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a strong magnetic field surrounds Earth. How does Earth’s magnetic field help humans?

a. it holds people’s belongings thigh to the Earth’s surface.

b. it’s responsible for cycling water through Earth’s four spheres.

c. it keeps the solar wind from reaching Earth.

d. it helps people navigate using a compass.

e. it helps cool Earths surface.

The answer is A 00.2

Answer:

8.5 m/s

Explanation:

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 =  36.95 m/s

The car was moving at a speed of 36.95 m/s when it began to accelerate.

To find the initial velocity of the car when it began to accelerate, we can use the equation for average velocity:

V = (U + V)/2

Where:

Given that the car changes velocity at a constant acceleration, we can use the kinematic equation:

V = U + at

Where:

Substituting the known values, we have:

43.7 m/s = U + (2.5 m/s²)(2.7 s)

Simplifying the equation, we find:

U = 43.7 m/s - (2.5 m/s²)(2.7 s)

U = 43.7 m/s - 6.75 m/s

U = 36.95 m/s

Therefore, the car was moving at a speed of 36.95 m/s when it began to accelerate.

Answer:

The statement is true. The vestibulo-ocular reflex ensures stable vision by moving the eyes in the opposite direction of the head.

Explanation:

The vestibulo-ocular reflex is a reflection of ocular movement that stabilizes the image in the retina during the movement of the head, producing an eye movement in the opposite direction to the movement of the head, conserving the image in the center of the visual field. For example, when the head moves to the right, the eyes move to the left, and vice versa. As there are slight movements of the head at all times, the RVO is very important to stabilize the vision: patients who have damaged RVO find it difficult to read printed media, because they can not stabilize the eyes during small tremors of the head.

Final answer:

The object's velocity remains constant as it covers the same distance each second, implying that the acceleration is zero (0 m/s^2). So the correct option is A.

Explanation:

The question relates to constant velocity and acceleration and to identifying the acceleration of an object moving 10 meters each second for three seconds. Since the object travels the same distance in each second, it means that its velocity remains constant. Therefore, the rate of change of velocity, which is acceleration, is zero. The correct answer to the question is A) 0 m/s2. To understand this, consider the definition of acceleration: a change in velocity over some time. Since there is no change in velocity here (the object keeps moving at a constant rate), there is no acceleration.

The part of the neuron that houses the nucleus and other cell parts is the cell body that is in Option c, as the cell body, also known as the soma, is a central part of a eukaryotic cell.

The cell body includes the nucleus and other organelles like the mitochondria, endoplasmic reticulum, Golgi apparatus, and cytoplasm. The cell body is the main site for the metabolic and biochemical processes of the cell, and it is responsible for maintaining the integrity of the cell and regulating its activities. In the case of neurons, the cell body also contains the genetic information of the neuron in the form of DNA, which is used to synthesize the proteins and other molecules needed for the growth, repair, and maintenance of the neuron.

Hence, the part of the neuron that houses the nucleus and other cell parts is the cell body that is in Option c.

Learn more about the cell body here.

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1. Some wooden rulers do not start with 0 at the edge, but have it set on a few millimeters because the cut of the woof alters the length of the first measures. Moreover, some objects are difficult to be aligned on the ruler, thus, in order to remove this difficulty, some wooden rulers do not start with 0 at the edge.

2. The micrometer has been badly bent which means that it will not measure properly and will not be perfectly parallel with the surface to be measured. This distortion will  change its precision. The device should be parallel with the surface to be measured in order to maintain its accuracy.

3. No, the parallax will not affect the precision of a measurement that we take as precision  is the closeness of the readings and the readings are already precise, thus taking the reading with the same parallax won't impair the precision but decreases the accuracy.

4. Length of box=18.1 cm, width= 19.2 cm and height=20.3 cm, then the a.volume=length×width×height

=[tex]18.1{\times}19.2{\times}20.3[/tex]

=[tex]7050cm^{3}[/tex]

b. The reading of the length is precise to 0.1 centimeters, while the volume's reading is precise to 10 cm³.

c. The stack of 12 of the boxes has height=[tex]12{\times}19.3[/tex]

=[tex]232cm[/tex]

d.The measure of one box is precise to 0.1 cm, while the measurement of the combined boxes is precise to 1 cm.

The time they will be 1200 mi apart is [tex]\(12:00 PM + \frac{1200}{2S + 30}\)[/tex] hours.


Let's denote the speed of the slower plane as [tex]\(S\) (in mi/h)[/tex]. The faster plane, which is flying east, will have a speed of [tex]\(S + 30\) mi/h[/tex].

The relative speed between the two planes is the sum of their individual speeds:

[tex]\[ \text{Relative speed} = S + (S + 30) = 2S + 30 \, \text{mi/h} \][/tex]

Now, we know that the planes leave at noon, and we want to find the time it takes for them to be 1200 mi apart. Let t be the time in hours.

The distance traveled by the slower plane in t hours is [tex]\(S \cdot t\)[/tex], and the distance traveled by the faster plane is [tex]\((S + 30) \cdot t\)[/tex]. The sum of these distances is the total distance between the planes:

[tex]\[ S \cdot t + (S + 30) \cdot t = 1200 \][/tex]

Combine like terms:

[tex]\[ 2S \cdot t + 30 \cdot t = 1200 \][/tex]

Factor out t:

[tex]\[ t(2S + 30) = 1200 \][/tex]

Now, solve for t:

[tex]\[ t = \frac{1200}{2S + 30} \][/tex]

We know that when t is the time, it should be in hours, and we want to find at what time they will be 1200 mi apart. If they leave at noon, the time will be [tex]\(12:00 PM\) + \(t\) hours[/tex].

So, the time they will be 1200 mi apart is [tex]\(12:00 PM + \frac{1200}{2S + 30}\)[/tex] hours.

the correct answer is A  *LOWER*

Answer:

X-Rays from space do not reach the ground.

Explanation:

Final answer:

The linear velocity of a tooth at the edge of the 14 inch diameter blade on a typical table saw can be calculated by converting the blade's rotational speed from revolutions per minute to radians per second and using the formula for linear velocity. The calculated value will give the speed of the tooth in miles per hour.

Explanation:

To calculate the linear velocity of a tooth at the edge of the blade, we need to first convert the blade's rotational speed from revolutions per minute to radians per second. We can do this by multiplying the revolutions per minute (rpm) by 2π (since there are 2π radians in one revolution) and dividing by 60 (to convert minutes to seconds). So, the angular velocity of the blade is:

ω = (3300 rpm) x (2π rad/rev) / (60 s/min)

Next, we can use the formula for linear velocity to find the speed of a point on the edge of the blade. The formula is:

v = ω x r

where v is the linear velocity, ω is the angular velocity, and r is the radius. In this case, the blade has a diameter of 14 inches, so the radius is half of that, which is 7 inches or 0.5833 feet. Converting this to miles, we get:

r = 0.5833 ft x (1 mile/5280 ft)

Finally, we can substitute the values into the formula:

v = (3300 rpm) x (2π rad/rev) / (60 s/min) x 0.5833 ft x (1 mile/5280 ft)

Calculating this equation gives the linear velocity of a tooth at the edge of the blade.

The density of water is 1.00 g/cm3, then in kilograms per centimeter cube unit, its density would be 1000 kg/m³.

It can be defined as the mass of any object or body per unit volume of the particular object or body. Generally, it is expressed as in gram per cm³ or kilogram per meter³.

As given in the problem we have to calculate the density of water in kilogram per centimeter cube,

1 cm³ = 1×10⁻⁶ m³

1 gram = 1×10⁻³ kg

The density of the water= mass of the water /volume of the water

                                 =1.00 g/cm3

                                 =1×10⁻³ kg /1×10⁻³ kg

                                 =1000 kg/m³

Thus, if the density of water is 1.00 g/cm3, then in kilograms per centimeter cube unit its density would be 1000 kg/m³.

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The landing speed of the airplane is 1.5 m/s.

The question is asking for the landing speed of an airplane given its landing time and the distance traveled on the airstrip. To find the landing speed, we can use the formula:

Speed = Distance / Time

Plugging in the values given:

Using the formula:

Landing Speed = 45 m/s / 30 s = 1.5 m/s

Therefore, the landing speed of the airplane is 1.5 m/s.

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The question pertains to the deceleration of an airplane after it lands, specifically calculating the final velocity given an initial velocity and a constant deceleration over a certain time span.

The student's question is related to the deceleration of an airplane after landing. An airplane that lands with an initial velocity of 70.0 m/s, and then decelerates at 1.50 m/s2 for 40.0 s, will have a final velocity that can be calculated using the following formula: final velocity = initial velocity + (acceleration × time). Here, the acceleration is negative because it is opposite the direction of motion (deceleration).

In the example given, if an airplane lands with a velocity of 70.0 m/s and decelerates at 1.50 m/s2, the final velocity after 40 seconds can be found by:

Final Velocity = 70.0 m/s - (1.50 m/s2 × 40.0 s) = 70.0 m/s - 60.0 m/s = 10.0 m/s. The plane would slow down to 10.0 m/s before heading to the terminal.

Answer:

range and angle of launch

Explanation:

Let the horizontal component of velocity be VCosθ = Vx and the vertical component of velocity is VSinθ = Vy.

Here , θ be the angle of projection.

So, if we divide the VSinθ by VCosθ, we get the value of Tanθ, i.e., we get the value of angle of launch.

Now we know that the formula fr the range is given by

[tex]R = \frac{V^{2}\times Sin2\Theta }{g}[/tex]

We can write it as

[tex]R = \frac{V^{2}\times 2\times Sin\Theta \times Cos\Theta }{g}[/tex]

Again we rewrite as

[tex]R = \frac{2\times Vx\times Vy}{g}[/tex]

It shows that the range also depends on the horizontal and vertical component of velocity.

We are given that there are two capacitors:

4.20 uf

8.50 uf

A. In parallel

The equivalent capacitance of capacitors is similar to calculating that of a current, they are added when in parallel. Therefore the equivalent capacitance is:

equivalent capacitance = C1 + C2 + C3 + ...

equivalent capacitance = 4.20 uf + 8.50 uf

equivalent capacitance = 12.70 uf

B. In series

When the capacitors are placed in series, the formula for the equivalent capacitance is:

equivalent capacitance = 1 / (1/C1 + 1/C2 + 1/C3 + ...)

equivalent capacitance = 1 / (1/4.20 + 1/8.50)

equivalent capacitance = 2.81 uf

For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances, so we get 12.70 µF. For capacitors in series, we calculate using the reciprocal formula and get approximately 2.79 µF.

To solve this problem, we need to know the formulas to calculate capacitance in a series and parallel circuit. For capacitors in parallel, the total equivalent capacitance (Ceq) is the sum of the capacitors, so Ceq = C1 + C2, where C1 and C2 are the capacitances of your individual capacitors. Here, C1 is 4.20 µF and C2 is 8.50 µF. So, Ceq = 4.20 µF + 8.50 µF which equals 12.70 µF.

For capacitors in a series connection, the total equivalent capacitance is calculated using the reciprocal formula: 1/Ceq = 1/C1 + 1/C2. So, 1/Ceq = 1/4.20 µF + 1/8.50 µF. Solving for Ceq gives us an equivalent capacitance of about 2.79 µF.

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Answer:

Instantaneous speed

Explanation:

It is given that, Henry was given this cool toy for his birthday. He can use it to tell how fast someone is going at that moment in time. The toy must calculate the person's instantaneous speed.

Instantaneous speed of an object is defined as the speed at a particular instant. Mathematically, it is given by :

[tex]v= \lim_{\Delta t \to 0}(\dfrac{\Delta x}{\Delta t})[/tex]

Instantaneous speed is a scalar quantity. It does not involve any direction.

So, Henry must calculate the person's instantaneous speed. Hence, the correct option is (d).

The maximum height reached by the arrow shot vertically at 60 meters per second is 183.67 meters. Upon returning to the ground, the arrow's velocity will be -60 meters per second.

The student is dealing with a problem that involves projectile motion, a topic within physics. To calculate the maximum height the arrow reached when shot vertically into the sky at a velocity of 60 meters per second, we use the kinematic equation that ignores air resistance:

Maximum height (h) = (v^2) / (2g)

where v is the initial velocity and g is the acceleration due to gravity (9.8 m/s^2). Substituting the given values:

h = (60^2) / (2*9.8) = 3600 / 19.6 = 183.67 meters

The maximum height reached by the arrow is 183.67 meters. When the arrow hits the ground, the velocity will be the same as the initial velocity in magnitude but in the opposite direction, so the velocity of the arrow when it hits the ground is -60 meters per second (the negative sign indicates the direction is downward).

The net force acting on the refrigerator is 400 N to the right

[tex]\texttt{ }[/tex]

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

[tex]\boxed {F = ma }[/tex]

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

mass of refrigerator = m = 200 kg

magnitude of applied force = F = 400 N

magnitude of frictional force = f = 0 N

Asked:

net force = ΣF = ?

Solution:

We will use Newton's Law of Motion to solve this problem as follows:

[tex]\Sigma F = F - f[/tex]

[tex]\Sigma F = 400 - 0[/tex]

[tex]\boxed{\Sigma F = 400 \texttt{ N} }[/tex]

[tex]\texttt{ }[/tex]

We could also calculate the acceleration of the refrigerator as follows:

[tex]\Sigma F = ma[/tex]

[tex]a = \Sigma F \div m[/tex]

[tex]a = 400 \div 200[/tex]

[tex]\boxed{a = 2 \texttt{ m/s}^2 }[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics