''In general, the normal force is not equal to the weight." Give an example where the two forces are equal in magnitude and at least two examples where they are not.

Answer:

A. The resonator behaves as a wave guide (a hollow pipe used as a transmission line). The characteristics of the pipe depend on the type of the wave to be transmitted.

Answer:

The slopes for each of the four line segments are \(a_{A} = 6\,\frac{m}{min^{2}}\), \(a_{B} = 0\,\frac{m}{min^{2}}\), \(a_{C} = -4\,\frac{m}{min^{2}}\) and \(a_{D} = 2.667\,\frac{m}{min^{2}}\), respectively.

Explanation:

There are four line segments:

(i) Line A: \(v(0\,min) = 0\,\frac{m}{min}\), \(v(10\,min) = 60\,\frac{m}{min}\)

(ii) Line B: \(v(10\,min) = 60\,\frac{m}{min}\), \(v(15\,min) = 60\,\frac{m}{min}\)

(iii) Line C: \(v(15\,min) = 60\,\frac{m}{min}\), \(v(40\,min) = -40\,\frac{m}{min}\)

(iv) Line D: \(v(40\,min) = -40\,\frac{m}{min}\), \(v(55\,min) = 0\,\frac{m}{min}\)

The slope of each line segment represents the acceleration of the particle, which can calculated by the geometrical concept of secant line. Hence, we proceed to determine the acceleration associated with each line segment:

Line A

\(a_{A} = \frac{v(10\,min)-v(0\,min)}{10\,min-0\,min}\)

\(a_{A} = \frac{60\,\frac{m}{min}-0\,\frac{m}{min}}{10\,min-0\,min}\)

\(a_{A} = 6\,\frac{m}{min^{2}}\)

Line B

\(a_{B} = \frac{v(15\,min)-v(10\,min)}{15\,min-10\,min}\)

\(a_{B} = \frac{60\,\frac{m}{min}-60\,\frac{m}{min} }{15\,min-10\,min}\)

\(a_{B} = 0\,\frac{m}{min^{2}}\)

Line C

\(a_{C} = \frac{v(40\,min)-v(15\,min)}{40\,min-15\,min}\)

\(a_{C} = \frac{-40\,\frac{m}{min}-60\,\frac{m}{min} }{40\min-15\,min}\)

\(a_{C} = -4\,\frac{m}{min^{2}}\)

Line D

\(a_{D} = \frac{v(55\,min)-v(40\,min)}{55\,min-40\,min}\)

\(a_{D} = \frac{0\,\frac{m}{min}-\left(-40\,\frac{m}{min} \right) }{55\,min-40\,min}\)

\(a_{D} = 2.667\,\frac{m}{min^{2}}\)

The slopes for each of the four line segments are \(a_{A} = 6\,\frac{m}{min^{2}}\), \(a_{B} = 0\,\frac{m}{min^{2}}\), \(a_{C} = -4\,\frac{m}{min^{2}}\) and \(a_{D} = 2.667\,\frac{m}{min^{2}}\), respectively.

Answer: the required mass is 1.7628 × 10²⁵ μ

Explanation:

Given that;

energy released in the fission of one ²³⁵U₉₂ is 206.6 MeV

power p = 28.7 Mega Watts = 28.7 × 10⁶ W = 28.7 × 10⁶/ 1.6× 10⁻¹³ = 17.9375 × 10¹⁹ MeV/s

now fission need per second will be;

⇒ power / energy released i  fission

= 17.9375 × 10¹⁹ MeV /  206.6 MeV = 8.68 × 10¹⁷ per second

now fission need per day will be;

⇒ ( 8.68 × 10¹⁷ × 360 × 24 ) = 7.5 × 10²² per day

hence mass of ²³⁵U₉₂ that is consumed in one day in this reactor will be;

⇒ (235 × 7.5 × 10²²)μ

= 1.7628 × 10²⁵ μ

Therefore the required mass is 1.7628 × 10²⁵ μ

Answer: please see below

Explanation:

A manned space exploration is defined as the exploration of individuals --- astronauts in space using a spacecraft as a vehicle and are  responsible for operating its controls

The extreme conditions in space that challenge manned space exploration is as follows.

1. extreme loud sound waves cause by the launch of spacecraft which can shatter the spacecraft

2. extreme Temperatures in space ranging from extreme hot temperatures  (near the sun) to extreme cold temperatures  ( below freezing point out of space.

3.micrometeorite showers responsible for sandblasting can  damage spacecraft.

4.Ultra violet Radiation which can alter the control unit of the spacecraft  

Because of theses extreme conditions that pose challenges to space explorations, necessary precautions should be taken into consideration to be able to overcome such challenges. These precautions include building the spacecraft and the control unit in such a way that can resist these harmful conditions, also taking in mind safe escape routes for the astronauts in case of failures.

Given:

The capacitor C1 = 1.29 F

The capacitor C2 = 3.17 F

The capacitor C3 =8.36 F

Capacitors C1 and C2 are connected in series while C3 is connected in parallel.

To find the capacitance of the total combination.

Explanation:

The capacitors connected in series can be calculated by the formula

So, the equivalent capacitance of C1 and C2 will be

The capacitors connected in parallel can be calculated by the formula

On substituting the values, the capacitance of the total combination will be

Thus, the total capacitance of the combination is 9.28 F

Answer:

(a) Area(small piston)/Area(large piston) = 0.037

(b) h = 1336.36 cm = 13.36 m

Explanation:

(a)

The stress on the smaller piston is equally transmitted to the larger piston, in a hydraulic lift. Therefore,

Stress (small piston) = Stress (large piston)

Force (small piston)/Area (small piston) = Force (Large Piston)/Area (Large Piston)

Area(small piston)/Area(large piston) = Force (small piston)/Force(Large piston)

Area(small piston)/Area(large piston) = 550 N/(1500 kg)(9.8 m/s²)

Area(small piston)/Area(large piston) = 0.037

(b)

The work is also transmitted equally to the large piston. So,

Work(small piston) = Work(Large Piston)

Force(small piston).Displacement(small piston) = Force(large piston).Displacement(small piston)

(550 N)(h) = (1500 kg)(9.8 m/s²)(50 cm)

h = 735000 N.cm/550 N

h = 1336.36 cm = 13.36 m

(a) The ratio of area smaller piston to area of larger piston is 0.037.

(b) The distance the smaller piston will be pushed down to lift the car at the given height is 13.36 m.

The given parameters;

let the area of the small piston = A₁

let the area of the large piston = A₂

Apply constant pressure principle as shown below;

\(P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\\frac{A_1}{A_2} = \frac{A_{small}}{A_{large}} = \frac{F_1}{F_2} = \frac{550}{mg} \\\\ \frac{A_{small}}{A_{large}} = \frac{550}{1500 \times 9.8} \\\\ \frac{A_{small}}{A_{large}} = 0.037\)

The distance the effort will be applied is calculated as follows;

\(550 d = mgh\\\\550d = (1500 \times 9.8 \times 0.5)\\\\550 d = 7350\\\\d = \frac{7350}{550} \\\\d = 13.36 \ m\)

Thus, the distance the smaller piston will be pushed down to lift the car at the given height is 13.36 m.

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Answer:

C. Y

Explanation:

From Newton's second law of motion, we know that:

       Force = mass x acceleration

So;

      acceleration  = \(\frac{Force }{mass}\)  

 Therefore, to have the highest acceleration at a constant force, the mass must be low. Acceleration is inversely proportional to mass.

Y has the least mass and it will have the highest acceleration

Additional mass (payload)  of a helium baloon lift is , \(M_{payload}=853.47kg\)

What is mass?

The amount of matter in a particle or object is represented by the dimensionless quantity mass (symbolised m). The kilogram is the International System's (SI) preferred unit of mass (kg).

The weight of the helium gas, the skin, other components of the balloon's construction, as well as the payload, must be balanced by the buoyant force. Using this, we determine the necessary payload mass.

Given pair=  \(1.29kg/m^{3}\)

buoyant force b = V*pair g

\(=(\frac{4}{3} *\pi R^{3} ) pair*g\\=20179.87N\)

mass of helium is,

\(m=(\frac{4}{3} * \pi R^{3} )(0.179)\\m=285.7kg\)

so total mass, \(M=m+920kg\)

\(M=1205.7kg\)

so, b = \((M+M_{payload}) g\)

\(M_{payload}=853.47kg\)

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Explanation:

We start by using the conservation law of energy:

\(\Delta{K} + \Delta{U} = 0\)

or

\(\dfrac{1}{2}mv^2 - G\dfrac{mM}{r} = 0\)

Simplifying the above equation, we get

\(v^2 = 2G\dfrac{M}{r}\)

We can rewrite this as

\(v^2 = 2\left(G\dfrac{M}{r^2}\right)r\)

Note that the expression inside the parenthesis is simply the acceleration due to gravity \(g\) so we can write

\( v^2 = 2gr\)

where \(v\) is the launch velocity.

To calculate the average deceleration while a person is under the water, we need to know the initial velocity, final velocity, and the time taken for deceleration.

Let's assume the person is initially moving with a velocity v0, and they come to a stop in a distance d underwater. We also need to know the time taken for this deceleration, denoted as t.

The average deceleration (a) can be calculated using the formula:

a = (v - v0) / t

where v is the final velocity, which is 0 since the person comes to a stop.

Since we don't have specific values for v0, v, and t, we cannot provide a numerical answer. However, if you provide the required values, I would be able to calculate the average deceleration for you.

Answer:

71.85 m/s

Explanation:

Given the following :

Length of skid marks left by jaguar (s) = 290 m

Skidding Acceleration (a) = - 8.90m/s²

Final velocity of jaguar (v) = 0

Speed of Jaguar before it Began to skid =?

Hence, initial speed of jaguar could be obtained using the formula :

v² = u² + 2as

Where

v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar

0² = u² + (2 × (-8.90) × 290)

0 = u² + (-5,162)

u² = 5162

Take the square root of both sides

u = √5162

u = 71.847 m/s

u = 71.85m/s

friction is the force that resists the sliding or rolling of an object over another.

Answer:

Because the truck is moving to the east faster than the ball is moving west.

Explanation:

In this problem the text reads:

Ball                                          Truck

Rolling west at 0.25 m/s Moving east at 1.0 m/s

You get asked if the observer in the truck would see the ball moving to the west, the answer is yes, cause the observer in the truck would be moving to the east along with the truck, but someone looking at the ball from the outside will see the ball moving east because the speed of the ball is less than that of the truck.

Answer:

An atom of sodium (Na) donates one of its electrons to an atom of chlorine (Cl) in a chemical reaction, and the resulting positive ion (Na+) and negative ion (Cl−) form a stable ionic compound (sodium chloride; common table salt) based on this ionic bond.

Explanation:

All the info i can give you

Explanation:

180 million km = 10 min

? = 1 min

180 million x 10 = 1,800,000,000 km

180 million km = 600s

? = l s

108,000,000,000km

The particle’s average acceleration is determined as 1.16 m/s².

The average acceleration refers to the rate at which the velocity changes.

Mathematically, average acceleration is given as;

a = Δv/Δt

where;

a = (V2 - V1) / (t)

Given,;

a = (7.81 - - 4.67)/10.8

a = (7.81 + 4.67) / 10.8

a = 1.16 m/s²

Thus, the particle’s average acceleration is determined as 1.16 m/s².

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The electric field at point B, located at the midpoint between the upper left and right corners of the square, can be approximated as 3.244 x \(10^6\) N/C in the upward direction.

To find the electric field at point B, we can consider the contributions from the two charges placed at the lower corners of the square. Since the charges are the same and the distance to point B is the same for both charges, the magnitudes of the electric fields produced by each charge will be equal.

First, let's calculate the magnitude of the electric field produced by one of the charges at point B using Coulomb's Law:

Electric field due to a point charge (E) = (k * q) / \(r^2\)

Where:

- k is the electrostatic constant, approximately equal to 8.99 x 10^9 \(10^9\)N \(m^2/C^2\)

- q is the charge of the object, 7.25 μC (7.25 x \(10^-^6\) C)

- r is the distance from the charge to point B, which is half the length of the square's side, 0.201 m / 2 = 0.1005 m

Plugging in the values, we have:

E = (8.99 x \(10^9 N m^2/C^2\) * 7.25 x \(10^-^6\) C) / (0.1005 \(m)^2\)

E ≈ 1.622 x \(10^6\) N/C

Since the electric fields from the two charges at the lower corners have equal magnitudes and point in the same direction (upward), the total electric field at point B is twice the magnitude of the individual electric field:

Electric field at point B = 2 * E ≈ 2 * 1.622 x \(10^6\) N/C

Electric field at point B ≈ 3.244 x \(10^6\) N/C

Therefore, the electric field at point B, midway between the upper left and right corners of the square, is approximately 3.244 x \(10^6\)N/C upward.

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Answer:

A homogeneous mixture is a type of mixture in which the composition is uniform and every part of the solution has the same properties. Example, air

Explanation:

The contact force (in N) between each pair of boxes is mathematically given as

F_{ab} = 3.56 N

F_{bc} = 1.2 N

Generally, Any force that is generated as a consequence of two objects coming into touch with one another is referred to as a contact force. Contact forces are present everywhere and are the cause of the vast majority of macroscopic groupings of matter's obvious interactions with one another.

In conclusion, The equation for is  Acceleration of the system is mathematically given as

a = 8.25 / (5.85 + 2.95 + 1.50)

a= 0.8 m/s^2

Therefore

F_{ab} = (Mb + Mc)*a

F_{ab} = (2.95 + 1.50) * 0.8

F_{ab} = 3.56 N

F_{bc} = Mc * a

F_{bc}= 1.5 * 0.8

F_{bc} = 1.2 N

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Answer:

6.53 m/s²

Explanation:

Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.

Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:

T = m₁a               (1)

m₂g - T = m₂a    (2)

substituting T = m₁a in equation 2:

m₂g - m₁a = m₂a

m₂a + m₁a = m₂g

a(m₁ + m₂) = m₂g

a = m₂g / (m₁ + m₂)

a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²

Both objects have an acceleration of 6.53 m/s²

The speed of the artillery shell when it hits its target is 100 m/s.

Given:

Initial vertical displacement (y) = 200 m

Vertical displacement at 100 m in the air (y') = 100 m

Final velocity in the vertical direction (vy') = 0 m/s (at the highest point of the trajectory)

Using the equation for vertical displacement in projectile motion:

y' = vy^2 / (2g),

where g is the acceleration due to gravity (approximately 9.8 m/s^2), we can solve for the initial vertical velocity (vy).

100 m = vy^2 / (2 * 9.8 m/s^2),

vy^2 = 100 m * 2 * 9.8 m/s^2,

vy^2 = 1960 m^2/s^2,

vy = sqrt(1960) m/s,

vy ≈ 44.27 m/s.

Now, since the horizontal motion is independent of the vertical motion, the horizontal speed of the shell remains constant throughout its trajectory. Therefore, the speed of the shell when it hits its target is 100 m/s.

Hence, the speed of the artillery shell when it hits its target is 100 m/s.

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The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².

To calculate the average acceleration of the motorcycle, we can use the formula:

Average acceleration = (final velocity - initial velocity) / time

First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.

Converting the final velocity:

Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s

Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:

Average acceleration = (21.67 m/s - 0 m/s) / time

To find the time taken to reach this velocity, we need to use the formula for average speed:

Average speed = total distance/time

Rearranging the formula:

time = total distance / average speed

Plugging in the values:

time = 50 m / 21.67 m/s ≈ 2.31 seconds

Now we can calculate the average acceleration:

Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²

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A 3m beam of negligible weight is balancing in equilibrium with a fulcrum placed 1m from its left end. If a force of 50N is applied on it's right end, The amount of force would need to be applied to the left end is (F)= 100N

Force is a physical phenomena that help of a object to change its motion and move its position from one end to another. It is a vector quantity. It can be measured in Newton.

To calculate the force we are using the formula,

F= f×(d₂-d₁)

Here we are given,

f= The initial force affect on the weight. = 50N

d₁= The primary distance from the weight.=1m

d₂= The secondary distance from the weight.=3m

We have to calculate the force applied on the left end.=F

Now we put the values in above equation, we get

F= f×(d₂-d₁)

Or, F= 50×(3-1)

Or, F= 100N

From the above discussion we can say that, The amount of force would need to be applied to the left end is (F)= 100N

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F = 980 N

h = 20 m

t = 25 s

P=? (power)

W=F*h   (work)

P=W*t  

P=F*h*t

P=980*20*25 =490000 W = 490 kW = 0.49 MW

Please i have asked a.question please help.me

Explanation:

Please i have asked a question in math please help me

Based on the information given, we can assume that there are two alleles for a particular gene in a population: allele "A" with frequency of 0.19 and allele "a" with frequency of 0.81.

What is the frequency?

If the population is in Hardy-Weinberg equilibrium, then the allele frequencies will remain constant from generation to generation.

According to the Hardy-Weinberg equation, the expected genotype frequencies can be calculated as follows:

What is genotype?

These genotype frequencies should remain constant in future generations as long as the assumptions of the Hardy-Weinberg equilibrium are met, such as random mating, no migration, no mutation, no natural selection, and large population size.

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