A ball is thrown upward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it moving 0.5 s later? In what direction? (b) How fast is it moving 2 s later? In what direction?​

Answer:

Quantum theory gives the concept of

Answer:

0.28 J

Explanation:

Let the mass of the object is 0.5 kg

The maximum velocity of the object is 1.06 m/s.

We need to find the kinetic energy at that time. It is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times 0.5\times (1.06)^2\\\\K=0.28\ J[/tex]

So, the required kinetic energy is equal to 0.28 J.

Answer:

[tex]t=0.50cm[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lamda=3c[/tex]m

Refraction Index [tex]n=1.50[/tex]

Generally the equation for Destructive interference for Normal incidence is mathematically given by

[tex]2nt=m(\frac{1}{2})\lambda[/tex]

Since  Minimum Thickness occurs at

At [tex]m=0[/tex]

Therefore

[tex]t=\frac{\lambda}{2}[/tex]

[tex]t=\frac{3}{4(1.50)}[/tex]

[tex]t=0.50cm[/tex]

Answer:

C. 16 m E

Explanation:

Applying,

The law of addition of vector: Vector in the same direction are added while vector in opposite direction are substracted

From the question above,

Step 1: Total distance covered towards east = 10+12 = 22 m E

Step2: Total distance covered towards west = 6 m W

Therefore, the resultant distance traveled = 22-6 = 16 m E

Hence the right option is C. 16 m E

Answer:

The correct answer is "[tex]21.344^{\circ}[/tex]" and "[tex]39.56^{\circ}[/tex]".

Explanation:

According to the question,

Slit width,

[tex]d=\frac{1}{910 \ lines/mm}[/tex]

  [tex]=\frac{1}{910\times 10^3}[/tex]

  [tex]=1.099\times 10^{-6} \ m[/tex]

The condition far first order maxima will be:

⇒ [tex]d Sin \theta = 1 \lambda[/tex]

Now,

⇒ [tex]\Theta_{min} = Sin^{-1} (\frac{\lambda}{d} )[/tex]

             [tex]=Sin^{-1} (\frac{400\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]

             [tex]=21.344^{\circ}[/tex]

⇒ [tex]\Theta_{max} = Sin^{-1} (\frac{\lambda}{d} )[/tex]

             [tex]=Sin^{-1} (\frac{700\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]

             [tex]=39.56^{\circ}[/tex]

Answer:

So the correct answer is letter e)

Explanation:

The electric field of an infinite yz-plane with a uniform surface charge density  (σ) is given by:

[tex]E=\frac{\sigma }{2\epsilon_{0}}[/tex]

Where ε₀ is the electric permitivity.

As we see, this electric field does not depend on distance, so the correct answer is letter e)

I hope it helps you!

Answer:

false?

Explanation:

The higher the modulus, the more stress is needed to create the same amount of strain; an idealized rigid body would have an infinite Young's modulus.

Answer:

I think the answer is False.

Answer:

In the given circuit, R

2

,R

6

and R

4

are in series. So,

R

1

′

=7+5+12=24Ω

Now R

1

′

and R

5

′

are in parallel. So,

R

2

′

1

=

8

1

+

24

1

=

24

3+1

=

24

4

=

6

1

R

2

′

=6ohm.

Now R

2

′

,R

1

and R

3

are in series. So,

R=R

2

′

+R

1

+R

3

=6+3+2=11ohm.

We know i=

R+r

E

=

11+1

6

=

12

6

=

2

1

i=0.5amp.

Answer:

speaker64

--------

34x

Explanation:

64-34

x

speaker

4

2

4

788

- circuit

voltage

100000

x.34

Sorry but you have no picture shown

Answer:

Well if you want to be sure you should just throw it to the ground so then when he lands he can catch it.

If the cannon throws the banana with the same force the monkey falls

(m.g=Fz <=> m.9,81N/kg=...N).

Then the throw will slow down because of the gravitational pull.

Because the banana cannon is selfmade you can choose what mass the bananas in question have, so let that be the same as the monkeys.

The monkey falls with the speed of 9,81m.s => so it takes the monkey 7,1s to land.

If the cannon can shoot the banana at the same speed the monkey falls then they would cross in the middle.

So to do so you need to throw the bananas with a speed of at least 9,81m.s

Soo ... throw them with a force of that is greater then the gravitational pull and things will work out.

I'm sorry I don't know why I wrote all of this irrelevant information it's 2:21 right now and I'm tired.

kind regards

Answer:

hehe

Explanation:

I dont know because I am a noob ant study

Answer:

34.64 cm

Explanation:

Given that:

The depth of the hole h = 10 cm

height of the water holding in the tank H = 40 cm

For a stream of flowing water, the distance (x) at which the stream strikes the floor can be  computed by using the formula;

[tex]x = 2 \sqrt{h(H-h)}[/tex]

[tex]x = 2 \sqrt{10(40-10)}[/tex]

[tex]x = 2 \sqrt{10(30)}[/tex]

[tex]x = 2 \sqrt{300}[/tex]

[tex]x = 2 \times 17.32[/tex]

x = 34.64 cm

Answer:

The covalent bond is formed by pairs of electrons that are shared between two atom

Explanation:

The covalent bond is formed by pairs of electrons that are shared between two atoms, in general the electrons must have opposite spins to have a lower energy state.

In this bond, the electrons are between the two atoms and are shared between them in such a way that there is a configuration of eight electrons in the orbit.

Answer:

the amount of energy needed is 1.8 x 10¹⁷ J.

Explanation:

Given;

mass of the object, m₀ = 1 kg

velocity of the object, v = 0.866 c

By physics convection, c is the speed of light = 3 x 10⁸ m/s

The energy needed is calculated as follows;

E = Mc²

As the object approaches the speed of light, the change in the mass of the object is given by Einstein's relativity formula;

[tex]M = \frac{M_0}{\sqrt{1- \frac{v^2}{c^2} } } \\\\ M = \frac{1}{\sqrt{1- \frac{(0.866c)^2}{c^2} } }\\\\ M = \frac{1}{\sqrt{1- \frac{0.74996c^2}{c^2} } }\\\\ M = \frac{1}{\sqrt{0.25} } \\\\ M = 2 \ kg[/tex]

The energy required is calculated as;

E = 2 x (3 x 10⁸)²

E = 1.8 x 10¹⁷ J

Therefore, the amount of energy needed is 1.8 x 10¹⁷ J.

Answer:

[tex]p.d' = 37.6 V[/tex]

Explanation:

From the question we are told that:

Potential difference [tex]p.d=18.8V[/tex]

New Capacitor [tex]C_1=C_2/2[/tex]

Generally the equation for Capacitor capacitance is mathematically given by

[tex]C=\frac{eA}{d}[/tex]

Generally the equation for New p.d' is mathematically given by

 [tex]C_2V=C_1*p.d'[/tex]

  [tex]p.d' = 2V[/tex]

 [tex]p.d'= 2 * 18.8[/tex]

 [tex]p.d' = 37.6 V[/tex]

Explanation:

Given:

[tex]A_1[/tex] = 4.5 cm[tex]^2[/tex]

[tex]v_1[/tex] = 40 cm/s

[tex]v_2[/tex] = 90 cm/s

[tex]A_2[/tex] = ?

a) The continuity equation is given by

[tex]A_1v_1 = A_2v_2[/tex]

Solving for [tex]A_2[/tex],

[tex]A_2 = \dfrac{v_1}{v_2}A1 = \left(\dfrac{40\:\text{cm/s}}{90\:\text{cm/s}}\right)(4.5\:\text{cm}^2)[/tex]

[tex]= 2\:\text{cm}^2[/tex]

b) If the cross-sectional area is reduced by 50%, its new area [tex]A_2'[/tex] now is only 1 cm^2, which gives us a radius of

[tex]r = \sqrt{\dfrac{A_2'}{\pi}} = 0.564\:\text{cm}[/tex]

Answer:

1) Law of Universal Gravitation     Gravity is an attractive force

5) General relativity               Gravity is due to the curvature of spacetime

Explanation:

In this exercise you are asked to relate the correct theory and its explanation

Theory Explanation

1) Law of Universal Gravitation              Gravity is an attractive force

2) Law of universal gravitation              Mass and distance affect force

3) Classical mechanics                           time and space are absolute

4) Special relativity                                 Time and space are relative

5) General relativity                                Gravity is due to the curvature of

                                                               spacetime

6) General relativity                                 Mass affects the curvature of space - time

Answer:

Explanation:

edge2022

Answer:

a)  I = 3.75 10⁴  kg  m², b)  w = 0.6 rad / s, c)  K = 6.75 10³ J, d) P = 6.75 10² W

Explanation:

This is a rotations exercise

a) the proper shape for a wheel is that of a rim where most of the weight is in the circumference plus the point weights of the people sitting on its periphery.

We are going to approximate the reda with a thin ring

           I = M r²

           I = 1500 5²

           I = 3.75 10⁴  kg  m²

b) angular and linear velocity are related

         v = w r

         w = v / r

         w = 3/5

         w = 0.6 rad / s

c) the expression for kinetic energy is

         K = ½ I w²

         K = ½ 3.75 10⁴   0.6²

         K = 6.75 10³ J

d) the power is

         P = W / t

to find the work we use the relationship between work and the variation of kinetic energy

         W = ΔK = K_f - K₀

the system part of rest wo = 0

         W = K_f

         W = 6.75 10³ J

we calculate

         P = 6.75 10³/10

         P = 6.75 10² W

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

[tex]W_{\rm total}=\Delta K[/tex]

or

[tex]W_{\rm friction}+W_{\rm spring}=0-K=-K[/tex]

where K is the block's kinetic energy at the equilibrium point,

[tex]K=\dfrac12\left(2.0\,\mathrm{kg}\right)\left(2.6\dfrac{\rm m}{\rm s}\right)^2=6.76\,\mathrm J[/tex]

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

[tex]W_{\rm spring}=-\dfrac12\left(250\dfrac{\rm N}{\rm m}\right)(0.20\,\mathrm m)^2=-5.00\,\mathrm J[/tex]

Compute the work performed by friction:

[tex]W_{\rm friction}-5.00\,\mathrm J=-6.76\,\mathrm J \implies W_{\rm friction}=-1.76\,\mathrm J[/tex]

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

∑ F = n - mg = 0   ==>   n = mg

where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.

So we have

[tex]W_{\rm friction}=-f(0.20\,\mathrm m)[/tex]

[tex]\implies -1.76\,\mathrm J=-\mu\left(2.0\,\mathrm{kg}\right)\left(9.8\dfrac{\rm m}{\mathrm s^2}\right)(0.20\,\mathrm m)[/tex]

[tex]\implies \boxed{\mu\approx0.45}[/tex]

The coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]

Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.

Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.

There are two forces performing work on the block: the restoring force of the spring and kinetic friction.

By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

[tex]W_{total}=\Delta K[/tex]

or

[tex]W_{friction}+W_{spring}=0-K=-K[/tex]

where K is the block's kinetic energy at the equilibrium point,

[tex]K=\dfrac{1}{2}(2)(2.6)^2=6.76 \ J[/tex]

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

[tex]W_{spring}=-\dfrac{1}{2}(250)(0.20)=-5\ J[/tex]

Compute the work performed by friction:

[tex]W_{friction}-5 =-6.76\ J=-1.76\ J[/tex]

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

∑ F = n - mg = 0   ==>   n = mg

where n is the magnitude of the normal force from the surface pushing up on the block. Then if f is the magnitude of kinetic friction, we have f = µmg, where µ is the coefficient of kinetic friction.

So we have

[tex]W_{friction}=-f(0.20)[/tex]

[tex]-1.76=\mu (2)(9.8)(0.2)[/tex]

[tex]\mu =0.45[/tex]

Thus the coefficient of kinetic friction between the block and the horizontal surface [tex]\mu =0.45[/tex]

To know more about Coefficient of friction follow

https://brainly.com/question/136431

Answer:

The answer is continuity ( D )

Explanation:

PLZ MARK AS BRAINLIEST

Answer:

Explanation:

Changing the resistance of a resistor means the resistance is either increased or decreased.

a. When the resistance of the resistor is increased, the value of current flowing through the circuit decreases.

Example: given voltage of 6V, and a resistance of 30 Ohm's. The value of current flowing in the circuit is;

V = IR

6 = I x 30

I = 0.2 A

If the resistance is changed to 50 Ohm's, then:

I = 0.12 A

(ii) When the resistance of the resistor is decreased, the value of the current flowing through the circuit increases.

In the previous example, if the resistance is changed to 5 Ohm's, then:

V = IR

6 = I x 5

I = 1.2 A

(b) The voltage of the battery does not change since it is directly proportional to the current flowing through the circuit. Consider the examples stated above.

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Answer:

a)   W = - 1.752 10⁻¹⁸ J,  b)    U = + 1.752 10⁻¹⁸ J

Explanation:

a) work is defined by

         W = F . x

the bold letters indicate vectors, in this case the force is electric

         F = q E

we substitute

         F = q E x

the charge of the electron is

         q = - e

         F = - e E x

let's calculate

         W = - 1.6 10⁻¹⁹  365  3 10⁻²

         W = - 1.752 10⁻¹⁸ J

b) the change in potential energy is

          U = q ΔV

the potential difference is

          ΔV = - E. Δs

we substitute

         U = - q E Δs

the charge of the electron is

           q = - e

          U = e E Δs

we calculate

           U = 1.6 10⁻¹⁹ 365  3 10⁻²

           U = + 1.752 10⁻¹⁸ J

Answer:

η = 0.5 = 50%

Explanation:

The efficiency of the power cycle is given by the following formula:

[tex]\eta = \frac{W}{Q_1}\\\\\eta = \frac{Q_1-Q_2}{Q_1}[/tex]

where,

where,

η = efficiency = ?

Q₁ = heat received from hot reservoir = 1000 KJ

Q₂ = heat discharged to cold reservoir = 500 KJ

Therefore,

[tex]\eta = \frac{1000\ KJ-500\ KJ}{1000\ KJ}[/tex]

η = 0.5 = 50%

length= metre

mass= kg

time= second

temperature = kelvin

current= ampere

luminous intensity= candela

Amount of substance = mole

etc

I hope this will help you

stay safe

Over the first 16.4 m, the person performs

W = (60.0 N) (16.4 m) = 984 J

of work.

Over the remaining 6.88 m, they perform a varying amount of work according to

F(x) ≈ 60.0 N + (-8.72 N/m) x

where x is in meters. (-8.72 is the slope of the line segment connecting the points (0, 60.0) and (6.88, 0).) The work done over this interval can be obtained by integrating F(x) over the interval [0, 6.88 m] :

W = ∫₀⁶˙⁸⁸ F(x) dx ≈ 206.4 J

(Alternatively, you can plot F(x) and see that it's a triangle with base 6.88 m and height 60.0 N, so the work done is the same, 1/2 (6.88 m) (60.0 N) = 206.4 J.)

So the total work performed by the person on the box is

984 J + 206.4 J = 1190.4 J ≈ 1190 J

Answer:

5g/cm

Explanation:

denisty=mass/volume

100/20

5g/cm

Answer:

The distance between sun & Earth at position A is less than the earth at position B. The gravitational force of two bodies is inversely proportional to the square of the distance. So At position A gravitational force is more & it decreases as it rotate towards position B.

Answer:

imma try and fail again and again

Answer:

Option A

Explanation:

From the question we are told that:

Mass [tex]m=0.20kg[/tex]

Velocity [tex]v=4m/s[/tex]

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

 [tex]M_{a1}=mV[/tex]

 [tex]M_{a1}=0.2*4[/tex]

 [tex]M_{a1}=0.8[/tex]

Final Momentum

 [tex]M_{a2}=-0.8kgm/s[/tex]

Therefore

 [tex]\triangle M_a=-1.6kgm/s[/tex]

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

 [tex]M_{b1}=mV[/tex]

 [tex]M_{b1}=0.2*4[/tex]

 [tex]M_{b1}=0.8[/tex]

Final Momentum

 [tex]M_{b2}=-0 kgm/s[/tex]

Therefore

 [tex]|\triangle M_a|>|\triangle Mb|[/tex]

Option A