Because the human body is a holistic organism, this means that



A. Each person reacts to stress the same way



B. Physical injuries do not affect mental processes



C. Religion plays a key role in how people react to stress



D. A negative effect in one area can affect others areas

The distances traveled by the boy will be 4 m. Newton's second equation of motion is applied.

Distance is a numerical representation of the distance between two objects or locations.

The distance can refer to a physical length or an estimate based on other factors in physics or common use. |AB| is a symbol for the distance between two points A and B.

Given data;

acceleration(a)=2m/s²

Time (t)=2 s

u(Initial velocity)=0

The distance traveled by the boy will be;

\(\rm S=ut+\frac{1}{2} at^2 \\\\ \rm S=0+\frac{1}{2} (2)(2)^2 \\\\ S=4 \ m\)

Hence, the distance traveled by the boy will be 4 m.

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Answer:

Option A. 500

Explanation:

From the question given, we obtained the following:

Numbers of home with pet dogs = 250

Percentage with pet dogs = 50%

Total number of homes =.?

Thus, we can obtain the total number of homes represented by the chart as follow:

Let T be the total number of Homes.

Number of homes with dog = percentage of homes with dog × total number

Numbers of home with pet dogs = 250

Percentage with pet dogs = 50%

Total number of homes = T

250 = 50% × T

Divide both side by 50%

T = 250 ÷ 50%

T = 250 ÷ 50/100

T = 250 × 100/50

T = 250 × 2

T = 500

Therefore, the total number of homes represented by the chart is 500.

As the vehicle turns, the tires on one side will leave marks on the ground while the tires on the other side do not. Therefore, only the tire marks from the turning side are visible.

When a vehicle travels in a curved path, typically only two tire marks are visible. This is because most vehicles have four tires, with two tires on each side. As the vehicle turns, the tires on one side will leave marks on the ground while the tires on the other side do not. Therefore, only the tire marks from the turning side are visible. therefore when a vehicle travels in a curved path, typically only two tire marks are visible.

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The equation of motion is given by:

x = e^(-t) [C1 cos 2t + C2 sin 2t] - 4/21 cos 2t + 8/21 sin 2t.

Mass, m = 1 kg

Spring constant, k = 5 N/m

Initial displacement, x = -1 m

Velocity, v = -7 m/s

The damping force is numerically equal to 2 times the instantaneous velocity.

So, damping coefficient, b = 2 m/s

For the given system, the equation of motion can be written as:

m d²x/dt² + b dx/dt + kx = f(t)

where,

m is the mass

b is the damping coefficient

k is the spring constant

f(t) is the external force

On substituting the given values in the equation of motion, we get:

1 d²x/dt² + 2 dx/dt + 5x = 16 cos 2t + 4 sin 2t ........ (i)

Let us assume a particular solution of the form:

x = Acos 2t + Bsin 2t

On substituting this in equation (i), we get:

-4A sin 2t + 4B cos 2t + 4A cos 2t + 4B sin 2t + 2(-2A sin 2t + 2B cos 2t) + 5(A cos 2t + B sin 2t) = 16 cos 2t + 4 sin 2t

Grouping the coefficients of cos 2t and sin 2t terms, we get:

-4A + 2B + 5A = 4..................(1)

4B - 2A + 5B = 0..................(2)

Solving the above two equations, we get:

A = -4/21B = 8/21

So, the particular solution is given by:

x = -4/21 cos 2t + 8/21 sin 2t

The general solution is given by the sum of the complementary function and the particular solution.

The complementary function can be obtained by solving the characteristic equation.

The characteristic equation is given by:

1 m² + 2 m + 5 = 0

On solving the above quadratic equation, we get:

m = -1 ± 2i

Therefore, the complementary function is given by:

x = e^(-t) [C1 cos 2t + C2 sin 2t]

where C1 and C2 are constants.

Substituting the values of A and B in the particular solution, we get:

x = -4/21 cos 2t + 8/21 sin 2t

The general solution is given by:

x = e^(-t) [C1 cos 2t + C2 sin 2t] - 4/21 cos 2t + 8/21 sin 2t

Hence, the equation of motion is given by:

x = e^(-t) [C1 cos 2t + C2 sin 2t] - 4/21 cos 2t + 8/21 sin 2t.

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To find the current in the resistor, we can use Ohm's Law and the concept of equivalent resistance. Thus, option A is correct.

First, let's calculate the equivalent resistance of the three cells connected in parallel. When resistors are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances:

1/Req = 1/R1 + 1/R2 + 1/R3

Given that R1 = R2 = R3 = 22 Ω (internal resistance of each cell), we can substitute the values:

1/Req = 1/22 + 1/22 + 1/22

1/Req = 3/22

Taking the reciprocal of both sides, we find:

Req = 22/3 Ω

Now we can use Ohm's Law to calculate the current (I) in the resistor. Ohm's Law states that the current flowing through a resistor is equal to the voltage across it divided by its resistance:

I = V/R

Given that V = 1.1 V (emf of each cell) and R = 32 Ω (resistance), we can substitute the values:

I = 1.1/32

Calculating this value, we find:

I ≈ 0.034375 A

Therefore, the current in the resistor is approximately 0.034375 A.

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No, magnetic fields do not exist on every planet. The magnetic fields of the four gas giants are extremely strong, while those of the Earth and Mercury are fairly strong and Venus and Mars have essentially no detectable magnetic fields.

Mars and Venus don't have much of a magnetic field, according to probes. The magnetic fields of Jupiter, Saturn, Uranus, and Neptune are all far stronger than those of the Earth. Jupiter is the winner since it has the biggest magnetic field. The magnetosphere grows in size as the magnetic field gets stronger. Jupiter's magnetic field, which is around 20,000 times stronger than Earth's magnetic field, generates a magnetosphere that extends over 3 million kilometers from Jupiter before it starts to deflect solar wind.

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The detection of quasars from such great distances is due to their extreme luminosity and the fact that they emit massive amounts of energy.

Quasars are the most energetic objects in the universe, and their bright emissions make them visible even at very large distances. On the other hand, even the biggest and most luminous galaxies cannot be seen from such distances because their emissions are not as powerful as those of quasars.

Quasars are actually supermassive black holes at the center of galaxies that are actively feeding on surrounding matter. As they consume matter, they emit large amounts of energy in the form of light, X-rays, and other types of radiation. This energy is what makes them visible from great distances, even beyond the limits of what other galaxies can achieve.

The fact that quasars can be detected from distances beyond the reach of other galaxies is a testament to their extreme power and luminosity. It also helps astronomers to study the universe at a deeper level and gain insight into the evolution of galaxies and supermassive black holes.

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The horizontal pulling force on the block is 22.5 N.

When a block is pulled by applying a force, friction also acts on it.

The net force is calculated by taking the net value of the two forces.

The free-body diagram(FBD) of the block,  is shown in the adjoining diagram.

Thus,

F - f = ma

Here F is the applied horizontal force and f is the frictional force.

Putting the given values in the above equation

F - 10 N = 5 x 2.5

F = 22.5 N

Thus, the horizontal pulling force on the block is 22.5 N.

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The horizontal pulling force on the block is 22.5 N.

When a block is pulled by applying a force, friction also acts on it.

The net force is calculated by taking the net value of the two forces.

The free-body diagram(FBD) of the block,  is shown in the adjoining diagram.

Thus,

F - f = ma

Here F is the applied horizontal force and f is the frictional force.

Putting the given values in the above equation

F - 10 N = 5 x 2.5

F = 22.5 N

Thus, the horizontal pulling force on the block is 22.5 N.

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Answer:

If the body is in equilibrium the two forces add up and the third is the opposite of the resultant.

F(1x)=F(1)=10 N

F(2x)=F(2)cos60=5•0.5=2.5 N

F(2y) =F(2)sin60 = 5•0.866= 4.33 N

F(3x) =- F(x)=- (10+2.5 )= -12.5 N

F(3y) =- F(2y)= - 4.33 N

F(3) = sqrt{ F(3x)²+F(3y)²} =13.23 N

tan φ = F(3y)/F(3x) =4.33/12.5=0.364

φ = 19.1⁰ (south-west)

Explanation:

High-mass stars are so luminous that their radiation strips gas from the outer layers and driving it into space whereas high-mass stars achieve much hotter temperatures in their cores and interior layers

Low mass stars are cooler and are reddish in color. High mass stars are hotter, and are white or blue white in color. Extremely high mass stars may also shine as a pale violet, which is more blue than blue white. High mass stars are brighter compared to low mass stars, because they produce more energy.

A low mass star is less than three solar masses whereas high mass star is above eight solar masses.

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F = m • a (I. Newton)

F = (4 kg) x (5 m/s^2)

F = 20 Newtons

Answer:

a low protein diet helps you lose weight because lack of protein can make you lose muscle mass

Explanation:

It cuts your strength it makes it harder to keep your balance and slows your metabolism and later you get tired easily it leads to muscle wasting overtime and sheds your fat

The total upward force before the skydiver opens her parachute is equal to the total upward force after she opens her parachute.

When a skydiver is falling at terminal velocity, the downward force due to gravity is balanced by the upward force of air resistance. At terminal velocity, the net force on the skydiver is zero, resulting in a constant velocity.

When the skydiver opens her parachute, the surface area exposed to air increases significantly. This leads to an increase in air resistance, which results in a reduced downward acceleration and a decrease in the terminal velocity.

However, the total upward force before and after opening the parachute remains the same. Both scenarios involve a balance between the downward force of gravity and the upward force of air resistance. The difference lies in the magnitude of the forces and the resulting velocity. Before opening the parachute, the forces balance at a higher terminal velocity, while after opening the parachute, they balance at a lower terminal velocity.

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