14 What is the weight of each of
the following masses on Eth?
a) 2 kg
b) 3.5kg
c) 5.25 kg
Newton’s

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When light strikes a black surface, it is reflected by the surface and

nothing is absorbed    This would be False because black absorbs the light not reflect

Mirrors are used in telescopes because they have the ability to reflect

True? Telescopes designed with mirrors avoid the problems of refracting telescopes. Because the light is reflected from the front surface only, flaws and bubbles within the glass do not affect the path of the light

Transmission in the passing of light through some materials​

False? Some materials allow much of the light that falls on them to be transmitted through the material without being reflected. Materials that allow the transmission of light waves through them are called optically transparent. ... Materials which do not allow the transmission of any light wave frequencies are called opaque.

Answer:

Alpha decay

Explanation:

Answer:

C

Explanation:

Alpha Decay

Question:

Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod:

(a) 0.5 cm, (b) 1.5 cm, (c) 2.5 cm, (d) 3.5 cm, (e) 7.0 cm

Answer:

(a) 0

(b) 4.795 N/C

(c) 2.877 N/C

(d) 2.055 N/C

(e) 1.027 N/C

Explanation:

The given parameters are;

The inner radius of the copper = 2 cm

The outer radius of the copper - 3 cm

The radius of the aluminum rod, r = 1 cm

The radius of the aluminum rod = 4 pC/m

a) The charge at the 0.5 cm from the center of the aluminum rod, E = 0

There is no charge inside the rod given that the aluminum road is a conductor of electricity, having all the charge on its surface

b) At 1.5 cm, we have;

[tex]E = \dfrac{2 \cdot \lambda}{4 \cdot \pi \cdot r \cdot \epsilon_0}[/tex]

Where;

[tex]\dfrac{ \lambda}{4 \cdot \pi \cdot \epsilon_0} = 8.99 \times 10^9 \times 4 \times 10^{-12}[/tex]

r = 1.5 cm = 0.015 m

∴ E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.015 ≈ 4.795

The magnitude of the electric field, E = 4.795 N/C

The direction of the charge depends on the charge sign on the rod

c) At 2.5 cm, we have;

r = 2.5 cm = 0.025 m

E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.025 ≈ 2.877

The magnitude of the electric field, E ≈ 2.877 N/C

d) At r = 3.5 cm, we have;

r = 3.5 cm =  0.035 m

E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.035 ≈ 2.055

The magnitude of the electric field, E ≈ 2.055 N/C

e) At

r = 7.0 cm = 0.070 m

E = (2 × 8.99 × 10⁹ × 4 × 10⁻¹²)/0.07 ≈ 1.027

The magnitude of the electric field, E ≈ 1.027 N/C

Answer:

The appropriate answer is "396 N/C".

Explanation:

The give values are:

Rectangular surface,

[tex]\Phi_e=396 \ N.m^2/C[/tex]

[tex]a = 2 \ m[/tex]

[tex]b = 1 \ m[/tex]

Angle,

[tex]\theta =\frac{\pi}{3} \ radian[/tex]

Now,

The area of rectangle (A) will be:

=  [tex]a\times b[/tex]

=  [tex]2\times 1[/tex]

=  [tex]2 \ m^2[/tex]

hence,

The electric field strength will be:

⇒  [tex]\Phi_e=E.A Cos \theta[/tex]

or,

⇒  [tex]E=\frac{\Phi_e}{ACos \theta}[/tex]

On putting the given values, we get

⇒      [tex]=\frac{396}{2\times Cos(\frac{\pi}{3} )}[/tex]

⇒      [tex]=\frac{396}{2\times 0.5}[/tex]

⇒      [tex]=\frac{396}{1}[/tex]

⇒      [tex]=396 \ N/C[/tex]

Answer:

5 kg

Explanation:

Answer:

Option C

Explanation:

Given

Pressure = Force /Area

For any gas sample, force is defined as basically the force exerted by the gas molecules when they strike the surface (with in which they are contained) and bounce back.

Hence, Option C is the correct

Answer:

The voltage across light bulb 1 and light bulb 2 is the the same i.e V

Explanation:

In a parallel circuit, the Voltage is same across all the components of the circuit and the current flowing through each component is added to get the total current across the circuit.

Let us say, the voltage across the circuit is V. The voltage across light bulb 1 and light bulb 2 is the the same i.e V

Answer:

it depends on a person's own weight

Answer:

The gravity is roughly the same. The inverse square law applies to gravity and that means that being 60 miles above Earth is about the same as being on Earth. The thing is, they are going really fast in an orbit and thus falling around the Earth. The weightlessness comes from a sort of cancellation of gravity pulling them down and their going perpendicular to that force. The forces are cancelling and making them relatively “weightless”. It’s not really weightlessness but just an equilibrium reached between the two accelerations of gravity and their motion against the pull of the Earth.

Explanation:

Answer:

4Al + 3O2 + heat ------> 2Al2O3

Explanation:

It is endothermic reaction.

Answer: you ever get the awnser?

Explanation:

Answer: 0.12 g

Explanation: I got it right on the quiz.

Answer:

Acceleration is the name we give to any process where the velocity changes. Since velocity is a speed and a direction, there are only two ways for you to accelerate: change your speed or change your direction—or change both.

Answer:

2.38 T

Explanation:

Since the number of coils and the change in time are both 1, I set it up as 3.57=1.50(x) and then solved for x to be 2.38 T.

Answer: A Graduated Cylinder or A Beaker ?

Explanation:

Answer:

Provides an estimate of how many planets there are in our galaxy that may harbor technically communicative life-forms.

Explanation:

A planet can be defined as a large celestial body having sufficient mass to allow for self-gravity and make it assume a nearly circular shape (hydrostatic equilibrium), revolves in an orbit around the Sun in the solar system and has a cleared neighborhood.

Basically, the planets are divided into two (2) main categories and these includes;

I. Outer planets: these planets are beyond the asteroid belt and comprises of jupiter, saturn, uranus and neptune, from left to right of the solar system.

II. Inner planets: these planets are the closest to the sun and comprises of mercury, venus, earth and mars.

These outer planets are made mostly of gases (hydrogen and helium) causing them to be less dense than the solid inner planets. These gases are generally known to be less dense in terms of physical properties.

Some examples of the planet are Mars, Venus, Earth, Mercury, Neptune, Jupiter, Saturn, Uranus, etc.

The Drake equation provides an estimate of how many planets there are in our galaxy that may harbor technically communicative life-forms. It is a probabilistic mathematical expression.

Answer:

The correct answer would be:

B.

the number of communicating civilizations in the cosmos

#PLATOFAM

Have a nice day!

Answer:

c and a

Explanation:

ape x

Answer:

A and d

Explanation:

Mixture and solid

Explanation:

Given:

m1 = 0.55 kg

m2 = 0.45 kg

vi1 = +5.8 m/a

vi2 = -3.9 m/a

Solve for V

m1vi1 + m2vi2 = (m1 + m2)V

(0.55 kg)(+5.8 m/s) + (0.45 kg)(-3.9 m/s) = (0.55 + 0.45)V

V = +1.46 m/a

This means that the combined mass after the collision is moving to the right at 1.46 m/s

Explanation:

this is the ans hope it works

Answer:

Forces make things speed up (or accelerate). When a force pushes or pulls the object, the object will move in the direction of the force. The bigger the force, and the lighter the object, the greater the acceleration.

Explanation:

The action from a force can cause an object to move or speed up (accelerate), to slow down (decelerate), to stop, or to change direction. Since any change in velocity is considered acceleration, it can be said that a force on an object results in the acceleration of an object.

A simple mathematical relationship exists between the mass of an object (m), the net force on the object (f) and its acceleration (a). The acceleration of an object is directly proportional to the net force and indirectly proportional to the object’s mass (a = f/m).

Answer:

hope it is helpful to you ☺️☺️✌️

Answer:

            T = reading (cm) time base (s / cm)

            f = 1 / T

Explanation:

An oscilloscope is a piece of equipment that allows you to visualize and measure a wave that reaches you, in the case of having a sonometer this transforms the sound wave into an electrical signal to be introduced through one of the voltage channels of the equipment, on the screen we will see the oscillating alternating signal, if it is fixed we can make the reading, if it is moving the time base and the trigger must be adjusted to stop it.

In the oscilloscope we can read the period of the signal, this is the time it takes for the signal to repeat itself with this value, we can calculate the frequency with the formula, for the reading of the period the distance is measured on the labeled screen and multiplied by the time base

            T = reading (cm) time base (s / cm)

            f = 1 / T

Answer:

They don’t ‘represent’ anything, they are properties of the wave.

Depending on the type of wave, we experience them as various phenomena. For example, with a sound wave we experience frequency (or wavelength, which is just another way to describe the same property) as the pitch of the sound. We experience amplitude as the loudness of the sound, although due to the characteristics of the ear, frequency also effects perceived loudness.

If the wave is a light wave, we experience the frequency (wavelength) as the colour of the light, and the amplitude as the brightness of the light.

For many waves, we don’t perceive them at all (e.g. radio waves).

For ocean waves, frequency is the time for each peak or trough to reach us, and amplitude is how tall the wave is.

Answer:

Explanation:

2) From F=ma

Force =15×40=600N or kgm/s2

3)From the same equation making acceleration the subject of the formula will give

a=f÷m

=24÷4=6m/s2

4)m=f÷a

=45÷15=3kg

Answer:

do you have the answer?

Explanation:

if so whats the answer?

Answer:

55N

Explanation:

Using Newton's second law of motion:

F=ma

Force=mass × acceleration

F=25×2.2

F=55N

So 55 Newtons are needed

Answer:

(i) Please find attached the required velocity time graphs plotted with MS Excel

(ii) The velocity of vehicle A at the 18th second = 20 m/s

The velocity of vehicle B at the 18th second = 0 m/s

(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters

Explanation:

The given parameters of the motion of vehicles A and B are;

The acceleration of vehicles A and B = Uniform acceleration starting from rest

The maximum velocity attained by vehicle A = 30 m/s

The time it takes vehicle A to attain maximum velocity = 10 s

The maximum velocity attained by vehicle B = 30 m/s

The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s

The time duration vehicle A maintains its maximum velocity = 6 s

The time duration vehicle B maintains its maximum velocity = 4 s

(i) From the question, we get the following table;

[tex]\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}[/tex]

From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here

(ii) The velocity of vehicle A at the start = 0 m/s

After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s

The maximum velocity is maintained for 6 seconds which gives;

At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s

The time it takes vehicle A to decelerate to rest = 6 s

The deceleration of vehicle A, [tex]a_A[/tex] = (30 m/s - 0 m/s)/(6 s) = 5 m/s²

Therefore, we get;

v = u - [tex]a_A[/tex]·t

At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s

u = 30 m/s

∴ v₁₈ = 30 - 5 × 2 = 20

The velocity of vehicle A at the 18th second, [tex]V_{18A}[/tex] = 20 m/s

For vehicle B, we have;

At the 14th second, the velocity of vehicle B = 40 m/s

Vehicle B decelerates to rest in, t = 4 s

The deceleration of vehicle B, [tex]a_B[/tex] = (40 m/s - 0 m/s)/(4 s) = 10 m/s²

For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s

∴ [tex]v_{18B}[/tex] = 40 m/s - 10 m/s² × 4 s = 0 m/s

The velocity of the vehicle B at 18th second, [tex]v_{18B}[/tex] = 0 m/s

(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;

The area triangle A₁ = (1/2) × 10 × 30 = 150

Area of rectangle, A₂ = 6 × 30 = 180

Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50

The distance covered in the 18th second by vehicle [tex]S_A[/tex] = A₁ + A₂ + A₃

∴ [tex]S_A[/tex] = 150 + 180 + 50 = 380

The distance covered in the 18th second by vehicle [tex]S_A[/tex] = 380 m

The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;

Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440

The distance covered by the trapezoid, [tex]S_B[/tex] = 440 m

The distance of the two vehicles apart at the 18t second, [tex]S_{AB}[/tex] = [tex]S_B[/tex] - [tex]S_A[/tex]

∴ [tex]S_{AB}[/tex] = 440 m - 380 m = 60 m

The distance of the two vehicles from one another at the 18th second, [tex]S_{AB}[/tex] = 60 m.

Answer: