The voltmeter reads 117 V and ammeter reads 0.13 A .If the resistance of ammeter and voltmeter are 9 kilo ohm and 0.015 ohm respectively ,find the value of R. 3marks

The electrostatic force is both attractive and repulsive and acts over long distances, while the strong nuclear force is a very powerful attractive force acting only over short distances within the atomic nucleus. The strong nuclear force is stronger than the electrostatic force and is responsible for the stability of the nucleus by binding protons and neutrons together.

Comparing the electrostatic force to the strong nuclear force within the nucleus of an atom, several key differences emerge. The electrostatic force can act over very large distances and can be either attractive or repulsive, depending on the charges involved. In contrast, the strong nuclear force is an extremely powerful attractive force that acts only over incredibly short distances, approximately 10-15 meters, which is about the scale of the atomic nucleus. This force is responsible for holding protons and neutrons together within the nucleus, overcoming the electrostatic repulsion between like-charged protons.

Examining the available choices:

The most accurate way to characterize these forces is that the strong nuclear force is a short-range but much stronger force, crucial for the stability of nuclei, while the electrostatic force, governed by Coulomb's law, has a longer range and is comparatively weaker than the strong nuclear force, but plays a critical role in interactions over larger distances and molecular bonding.

Final answer:

Real images are formed by convergence of light rays and can be projected onto a screen, while virtual images are formed by apparent divergence of light rays and cannot be projected onto a screen. You can tell if an image is real or virtual by its position relative to the lens/mirror.

Explanation:

Differences between Real and Virtual Images:

A real image is formed by the convergence of actual light rays and can be projected onto a screen. It is always inverted. On the other hand, a virtual image is formed by the apparent divergence of light rays and cannot be projected onto a screen. It is usually upright.

Determining if an Image is Real or Virtual:

By looking at an image, you can determine whether it is real or virtual based on its characteristics. If the image is formed on the opposite side of the lens/mirror from the object, it is virtual. If the image is formed on the same side as the object, it is real.

Answer:

B

A meteorologist studies weather, such as a weather forecaster. :)

Answer: B. Weather

Explanation: A meteorologist studies weather, and can be a weatherman.

Answer:

81.1 m

Explanation:

Given:

x₀ = 0 m

y₀ = y = 0 m

v₀ₓ = 29.5 cos 33.0° m/s

v₀ᵧ = 29.5 sin 33.0° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

Find:

x

First, find the time it takes to land.

y = y₀ + v₀ᵧ t + ½ aᵧt²

0 = 0 + (29.5 sin 33.0) t - 4.9 t²

t = (29.5 sin 33.0) / 4.9

t = 3.28 s

Now in the x direction:

x = x₀ + v₀ₓ t + ½ aₓt²

x = 0 + (29.5 cos 33.0) (3.28) + 0

x = 81.1 m

The outfielder was 81.1 meters away.

Answer:

Explanation:

First you should remember that the images produced by convex mirror is virtual.

In this question, you should find the focal length using the radius of curvature, then apply the focal length relationship to find the image distance which is a virtual distance and finally use the magnification relationship to find the image height.

Let;

Find the focal length

f=focal length is half the radius of curvature

f=r/2=34/2 =17cm =0.17m (This distance is virtual thus include a -ve sign)

f= -0.17m

Apply the focal length relationship

[tex]\frac{1}{o} +\frac{1}{i} =\frac{1}{f} \\\\\\\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17} \\\\\\2+\frac{1}{i} =-5.88\\\\\\\frac{1}{i} =-5.88-2\\\\[/tex]

Solve for the reciprocal

[tex]\frac{1}{i} =-7.88\\\\i=-0.13m[/tex]

This is a virtual distance for the virtual image formed

Apply the magnification relationship

Magnification = height of image÷height of object

or

Magnification= - image distance÷object distance

[tex]\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\0.5h'=0.13*0.20\\\\\\h'=\frac{0.13*0.20}{0.5} =0.052m=5.2cm[/tex]

The image formed by the convex mirror will be located at -50.0 cm and will have a height five times larger than the object.

To find the location and height of the image formed by a convex mirror, we can use the mirror equation / + /ᵢ = /, where is the image distance, is the object distance, and is the focal length of the mirror. Plugging in the given values, we have / + /ᵢ = /. The focal length of a convex mirror is negative, so we have / + /ᵢ = −/.

Using the given values of ᵢ = −50.0 cm and = −34.0 cm, we can solve for . The image will be located at ᵢ = −50.0 cm and will be virtual, as the image distance is negative. The height of the image can be determined using the magnification formula: = −ᵢ / . Plugging in the values of ᵢ = −50.0 cm and = −10.0 cm, we get = −(−50.0 cm) / (−10.0 cm) = 5.00.

Therefore, the image will be located at −50.0 cm and will have a height five times larger than the object.

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Answer: I think, the number of oscillations in a given period of time.

Explanation: Well I guess because in a period time is known as the rate of occurrence of the wave. Hope this helps!

To determine the frequency of a wave you must measure by counting the number of high points of waves that pass the fixed point in 1 second or some other time period.

Wave frequency is the number of waves that pass a fixed point at a particular time. The SI unit of wave frequency is hertz (Hz). Here, 1 hertz corresponds to a wave passing through a fixed point every second. High frequencies have more energy than low frequencies of the same amplitude.

The relationship between wavelength and frequency is called the reciprocal because the higher the frequency, the shorter the wavelength.

In addition to the defined speed, each wave has a frequency. This is the number of wavelengths that pass a point per second. Frequency is measured in Hertz (wavenumber per second).

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Answer:625

Explanation:

The gravitational force between two object sets is onversly proportional to the square of the distance between the objects.if the distance reduces by a factor of 5 the force will increase by a factor of 5^2=25

The changing force of attraction between them is 625 Newtons

The answer is option D.

The gravitational force between two object sets is inversely proportional to the square of the distance between the objects. If the distance reduces by a factor of 5 the force will increase by a factor of 5^2=25

The gravitational force is a pressure that draws any two objects with mass. We call the gravitational pressure attractive as it usually attempts to pull masses together, it in no way pushes them aside. In reality, every object, including you, is pulling on every other item inside the complete universe.

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1. The force of gravity as a satellite orbits a planet.

According to Newton's first law, an object  stays at rest (or continue its motion with constant velocity) if the net force acting on it is zero.

This means that in order to have a circular motion (where the velocity changes, because the direction of motion constantly changes), we must have a net force acting on the object. This force is called centripetal force, and it always acts towards the centre of the circular trajectory.

An example of circular motion is a satellite orbiting a planet - in this case, the force of gravitational attraction between the planet and the satellite acts as centripetal force, keeping the satellite in circular motion.

2.  13.7 m/s

The centripetal force acting on an object is given by:

[tex]F=m\frac{v^2}{r}[/tex]

where

v is the speed of the object

m is its mass

r is the radius of the trajectory

In this situation, the centripetal force is provided by the frictional force, which is given by

[tex]F_f = \mu mg[/tex]

where

[tex]\mu[/tex] is the coefficient of friction between the tire and the road

g = 9.8 m/s^2 is the acceleration due to gravity

Equalizing the two forces,

[tex]\frac{v^2}{r}=\mu g[/tex]

where we have:

r = 55.0 m is the radius of the track

[tex]\mu=0.350[/tex] is the coefficient of friction

Solving for v, we find the maximum speed that the car can substain:

[tex]v=\sqrt{\mu g r}=\sqrt{(0.350)(9.8)(55.0)}=13.7 m/s[/tex]

3. 1.2 N, toward the center of the circle

In this situation, the tension in the string provides the centripetal force that keeps the ball in circular motion, so we can write:

[tex]T=m\frac{v^2}{r}[/tex]

where

T is the tension in the string

m = 0.015 kg is the mass of the ball

r = 0.70 m is the radius of the circle

v is the speed of the ball

Since the ball takes t = 0.60 s to complete one full circle, its speed is

[tex]v=\frac{2\pi r}{t}=\frac{2\pi (0.70)}{0.60}=7.33 m/s[/tex]

So now we can calculate the tension in the string:

[tex]T=m\frac{v^2}{r}=(0.015)\frac{(7.33)^2}{0.70}=1.15 N\sim 1.2 N[/tex]

And we said previously, since this force acts as centripetal force, its direction is towards the centre of the circle.

Answer:

The answer is A.  Spokespersons are paid for their testimony.  Psueoscience is NOT a real science.

Explanation:

We know that frequency is an inverse value of time [tex]f=\dfrac{1}{t}[/tex] this implies that time is the inverse value of frequency [tex]t=\dfrac{1}{f}[/tex].

Now since f is 40kHz, we can calculate period or duration for that matter time.

[tex]t=\dfrac{1}{40\mathbf{kHz}}=0.000025\mathbf{s}=25\mathbf{\mu s}[/tex]

Hope this helps.

r3t40

Explanation:

8 m × (1000 mm / m) = 8000 mm

Explanation:

Refracting telescopes have an optical system consisting of a set of lenses that have the property of deflecting (refracting) the light that passes through them.

In this way, the images captured from distant objects converge on a point in the focal plane of the telescope.

However, the main disadvantage of this type of telescope is the chromatic aberration, which produces annoying color halos (mainly red and blue) around the image.

It should be noted that the chromatic aberration occurs because the different wavelengths that make up the light (the colors of the light) that passes through the objective lens deviate (refract) in different ways, causing them not to focus on the same point. For example,  blue wavelengths are focused before the green wavelengths and these before the red ones. And this problem is stronger the shorter the focal length of the lens is.

Answer:

C

Chromatic aberration affected:

spectrographs

radio telescopes

refracting telescopes

reflecting telescopes

Answer:

4200 kJ

10 kg/min

Explanation:

70 kW means 70 kJ per second.  So in a minute, the amount of heat generated at the brake is:

70 kJ/s × 60 s/min = 4200 kJ

Heat = mass × specific heat capacity × increase in temperature

4200 kJ = m × 4.184 kJ/kg/C × 100 C

m = 10 kg

So every minute, 10 kg of water is heated.  So the mass flow is 10 kg/min.

Answer:

43.1

Explanation:

Mv=Mv

0.07*800=1.3*V

V=56/1.3

V=43.07

then 43.1

Answer:

[tex]v = 4.28 m/s[/tex]

[tex]Impulse = 5.57 kg m/s[/tex]

Explanation:

Here we can say that bullet + block system is an isolated system and there is no external force on it

So Net momentum of bullet + block system will remains conserved

So we will say

[tex]m v_o = (M + m) v[/tex]

so we will have

[tex]v = \frac{m}{M + m} v_o[/tex]

[tex]v = \frac{0.007 (800)}{1.3 + 0.007}[/tex]

[tex]v = 4.28 m/s[/tex]

Also in order to find the impulse on the block we know that

impulse = change in the momentum of the block

[tex]Impulse = m(v_f - v_i)[/tex]

[tex]Impulse = 1.3(4.28 - 0)[/tex]

[tex]Impulse = 5.57 kg m/s[/tex]

D

hydrosphere-- anything to do with water

geosphere -- the rocks and soil

topsoil (geosphere) is eroded by rainwater(hydrosphere)

Answer:

D

Explanation:

Ans:

Low

Explanation:

Activities in physics are divided into 3 levels, depending on the distance from the ground (or the gravitational pull on the object)

Low level (Low Plane) activities, as the name suggests, are performed at a lesser distance from the ground. Since the given statement mentions that the activity is performed closer to the ground, it will be the low level activity.

When an activity is described as being performed at a low level,

it means that it is taking place close to the ground or at a height that is not significantly above the ground.

This term is often used in contexts such as sports, exercise, or certain job tasks that require proximity to the ground.

For example, a low-level yoga pose would be one where the body is near the floor, or a low-level aerial maneuver would be performed close to the ground.

The antonym of this would be a high level, which indicates an activity performed at a considerable height above the ground.