be:
T=200-125
= 75
Calculate area moment of inertia for a circular cross-section with 3 mm diameter:
Answer:
circles
A=7.07multiple 10-6 m2
I hope you understand and help
The area of the circular cross-section will be 7068 × [tex]10^{-6}m^{2}[/tex].
What is the area?The measurement that represents the size of a region on a plane or curved surface is called an area.
What is cross-section?A cross-section would be the non-empty point where a solid body intersects a plane in three dimensions or its equivalent in higher dimensions.
Given data:
Diameter = 3 × [tex]10^{-3} m[/tex].
It is known that. Diameter = 2 radius.
The area can be calculated by using the formula:
A = 1/4 [tex]\pi[/tex][tex]d^{2}[/tex] = 1/4 (3.14) [tex](3 * 10^{-3})^{2}[/tex]= 7068 × [tex]10^{-6}m^{2}[/tex].
To know more about area and cross-section
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write a use of magnetic force and frictional force each
write state of matter with 5 example of each
There are broadly 3 states of matter (there are 5, but they don't teach 2 of them at school).
1. Solid
Examples: Iron, wood, steel, ice, paper
2. Liquid
Examples: Water, mercury, milk, soup, juice
3. Gas
Examples: Oxygen, Chlorine, Carbon dioxide, Sulphur dioxide, Nitrogen
With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how many grid marks is the wavelength of the wave (use the pause button and step button as you need to in order to get a good measure, and round to the nearest whole grid mark)?
Answer:
The wavelength stays the same.
Explanation:
When the amplitude is increased, the wavelength stays the same.
Here the wavelength doesn't depend upon the amplitude.
What is the feature known as the "Great Dark Spot" of Neptune? It is an apparently permanent feature about five times the size of Earth, similar to the Great Red Spot of Jupiter, near Neptune's south pole. It was a dark hole in the upper atmosphere left by the collision of the comet Shoemaker-Levy 9. It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years. It is a dark surface feature on the surface snow layers caused by radiation discoloration of the older layers. It is a permanent discoloration of the north polar region of Neptune caused by locally prevailing lower surface temperatures there.
Answer:
It was an apparently temporary feature about the size of Earth, similar to the Great Red Spot of Jupiter, but disappeared within a few years.
Explanation:
The Great Dark Spot of Neptune was an immense spinning storm in the southern atmosphere of Neptune. The size of the entire Earth, it had the strongest winds ever recorded on any planet in the solar system. It was discovered by the Voyager 2 spacecraft in 1989, but by 1994 the Hubble Space Telescope saw it was gone.
The Great Red Spot is a storm found in Jupiter's southern hemisphere, with similar characteristics to the Great Dark Spot.
Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are doubled and their separation is halved, what happens to the electrical force that each charge exerts on the other one
Answer:
F' = 16 F
Hence, the electric force between charges becomes sixteen times its initial value.
Explanation:
The electric force between the two charges is given by the Colomb's Law:
[tex]F = \frac{KQ_1Q_2}{R^2}[/tex] ------------------- eq(1)
where
F = electric force
K = Colomb's Constant
Q₁ = magnitude of the first charge
Q₂ = magnitude of the second charge
R = Distance between charges
Now the magnitudes of the charges are doubled and the distance between them is halved. Therefore:
[tex]F' = \frac{K(2Q_!)(2Q_2)}{(\frac{R}{2})^2}\\\\F' = 16 \frac{KQ_1Q_2}{R^2}[/tex]
using equation (1):
F' = 16 F
Hence, the electric force between charges becomes sixteen times of its initial value.
which option is correct n why?
6. The projectile motion is a good example of
A. one dimensional motion.
B. two dimensional motion.
C. three dimensional motion.
D. four dimensional motion.
2. two dimensional motion
Because it has just 2 dimensions x and y
A mass of 240 grams oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm.
a. What is the effective spring constant for this motion?
b. How much energy is involved in this motion?
Answer:
(a) The spring constant is 59.23 N/m
(b) The total energy involved in the motion is 0.06 J
Explanation:
Given;
mass, m = 240 g = 0.24 kg
frequency, f = 2.5 Hz
amplitude of the oscillation, A = 4.5 cm = 0.045 m
The angular speed is calculated as;
ω = 2πf
ω = 2 x π x 2.5
ω = 15.71 rad/s
(a) The spring constant is calculated as;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m[/tex]
(b) The total energy involved in the motion;
E = ¹/₂kA²
E = (0.5) x (59.23) x (0.045)²
E = 0.06 J
The paper dielectric in a paper-and-foil capacitor is 0.0785 mm thick. Its dielectric constant is 2.35, and its dielectric strength is 49.5 MV/m. Assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.
Required:
a. What area of each plate is required for for a 0.300 uF capacitor?
b. If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor?
Answer:
a) required area is 1.1318 m²
b) the maximum potential difference that can be applied across the compactor is 1931.1 V
Explanation:
Given the data in the question;
dielectric constant εr = 2.35
distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m
dielectric strength = 49.5 MV/m
a)
given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F
To find the Area, we use the following the expression.
C = ε₀εrA / d
we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹² (F/m)
we substitute
0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A ] / 7.85 × 10⁻⁵
A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]
A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹
A = 1.1318 m²
Therefore, required area is 1.1318 m²
b)
the maximum potential difference that can be applied across the compactor.
We use the following expression;
⇒ 1/2 × dielectric strength × thickness d
we substitute
⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )
⇒ 1931.1 V
Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V
The speed of a car decreases uniformly as it passes a curve point where normal component of acceleration is 4 ft/sec2. If the car total acceleration of 5ft/sec2 is the same as it passes a hump, the tangential component of acceleration is _______________ ft/sec2.
Answer:
45
Explanation:
ft/sec2
If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be
Hz 50
Hz 70.7
Hz 100
Hz 25
Answer:
100Hz
Explanation:
In a full wave rectifier, the fundamental frequency of the ripple is twice that of input frequency. Given the input frequency of 50 Hz, the fundamental frequency will be 2 × 50 = 100Hz
Answer:
HZ 100 is the right answer hope you like it
Outline five ways of varying the force on a current-carrying conductor in a magnetic field. (7 marks)
1. A flywheel begins rotating from rest, with an angular acceleration of 0.40 rad/s. a) What will its angular velocity be 3 seconds later? b) What angle will it have turned through in that time?
Answer:
(a) 1.2 rad/s
(b) 1.8 rad
Explanation:
Applying,
(a) α = (ω-ω')/t................ Equation 1
Where α = angular acceleration, ω = final angular velocity, ω' = initial angular velocity, t = time.
From the question,
Given: α = 0.40 rad/s², t = 3 seconds, ω' = 0 rad/s (from rest)
Substitute these values into equation 1
0.40 = (ω-0)/3
ω = 0.4×3
ω = 1.2 rad/s
(b) Using,
∅ = ω't+αt²/2.................. Equation 2
Where ∅ = angle turned.
Substitutting the values above into equation 2
∅ = (0×3)+(0.4×3²)/2
∅ = 1.8 rad.
I NEED YOUR HELP!!
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.
52 cm4
72 cm4
32 cm4
24 cm4
2 cm4
Answer:
Minimum area of rectangle = 24 centimeter²
Explanation:
Given:
Length of rectangle = 6 centimeter
Width of rectangle = 4 centimeter
Find:
Minimum area of rectangle
Computation :
Minimum area of rectangle = Length of rectangle x Width of rectangle
Minimum area of rectangle = 6 x 4
Minimum area of rectangle = 24 centimeter²
One ball is aprojected in the uapward directon with a certain velocity ‘v’ and other
is thrown downwards with the same velocity
Complete question is;
One ball is projected in the upward direction with a certain velocity ‘v’ and other is thrown downwards with the same velocity at an angle θ.
The ratio of their potential energies at highest points of their journey, will be:
Answer:
u² : (u cos θ)²
Explanation:
Maximum potential energy for the first ball will be at a maximum height of;
H = u²/2g
Thus;
PE = mg(u²/2g)
For second ball at an angle θ, maximum PE will occur at a max height of (u cos θ)²/2g
PE = mg((u cos θ)²/2g)
The ratios of the potential energies are;
mg(u²/2g) : mg((u cos θ)²/2g)
mg will will cancel out since they are of same mass.
Thus;
(u²/2g) : (u cos θ)²/2g
Again 2g will cancel out to give;
u² : (u cos θ)²
A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a magnetic force of 0.16 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 72.0° with respect to the wire.
Answer:
7.28×10⁻⁵ T
Explanation:
Applying,
F = BILsin∅............. Equation 1
Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°
Substitute these values into equation 2
B = 0.16/(68×34×sin72°)
B = 0.16/(68×34×0.95)
B = 0.16/2196.4
B = 7.28×10⁻⁵ T
distance of distinct vision.
is placed at a distance less than the distance of near point, its image o
will be blurred. Hence human eye can not see such object clearly.
ADDITIONAL INFORMATION
distance of distinct vision for a normal eye of different age groups
Babies = 7 cm
Adults = 25 cm
erson of age 55 years and above = 100 cm
ever, in our discussion we are concerned with a normal eye of an adult so least
The foulart position of an ahiect from a human eve so that the sh
The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.
Answer:
25 you said ? thats incorecct
Explanation:
How much work does the electric field do in moving a proton from a point with a potential of 170 V to a point where it is -50 V
Urgent please help me
1433 km
Explanation:
Let g' = the gravitational field strength at an altitude h
[tex]g' = G\dfrac{M_E}{(R_E + h)^2}[/tex]
We also know that g at the earth's surface is
[tex]g = G\dfrac{M_E}{R_E^2}[/tex]
Since g' = (2/3)g, we can write
[tex]G\dfrac{M_E}{(R_E + h)^2} = \dfrac{2}{3}\left(G\dfrac{M_E}{R_E^2} \right)[/tex]
Simplifying the above expression by cancelling out common factors, we get
[tex](R_E + h)^2 = \dfrac{3}{2} R_E^2[/tex]
Taking the square root of both sides, this becomes
[tex]R_E + h = \left(\!\sqrt{\dfrac{3}{2}}\right) R_E[/tex]
Solving for h, we get
[tex]h = \left(\!\sqrt{\dfrac{3}{2}} - 1\right) R_E= 0.225(6.371×10^2\:\text{km})[/tex]
[tex]\:\:\:\:\:= 1433\:\text{km}[/tex]
Electricity is distributed from electrical substations to neighborhoods at 13000 V. This is a 60 Hz oscillating (AC) voltage. Neighborhood transformers, seen on utility poles, step this voltage down to the 120 V that is delivered to your house.
Required:
a. How many turns does the primary coil on the transformer have if the secondary coil has 130 turns?
b. No energy is lost in an ideal transformer, so the output power Pout from the secondary coil equals the input power Pin to the primary coil. Suppose a neighborhood transformer delivers 280 A at 120 V. What is the current in the 1.3×10^4 V line from the substation?
Answer:
a) N₁ = 14083 turns, b) I₁ = 2.58 A
Explanation:
The relationship that describes the relationship between the primary and secondary of the transformer is
[tex]\frac{V_2}{N_2} = \frac{V_1}{N_1}[/tex]
a) They indicate that the secondary has N2 = 130 turns, the turns of the primary are
N₁ = [tex]N_2 \frac{V_1}{V_2}[/tex]
N₁ = [tex]130 \ \frac{13000}{120}[/tex]
N₁ = 14083 turns
b) since there are no losses, the power of the neighboring transformer is
P = V I
P = 120 280
P = 33600 W
this is the same power of the substation
P = V₁ I₁
I₁ = P / V₁
I₁ = 33600/13000
I₁ = 2.58 A
What is the meant of by renewable energy and non-renewrable with example of each.
Answer:
Renewable energy is a type of energy that can be renewed easily, such as sunlight. By using Solar panels to collect the suns energy, we are not depleting it, so this source is renewable.
Non-renewable energy is something that cannot easily be replenished. An example would be oil because oil takes millions of years to form and cannot be renewed easily.
Me Ayudan con este ejercicio por favor !!!
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?
Answer:
The correct answer would be - Low pitch.
Explanation:
As it is known that if frequency increases then pitch will be increase as well as pitch depends on frequency, Now for the question it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases. As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
A car travelling at 14.0 m/s approaches a traffic light. The driver applies the brakes and is able to come to halt in 5.6 s. Determine the average acceleration of the car during this time interval.
Answer:
[tex]a=2.5\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the car, u = 14 m/s
Finally, it comes to rest, v = 0
Time, t = 5.6 s
We need to find the average acceleration of the car during this time interval. We know that,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-14}{5.6}\\\\a=-2.5\ m/s^2[/tex]
So, the acceleration of the car is [tex]2.5\ m/s^2[/tex] in the opposite direction of motion.
A 1500kg car start from rest and increases it velocity to 30mls in a time of 25sec. calculate the distance the car travel, how much force was use, how much work was done.
Answer:
workdone= 1/2mv^2
1/2×1500×30^2
675000J
In a similar rolling race (no slipping), the two objects are a solid cylinder and hollow cylinder of the same radius and mass. Which reaches the bottom first
Answer:
solid cylinder
Explanation:
the object that arrives first is the object that has more speed, let's use the concepts of energy
starting point. Highest point
Em₀ = U = m g h
final point. Lowest point
Em_f = K = ½ mv² + ½ I w²
since the body has rotational and translational movement
how energy is conserved
m g h = ½ mv² + ½ I w²
linear and angular velocity are related
v = w r
w = v / r
we substitute
m g h = ½ mv² + ½ I (v/r) ²
mg h = ½ v² (m + I /r²)
v = [tex]\sqrt{2gh \ \frac{m}{m + \frac{I}{r^2} } }[/tex]
the tabulated moments of inertia for the bodies are
solid cylinder I = ½ m r²
hollow cylinder I = m r²
we look for the speed for each body
solid cylinder
v₁ = [tex]\sqrt {2gh} \ \sqrt{\frac{m}{m + m/2} }[/tex]
v₁ = [tex]\sqrt{2gh} \ \sqrt{2/3}[/tex]
let's call v₀ = [tex]\sqrt{2gh}[/tex]
v₁ = 0.816 v₀
hollow cylinder
v₂ = [tex]\sqrt{2gh } \ \sqrt{\frac{m}{m+ m} }[/tex]
v₂ = v₀ √½
v₂ = 0.707 v₀
Therefore, the body that has the highest speed is the solid cylinder and since time is the inverse of speed, this is the body that spends less time to reach the bottom, that is, it is the first to arrive
derive expression for pressure exerted by gas
A car driving down a road runs of gas and will eventually stop because of:
A. Friction
B. Thrust
C. It will remain in motion forever
OD. Gravity
A wire carrying a 23.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's field, and there is a 2.45 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet?
Answer:
3.55 T
Explanation:
Applying,
F = BILsin∅.............. Equation 1
Where F = Force, B = magnetic Field, I = current, L = Length of the wire, ∅ = Angle between the wire and the magnetic field
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 2.45 N, L = 3.00 cm = 0.03 m, I = 23.0 A, ∅ = 90° (Perpendicular)
Substitute these values into equation 2
B = 2.45/(0.03×23×sin90)
B = 2.45/0.69
B = 3.55 T
what can you do to keep you BMI under weight
Here are some healthy ways to gain weight when you're underweight:
Eat more frequently. When you're underweight, you may feel full faster. Eat five to six smaller meals during the day rather than two or three large meals.
Choose nutrient-rich foods. As part of an overall healthy diet, choose whole-grain breads, pastas and cereals; fruits and vegetables; dairy products; lean protein sources; and nuts and seeds.
Try smoothies and shakes. Don't fill up on diet soda, coffee and other drinks with few calories and little nutritional value. Instead, drink smoothies or healthy shakes made with milk and fresh or frozen fruit, and sprinkle in some ground flaxseed. In some cases, a liquid meal replacement may be recommended.
Watch when you drink. Some people find that drinking fluids before meals blunts their appetite. In that case, it may be better to sip higher calorie beverages along with a meal or snack. For others, drinking 30 minutes after a meal, not with it, may work.
Make every bite count. Snack on nuts, peanut butter, cheese, dried fruits and avocados. Have a bedtime snack, such as a peanut butter and jelly sandwich, or a wrap sandwich with avocado, sliced vegetables, and lean meat or cheese.
Top it off. Add extras to your dishes for more calories — such as cheese in casseroles and scrambled eggs, and fat-free dried milk in soups and stews.
Have an occasional treat. Even when you're underweight, be mindful of excess sugar and fat. An occasional slice of pie with ice cream is OK. But most treats should be healthy and provide nutrients in addition to calories. Bran muffins, yogurt and granola bars are good choices.
Exercise. Exercise, especially strength training, can help you gain weight by building up your muscles. Exercise may also stimulate your appetite.
