The potential at point P is the work required to bring a one-coulomb test charge from far
away to the point P?

True or false ?

Answer:

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

I hope this helped!+*

Answer:

The tension of the string is 41.876 N

Explanation:

Given;

length of the string, L = 2.11 m

mass of the string, m = 19.5 g = 0.0195 kg

frequency of the wave, f = 440 Hz

wavelength, λ = 15.3 cm = 0.153 m

The velocity of the wave is given by;

v = fλ

v = 440 x 0.153

v = 67.32 m/s

Also the velocity of the wave is given by

\(v = \sqrt{\frac{T}{\mu} }\)

where;

μ is mass per unit length = 0.0195 / 2.11 = 0.00924 kg/m

T is the tension of the string

T = v²μ

T = (67.32)²(0.00924)

T = 41.876 N

Therefore, the tension of the string is 41.876 N

Answer:

options not given

Let r = height

A = h/2 * r = r h / 2       area of triangle

Centroid = r / 3       from calculus

2 π * r / 3 = 2 π r / 3     distance traveled by centroid in one revolution

r h / 2 * 2 π r / 3 = when revolving about x-axis

V = π r^2 h / 3

Answer:

1- t^3

2- t^2

3- t1

Explanation:

The acceleration produced in a body, while travelling in a circular motion, due to change in direction of motion is called centripetal acceleration. The formula of the centripetal acceleration is as follows:

ac = v²/r

where,

ac = centripetal acceleration

v = speed

r = radius

for a constant radius the centripetal acceleration will be directly proportional to the speed of object. The speed of pendulum will be lowest at t1 due to zero speed initially. Then the speed will increase gradually having greater speed at t^2 and the highest speed and centripetal acceleration at t^3. Therefore, the three instants in tie can be written in following order from greatest centripetal acceleration to lowest:

1- t^3

2- t^2

3- t1

The orientation of an mRNA codon corresponding to a tRNA anticodon is best described as A: antiparallel.

An anticodon is a three-base sequence, paired with a particular amino acid, that a tRNA molecule brings to the associated codon of the mRNA during translation. The anticodon sequence is said to be complementary to the mRNA, using base pairs in the anti-parallel direction.  The tRNA anticodon 3-UAC-5 will make pair with the mRNA codon.

Thus, the orientation of an mRNA codon relative to a tRNA anticodon is said to be antiparallel.

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Explanation:

Height is the x-axis, and gravitational potential energy is the y-axis.  As the height increases, the gravitational potential energy increases linearly.

Part 1

C. Its frequency remains the same.

Part 2

A. A wave's frequency always remains the same.

The relationship between wavelength, frequency, and speed is given as:

When the frequency of a wave increases, its wavelength also increases.

The speed of a wave does not change as its frequency changes, and vice-versa.

Therefore, as the speed of the wave decreases, its frequency remains the same

A wave's frequency always remains the same regardless of the speed and wavelength

The unit vector notation in the given direction is \(0.58 \hat x + 0.81 \hat y\).

A physical quantity that has both directions and magnitude is referred to as a vector quantity.

A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.

In terms of subtracting a vector, it works the same as adding the vector's opposite.

Given that:

\(\vec A = 22 \hat x + (-14)\hat y,\\\vec B = 2.5\hat x + 13\hat y.\)

( two given vectors are written here.)

Therefore: the subtraction of these two vectors will be:

\(\vec A - \vec B = (= 22 \hat x + (-14)\hat y) - (2.5\hat x + 13\hat y)\\= 19.5\hat x - 27 \hat y\)

Unit vector in that direction of vectors  is =

\(= \frac{19.5\hat x - 27 \hat y}{\sqrt{19.5^2 + (-27)^2} } \\= 0.58 \hat x + 0.81 \hat y\)

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Answer:

the Indian Ocean on 26 December 2004. This event claimed 227,898 dead and missing from 14 countries. The difference in mortality rates between these tsunamis reflects, in part, the benefits of understanding how tsunami waves are generated and move, and educating citizens to make scientifically

sound and potentially life-saving decisions.

A tsunami is a series of rapidly propagating, shallow-water ocean waves that develops when a submarine earthquake, landslide, or volcanic eruption displaces a large volume of water. Powerful earthquakes, with magnitudes of 9 or greater, caused both the 2004 and 2011 tsunamis. The earthquakes resulted from the movement of large tectonic plates. The 11 March 2011 earthquake occurred at 32 km (20

mi.) deep in Earth’s crust about 130 km (81 mi.) east of the city of Sendai. This location is on the boundary between two tectonic plates—the Pacific plate to the east and North American plate to the west. This

boundary fractured, releasing energy that was transmitted through the rocks and elevated portions of the

ocean floor. This drastic movement transmitted energy to the overlying ocean water, which generated

tsunami waves that radiated outward. The waves washed over the nearby coastlines and were felt around

the globe within hours (Figure 1.1).  

Explanation:

Answer:

I believe Geosphere (lithosphere) and Hydrosphere

Explanation:

I hope it's right if not please notify me.

Answer:

D

Explanation:

The motion of an object can be described by its speed, position, and direction so its all of the above

Answer:

C

Explanation:

Some factors determine the amount of UV radiation that reach certain part of Earth's surface. They are listed and briefly explained below:

• Cloud coverage. Water molecules on clouds scatter the radiation, hence the more clouds the less UV radiation.

• Ozone. Similarly as cloud coverage, the more concentration of ozone the less UV radiation that reaches the surface of the Earth.

• Angle of incidence. If the angle of incidence of the UV light is oblique the light will spread in a wider area, and hence the intensity is spread across this area.

• Aerosols. The molecules of aerosols also scatter the UV light.

• Elevation. The more the elevation the greater the amount of UV light.

(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

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Answer:

They both hit the ground at the same time.

Explanation:

if you drop a 100 g and 200 g mass simultaneously from the same height they’ll hit the ground at the same time.

Yes, there is a role of a very simple equation of kinematics i.e.

s=ut+1/2(at^2)

so, s means the height is the same for both masses(h), u(initial velocity) is obviously zero cause you are dropping it, a (acceleration) is due to the gravity here only ‘g’.

time is taken for a particular object to hit the ground,

t= (2h/g)^1/2 .

We can see that there is no term for the mass of objects, meanwhile, the mass of the object has nothing to do with the time taken in hitting the ground in this case.

The initial velocity of the train  is 12.56 m/s.

The initial velocity of the train can be determined by using the first kinematic equation as shown below;

v = u + at

u = v - at

where;

u = 30.56 - (36 x 0.5)

u = 12.56 m/s

Thus, the initial velocity of the train  is 12.56 m/s.

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4.44×\(10^{9}\) kg/s is the rate at which the sun mass is decreasing.

The Sun radiates energy through a process called nuclear fusion, where hydrogen atoms combine to form helium, releasing a tremendous amount of energy in the process. According to Einstein's mass-energy equivalence principle (E=mc²), this energy release corresponds to a decrease in mass.

To calculate the rate at which the Sun's mass is decreasing, we can use the formula ΔE = Δmc², where ΔE is the change in energy, Δm is the change in mass, and c is the speed of light.

Given that the Sun radiates energy at a rate of 4×10^26 W, we can substitute this value into the equation as ΔE and solve for Δm.

ΔE = 4×10^26 W

c = 3×10^8 m/s (speed of light)

Using the equation ΔE = Δmc² and rearranging it, we get Δm = ΔE / c².

Substituting the values, we have:

Δm = (4×10^26 W) / (3×10^8 m/s)²

Evaluating this expression, we find that the rate at which the Sun's mass is decreasing is approximately 4.44×10^9 kg/s.

This calculation demonstrates that the Sun's mass is gradually decreasing as it continuously radiates energy into space, primarily through the process of nuclear fusion in its core.

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Answer:

Explanation:

Let the velocity of projectile be v and angle of throw be θ.

The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m

considering its vertical displacement

h = - ut +1/2 g t²

100 = - vsinθ x 5 + .5 x 9.8 x 5²

5vsinθ =  222.5

vsinθ = 44.5

It covers 160 horizontally in 5 s

vcosθ x 5 = 160

v cosθ = 32

squaring and adding

v²sin²θ +v² cos²θ = 44.4² + 32²

v² = 1971.36 + 1024

v = 54.73 m /s

Answer:

55.42 m/s

Explanation:

Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by

u_x = d/t

Where

d = 160 m is the distance covered

t = 5.0 s is the time taken

Substituting, we get

u_x =160/5 = 32 m/s.

Along the vertical direction, the rock is in free-fall - so its motion is a uniform accelerated motion with constant acceleration g = -9.8 m/s^2 (downward). Therefore, the vertical distance covered is given by the

\(S=u_yt+\frac{1}{2}at^2\)

where

S = -100 m is the vertical displacement

u_y is the initial vertical velocity

Replacing t = 5.0 s and solving the equation for u_y, we find

-100 = u_y(5) + (-9.81)(5)^2/2

u_y = 45.25 m/s

Therefore, the speed with which the rock was thrown u

\(u= \sqrt{u_x^2+u_y^2} \\=\sqrt{32^2+45.25^2}\\ = 55.42 m/s\)

The magnitude of the vector sum = 0.149 m

The direction = 321.8 degrees

Find the x and y components of vector E

Find the x and y components of vector F

The vector sum of E and F is:

The magnitude of the vector sum is given as:

The magnitude of the vector sum = 0.149 m

The direction is given as:

Answer:

The gravitational force between Royce and Earth would be doubled at 16 years.

Explanation:

"Newton's law of universal gravitation states that gravitation force between two masses is proportional to the magnitude of their masses and inverse-squared of their distance".

Royce Scenario

At the age of 10 Royce's mass = 30kg

At the age of 16 Royce's mass = 60kg

From Newton's law of universal gravitation, an Increase in the mass of a body would amount to a corresponding increase in the gravitational force.

In the case of Royce, the mass double between the age of 10 and 16, so there would be an increase of the gravitation force by double.

The gravitational force at the age of 60 is doubled of the gravitational force at the age of 10. A complete solution is below.

Given:

At 10 years old,

Mass,

now,

→ \(Gravitational \ force (F_1) = \frac{GMm}{r^2}\)

                                          \(= mg\)

or,

→ \(g = \frac{GM}{r^2}\)

then,

Gravitational force = 30 g

Now,

At 16 years old,

Mass,

Now,

→ \(\frac{F_1}{F_2} = \frac{30}{60}\)

       \(=\frac{1}{2}\)

  \(F_2 = 2F_1\)

Thus the above response is right.

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Answer:

\(4.77\ \text{A}\)

Explanation:

F = Magnetic force = 4.11 N

\(I_n\) = Net current

\(I_2\) = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

\(\theta\) = Angle between current and magnetic field = \(65^{\circ}\)

\(l\) = Length of wires = 2.64 m

\(I\) = Current in the other wire

Magnetic force is given by

\(F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}\)

Net current is given by

\(I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}\)

The current I is \(4.77\ \text{A}\).

Answer:

A

Explanation:

The sun is an emits energy in the form of light and heat.

The planets absorb this energy which helps keep them in motion. The closer they are to the sun, the more energy, the more speed and less the time it takes a planet to orbit the sun.

The gravitational pull of the Sun contributes motion of the planets because here the gravitational pull of the sun is balanced by the centrifugal force produced by the revolution of planets. The correct option Is A.

Centrifugal force is the pseudo-force that occurs only in the rotating frame of reference (non-inertial frame). it is directed away from the center. Its magnitude is,

F=\(\frac{mv^{2} }{r}\)

where m= mass of the body in kg.

v= velocity f the body in m/s.

r = radius in m.

The star sun has a  gravitational pull that is about 274 m/s²  which is greater than the other planets. Due to this pull the planets tend to pull toward the sun but it does not happen because the pull force is balanced by the centrifugal force that is produced by the revolution of the planets around the sun.

Hence The Sun's gravitational pull contributes to the motion of the planets.

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The magnitude of the resultant of the two vectors when the position of vector A is known is \(R = \sqrt{92.25 - 90 cos(\theta)} \ \ \ units\).

The given parameters;

The resultant of the two vectors is calculated by applying parallelogram law as shown below;

\(R^2 = a^2 + b^2 - 2ab \ cos(\theta)\\\\R^2 = (6)^2 + (7.5)^2 - 2(6 \times 7.5) cos(\theta)\\\\R^2 = 92.25 - 90 cos(\theta)\\\\R = \sqrt{92.25 - 90 cos(\theta)}\)

Thus, the magnitude of the resultant of the two vectors when the position of vector A is known is \(R = \sqrt{92.25 - 90 cos(\theta)} \ \ \ units\).

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Explanation:

1. 500k

2. 90km/h

i don't have time for diagram

Answer:

Explanation:

1) 300 - 200 + 400 = 500 km S

2) 500/10 = 50 km/hr S

Answer:

A force is a push or a pull and it affects our daily lives because without force,people would not be able to open and close stuff or lift up our arms or legs .....or anything, for that matter.

Explanation:

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2m/s

300m+200m=500m

100s+150s=250s

500m/250s= 2m/s

The average speed of tina will be 2m/s.

We have Tina who went for a walk.

We have to determine her average speed.

Average speed is the rate of change of displacement with respect to time.

v(avg) = Δx/Δt

According to the question -

Total Distance covered = d = 500 m

Total time taken = t = 250 s

Average speed = 500/250 = 2 m/s

Hence, the average speed of tina will be 2m/s.

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Answer:

C

Explanation:

edge 2020... Using elimination it's the only one that makes sense.

The statement third "an increase in the distance between the objects causes a greater change in the gravitational force than the same increase in mass" is correct.

The gravitational force is a force that attracts all mass-bearing objects. The gravitational force is referred to as attractive because it always strives to pull masses together rather than pushing them apart.

As we know, the gravitational force is given by:

\(\rm F = \dfrac{Gm_1m_2}{r^2}\)

Where, G is the gravitational constant.

m1 and m2 are masses.

r is the distance between the masses.

From the data given in the table, shows that:

The gravitational force is indirectly proportional to the square of the distance.

Thus, the statement third "an increase in the distance between the objects causes a greater change in the gravitational force than the same increase in mass" is correct.

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