QUESTION 5 When 235U is bombarded with one neutron, fission occurs and the products are three neutrons, 94Kr and?

a. 139Ba
b. 141Ba
c. 139Ce
d. 139Xe​

Answers

Answer 1
The answer is of the question is A

Related Questions

what is 4 differences between saturated unsaturated and supersaturated solutions

Answers

Answer:

Unsaturated Solution: Less amount of salt in water, clear solution, no precipitation. Saturated Solution: The maximum amount of salt is dissolved in water, Colour of the solution slightly changes, but no precipitation. Supersaturated Solution: More salt is dissolved in water, Cloudy solution, precipitation is visible.

1. As the angle of the ramp is increased, the normal force increases /decreases / remains the same and the friction-force increases /decreases / remains the same. [1 Point] 2. As the angle of the ramp is increased, the force parallel increases /decreases / remains the same. [1 Point] 3. The angle at which the force down the plane was equal to the force of friction (for the cabinet) was _____________. [1 Point] 4. Consider a very low (~ zero) friction, 5.0 kg skateboard on a ramp at an angle of 15o to the horizontal. What would be the net force that would cause acceleration when the skateboard is allowed to move

Answers

(1) As the angle of the ramp is increased, the normal force decreases.

(2) As the angle of the ramp is increased, the parallel force increases.

(3) The angle at which the force down the plane was equal to the force of friction is zero degree.

(4) The net force that would cause acceleration is 47.33 N.

Let the angle of inclination of the ramp = θ

(1)

The normal force on an object on the ramp inclined to the ramp is calculated as follows;

[tex]F_n = mgcos (\theta)[/tex]

when θ is 0;

[tex]F_n = mgcos (0)\\\\F_n = mg[/tex]

when θ is 90;

[tex]F_n = mgcos(90)\\\\F_n = 0[/tex]

Thus, as the angle of the ramp is increased, the normal force decreases.

(2)

The parallel force on an object on the ramp inclined to the ramp is calculated as follows;

[tex]F_x = mgsin(\theta)\\\\[/tex]

when θ is 0;

[tex]F_x = mgsin(\theta)\\\\F_x = mgsin(0) \\\\F_x = 0[/tex]

when θ is 90;

[tex]F_x = mgsin(90)\\\\F_x = mg[/tex]

Thus, as the angle of the ramp is increased, the parallel force increases.

(3)

The force of friction is calculated as follows;

[tex]F_n = \mu F_n[/tex]

[tex]F_k = \mu mgcos(\theta)[/tex]

[tex]F_k = \mu mg cos(0)\\\\F_k = \mu mg[/tex]

Thus, the angle is zero degree

(4)

The net force that would cause acceleration is calculated as follows;

[tex]F_k = Fn\\\\F_k = mg cos(\theta)\\\\F_k = 5 \times 9.8 \times cos(15)\\\\F_k = 47.33 \ N[/tex]

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define potential difference as used in electricity​

Answers

Answer:

This ability of charged particles to do work is called an electric potential.

Explanation:

When two negative charges are brought close to each other, they also repel. But when a positive and a negative charge are brought close together, they attract each other. When these two opposite charges are combined, they can be used to work. This is why we need a positive (+) and a negative (–) to light a bulb or run any electrical tool, equipment, mobile phone or home appliance.

Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit. ... The energy is transferred to the electrical components in a circuit when the charge carriers pass through them. We use a voltmeter to measure potential difference (or voltage).

A diffraction grating, ruled with 300 lines per mm, is illuminated with a white light source at normal incidence.
(i) What is the angular separation, in the third-order spectrum, between the 400 nm and 600 nm lines? [5]
(ii) Water (of refractive index 1.33) now fills the whole space between the grating and the screen. What is the angular separation, in the first-order spectrum, between the 400 nm and 600 nm lines? [5]​

Answers

the expression for diffraction grating allows to find the results for the questions for the angular separation are:

i) The third order is Δθ = 0.203 rad.

ii) The first order with water is Δθ = 0.046 rad.

The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.

          d sin θ = m λ

Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.

i) Let's start by looking for the separation between two lines

Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.

         d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm

         d = 3.333 10⁻⁶ m

Let's find the angle of diffraction for the third order (m = 3) for each wavelength.

λ₁ = 400 nm = 400 10⁻⁹ m

         sin θ₁ = [tex]\frac{m \ \lambda }{d}[/tex]m λ/ d

         sin θ₁ = [tex]\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]  

         θ₁ = sin⁻¹ 0.3600

         θ₁ = 0.368 rad

λ₂ = 600 nm = 600 10⁻⁹ m

         sin θ₂ = [tex]\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]  

         θ₂ = sin⁻¹ 0.5401

         θ₂ = 0.571 rad

The angular separation is

         Δθ = θ₂ - θ₁

         Δθ = 0.571 - 0.368

         Δθ = 0.203 rad

ii) In this case, the separation between the network and the observation screen is filled with water.

When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.

           [tex]n_i \ sin \theta_1 = n_r \ sin \theta_r[/tex]

The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.

Let's start looking for the incident angles for the first order of diffraction.

      m = 1

λ₁ = 400 nm

         θ₁ = sin⁻¹  [tex]\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}[/tex]

         θ₁ = 0.120 rad

λ₂ = 600 nm

        θ₂ = sin⁻¹¹ [tex]\frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}[/tex]

        θ₂ = 0.181 rad

we use the equation of refraction.

         [tex]\theta_r[/tex]  = sin⁻¹ ([tex]\frac{n_i}{n_r} \ sin \ \theta_i[/tex] )

λ₁ = 400 nm  

       θ₁ = sin¹ ([tex]\frac{1 sin 0.120}{1.33}[/tex]

       θ₁ = 0.090 rad

λ₂ = 600 nm

        θ₂ =sin⁻¹  [tex]\frac{1 sin 0.181}{1.33}[/tex]

        θ₂ = 0.1358 rad

The angular separation is

          Δθ = 0.1358 - 0.090

          Δθ = 0.046 rad.

In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:

       i) The third order is Δθ = 0.203 rad.

      ii) The first order with water is Δθ = 0.046 rad.

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A football is kicked with a velocity of 5 m/s at an angle of 53° above the horizontal. What is its speed at the
maximum height?
A) 3 m/s
B) 6 m/s
C) 9 m/s
D) 12 m/s
E) 15 m/s

Answers

A is the correct answer

A string is wrapped around a solid cylinder with mass M and radius R. The free end of the string is held in place, allowing the cylinder to fall. Recall that the moment of inertia of a solid cylinder rotated about its center is given by MR2/2. All answers to this problem should be symbolic, purely in terms of the variables M, R, and g. (a) Find the linear acceleration (in m/s2) of the cylinder and the tension in the string (in Newtons) as the cylinder falls. (b) Now suppose the cylinder is hollow instead of solid. The moment of inertia of a hollow cylinder rotated about its center is given by MR2. What is the acceleration and tension in this case?

Answers

Answer:

I will use (a / R) for alpha the angular acceleration

T R = I a / R     torque equals angular acceleration for cylinder

M g - T = M a    linear acceleration of center of mass

T = M (g - a) = I a / R^2 from first equation

If I = 1/2 M R^2 then M ( g - a) = M a / 2    from above

or g = 3 a / 2  and a = 2 g / 3

Also we have T = M (g - a) = M (g - 2 g / 3) = g / 3

Substitute I = M R^2 for the hollow cylinder

Looks like a = g/2 for hollow cylinder

Technician a says that the piston engine starts and stops several times per 2nd technician B says that with the rotary engine each rotor has 3 working chambers so it acts like a 3 cylinder engine while it is at operating

Answers

Answer:

Explanation:

trfbehherebgqerfherbfhyqerwbfvhrbhferhgvr

An astronaut standing on a platform on a foreign planet drops a hammer. If the hammer falls 9.0 meters vertically in 2.5 seconds, what is the acceleration due to gravity on that planet?

Answers

Answer: 1.646 m/s²

Explanation:

The distance that is traveled by the astronaut given that the motion is free-fall can be calculated through the equation,

   d = Vot + 0.5at²

where d is the distance, Vo is the initial velocity, t is the time, and a is the acceleration. Substituting the known,

  6 = (0 m/s)(2.7 s) + 0.5(a)(2.7 s)²

Determining the value of a,

   a = 1.646 m/s²

what is an electrostatic phenomenon?​

Answers

Answer:

Electrostatic phenomenon is a from the forces that electric charges exert on each other.

i just need help plzzzz

Answers

Answer:

I think D it's the correct answer!! Let me know if i'm wrong

Explanation:

Content from magnets

Answers

Answer:

magnets are magnetic :)

Explanation:

Need help on this thank you

Answers

Answer:

TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.

The lantern rises up because the __________ of the _________inside the lantern is _________ compared to the air around it

liquid

Gas

Density

fire

less dense

the same density

more dense

paper

Answers

density, fire, less dense

Answer:

gass fire and the same density

Explanation:

A brick is thrown upward from the top of a building at an angle of 25 degrees to the horizontal with an initial speed of 15 m/s. If the brick is in flight for 4 s before it hits the ground, how tall is the building?

Answers

Answer:

First, find the maximum height, which according to the values given, can be stated as:

H=(u²sin²theta)/2g

u=15m/s, theta=25 degrees, g=9.8m/s²

H= (15² * (sin 25)²))/2*9.8

H= (225*0.179)/19.6

H= 40.275/19.6

=2.06m

To find the velocity at maximum height:

Use the formula

v²=u²-2gH

It's minus because the brick was thrown upwards

So plugging everything into the above formula:

v²=15²-2*9.8*2.06

v²=225–40.376

v²=184.624

v=√184.624

v=13.59m/s

what force is represented by the vector​

Answers

Answer/explanation: A force vector is a representation of a force that has both magnitude and direction.

Which of the following statements about migration is true?

a. Migration is always from one region to another.
b. Animals always migrate within a region.
c. Migration always negatively impacts an ecosystem.
d. Animals migrate for various reasons.

Answers

Answer:

A and D

b and C are false

Answer:

D. Animals migrate for various reasons.

Explanation:

Got it right on Edge 2023. : )

Light of 650 nm wavelength illuminates two slits that are 0.20 mm
apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?

PLEAsE URGENT. thank you

Answers

We have that for the Question "Light of 650 nm wavelength illuminates two slits that are 0.20 mm  apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.

What is the distance to the screen?" it can be said that  the distance to the screen

d=1.168m

From the question we are told

Light of 650 nm wavelength illuminates two slits that are 0.20 mm

apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.

What is the distance to the screen?

Generally the equation for the distance  is mathematically given as

[tex]d=\frac{0.33*10^{-2}*0.23*10^{-3}}{650*10^{-9}}\\\\d=1.160m[/tex]

d=1.168m

Therefore

The distance to the screen

d=1.168m

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When people do a bungee jump theyare asked how much they weigh. Why?​

Answers

Answer:

The elasticity of the bungee cord reduces the gravitational forces applied on the body during bungee jumping. For example, if a 100-pound individual jumps from a building and encounters 900 pounds of deceleration force, they will feel 9 "G's" of force.

hopefully this'll help

have a nice day!!! :D

Please HELP!!

Diagram's BELOW↓

1. What is the mass of the object experimented on in this situation?

a. 10 kg
b. 15 kg
c. 20 kg
d. 25 kg


2. What is the net force on this object?

a. 20 N to the left
b. 10 N up
c. 10 N down
d. 35 N to the right


Thank you in advance! Very appreciated! (I will mark brainliest) :)

Answers

1. 20
2. 35n to the right

The mass of the object experimented in the situation will be equal to 20 kg in each case. Hence, option C is correct.

What is Acceleration?

Acceleration is the rate at which the speed and orientation of a moving object change over time. When a spot or object moves faster or slower along a straight line, it is said to be accelerated. Even though the speed is constant, motion on a circle accelerates so because the direction is always shifting. Both by and to acceleration for all other types of motion.

It is a vector quantity because it has both magnitude and direction.

The given data as per the question is,

Force, F = 30 N

Acceleration, a = 1.5 m/s²

[tex]Force = mass*acceleration[/tex]

30 N = m × 1.5

m = 30/1.5

m = 20 kg.

Therefore, the mass of the object is 20 kg.

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#SPJ2

1.Which term describes the sum of the number of protons and neutrons in an atom?

Answers

Your answer is → Mass number

PLEASE I NEED HELP WITH 6 and 8 !!!! I will love u sm

Answers

Answer:

Explanation:

006

They are acting in opposite directions. Therefore the net force is found by subtraction. The sign is the same as the larger number.

Net Force = 99.6 - 52.8 = 46.8 N acting in the same direction as the 99.6  which is upward.

008

If the two forces act in the same direction, the net force is found by addition.

Net Force = 99.6 + 52.8  = 152.4 N downward.

They’re acting in opposite directions o

A mover pushes a 45 kg crate across a level floor with a force of 300 N, but the crate accelerates at a rate of only 4.44 m/s2 because a friction force opposes the crate's motion. What is the magnitude of this force of friction? 300 N Force of Friction O A. 50 N N OB. 25 N C. 15 N 0 D. 100 N​

Answers

By Newton's second law, the net force on the crate is

F = 300 N - f = (45 kg) (4.44 m/s²)

where f is the magnitude of friction. Solve for f :

300 N - f = (45 kg) (4.44 m/s²)

f = 300 N - (45 kg) (4.44 m/s²)

f = 100.2 N ≈ 100 N (D)

(c) Changing water into vapour is condensation true or false​

Answers

Answer:

true

Explanation:

The first motor abilities a new born exhibits are

Answers

Answer:

here your answer

i am sorry if wrong

Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ tốc độ của vật lúc chạm đất?

Answers

Answer:

Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ cao cực đại của vật so với mặt đất?

Explanation:

in order to keep heat in or out, you need a(n)

Answers

Answer:

What is instillation

Explanation:

The displacement of an object in SHM is described by the equation
[tex] x = cos\binom{2\pi}{3}t[/tex]
where x is in meters and t in seconds. Determine the velocity of the object at t = 0.6 s. ​

Answers

Answer:

[tex]-1.99\:\mathrm{m/s}[/tex]

Explanation:

Assuming that the equation is intended to be [tex]\displaystyle x=\cos\left(\frac{2\pi}{3}t\right)[/tex], we can find the velocity vs. time equation by taking the first derivative with respect to [tex]t[/tex]:

[tex]\displaystyle \frac{dx}{dt}=\frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)[/tex]

Recall the chain rule:

[tex]\displaystyle f(g(x))'=f'(g(x))\cdot g'(x)[/tex]

Therefore,

[tex]\displaystyle \frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex]

Therefore, the velocity vs. time equation of the object is [tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex].

Substitute [tex]t=0.6\text{ s}[/tex] into this equation to find the velocity at that given time:

[tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}(0.6)\right)\cdot \frac{2\pi}{3}\approx \boxed{-1.99\text{ m/s}}[/tex]

The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 14900 N and acts as shown in the figure. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.

Answers

We have that for the Question, it can be said that With respect to this axis, the magnitude of the torque due to the weight and ,the thrust is

TW=19740N-mTT=130387.39N-m

From the question we are told

The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 14900 N and acts as shown in the figure. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.

a)

Generally the equation for the Torque due to weight  is mathematically given as

[tex]TW=Engine weight*2.50*sin32\\\\TW=14900*2.50*sin32[/tex]

TW=19740N-m

b)

Generally the equation for the Torque due to thrust  is mathematically given as

[tex]TT=Engine thrust*2.50*cos32\\\\TT=61500*2.50*cos32\\\\[/tex]

TT=130387.39N-m

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relationship between the length (l) and ohmic conductor, its thickness (d) and the opposition (R) it offers a current passage​

Answers

Answer:

Ohm’s Law states that current is proportional to voltage; circuits are ohmic if they obey the relation V=IR.

Explanation:

LEARNING OBJECTIVES

Contrast shape of current-voltage plots for ohmic and non-ohmic circuits

A 0.750 kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board. (a) Calculate the duration of the impact in seconds. (Enter a number.) s (b) What was the average force in newtons exerted downward on the nail

Answers

Answer:nh

0.750 kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board.

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