A rifle is aimed at a target 40m away. The bullet hits the target 2.2cm below the horizontal plane (ignore wind and rotational effects).
(a) What is the time of flight for bullet?
(b) What is the muzzle velocity (velocity of bullet when it leaves the rifle)?

Answer:

Send the pic so I can see

Answer:

B

Explanation:

I'm learning it in science.

Answer:

its not b i just took the test and b was wrong

Explanation:

The Box's Acceleration : g sin θ

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object  

∑F = m. a  

F = force, N  

m = mass = kg  

a = acceleration due to gravity, m / s²  

We plot the forces acting on the block (picture attached) according to the y-axis and the x-axis.

Because the motion of the block is in the same direction as the x-axis, ignoring the friction force with the inclined plane, then

[tex]\tt \sum F_x=m.a\\\\W.sin\theta=m.a\\\\mgsin\theta=m.a\\\\a=gsin\thet\theta[/tex]

Answer:

spirit of the team / game

Answer:

True  b and c

Explanation:

In an RLC circuit the impedance is

         [tex]Z = \sqrt{[R^{2} + ( (wL)^{2} + (\frac{1}{wC})^{2} ] }[/tex]

examine the different phrases..

a) False. The maximum impedance is the value of the resistance  

b) True. Resonance occurs when  

              (wL)² + (1 / wC)² = 0

               w² = 1 / LC

c) True. In resonance the impedance is the resistive part and the power is maximum  

d) False. In resonance the inductive and capacitive part cancel each other out  

e) False. The impedance is always greater outside of resonance, but at the resonance point they are equal

Answer:

a) v = 5.59x10³ m/s

b) T = 4 h

c) F = 1.92x10³ N

Explanation:

a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:

[tex] F_{c} = F_{G} [/tex]

[tex]\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}[/tex]

[tex] v = \sqrt{\frac{gr^{2}}{r+h} [/tex]          

Where:

g is the gravity = 9.81 m/s²        

r: is the Earth's radius = 6371 km

h: is the satellite's height = r = 6371 km      

[tex]v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s[/tex]                                      

b) The period of its revolution is:

[tex] T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h [/tex]

c) The gravitational force acting on it is given by:

[tex] F = \frac{GMm}{(r + h)^{2}} [/tex]

Where:

M is the Earth's mass =  5.97x10²⁴ kg    

m is the satellite's mass = 782 kg

G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²

[tex] F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N [/tex]

I hope it helps you!

Answer:

Acceleration

Explanation;

Answer:

C. Some of the mass of the four hydrogen nuclei was converted into

energy

Explanation:

a p e x , just took the quiz

Some of the mass of the four hydrogen nuclei was converted into energy.

When four hydrogen nuclei fuse to form one helium nuclei, mass defect, some mass is converted into energy, that is the reason of energy of the sun.

"When two nuclei form a big nuclei, the phenomenon is known as fusion."

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#SPJ2

By conservation of momentum,

Pinitial = Pfinal

m1v1 + m2v2 = (m1 + m2)*vf

m1 = mass of skateboard = 1.13 kg

m2 = mass of cat = 0.93 kg

v1 = initial velocity of skateboard = 4.28 m/s

v2 = initial velocity of cat = 0 m/s

vf = final velocity of skateboard-cat combo

So plug in the values and solve for vf,

1.13(4.28) + 0.93(0) = (1.13 + 0.93)vf

vf = 2.35 m/s

The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

• net vertical force:

∑ F = N - W = 0

• net horizontal force:

∑ F = Fs = m a

where

N = magnitude of normal force

W = car's weight

Fs = mag. of static friction

m = car's mass

a = v ²/R = mag. of the centripetal acceleration

v = car's speed

R = radius of curve

Now,

• compute the car's weight:

W = m g = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

N = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

Fs = 0.5 N = 7350 N

• solve for the (maximum) acceleration:

7350 N = (1500 kg) a   →   a = 4.9 m/s²

• solve for the (maximum) speed:

4.9 m/s² = v ²/ (35 m)   →   v ≈ 13 m/s

Answer:

a) The velocity when it passes the equilibrium point is [tex]\pm 2.451[/tex] meters per second.

b) The velocity when it is 0.10 meters from equilibrium is [tex]\pm 1.567[/tex] meters per second.

c) The total energy of the system is 1.802 joules.

d) The equation describing the motion of the mass, assuming that initial position is a maximum is [tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex].

Explanation:

a) If all non-conservative forces can be neglected and spring has no mass, then the mass-spring system exhibits a simple harmonic motion (SHM). The kinematic formula for the position of the system ([tex]x(t)[/tex]), measured in meters, is:

[tex]x(t) = A\cdot \sin(\omega \cdot t +\phi)[/tex] (1)

Where:

[tex]A[/tex] - Amplitude, measured in meters.

[tex]\omega[/tex] - Angular frequency, measured in radians per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]\phi[/tex] - Phase, measured in radians.

The kinematic equation for the velocity formula of the system ([tex]v(t)[/tex]), measured in meters per second, is derived from (1) by deriving it in time:

[tex]v(t) = \omega\cdot A\cdot \cos (\omega\cdot t+\phi)[/tex] (2)

The velocity when it passes the equilibrium point occurs when the cosine function is equal to 1 or -1. Then, that velocity is determined by following formula:

[tex]v = \pm \omega\cdot A[/tex] (3)

The angular frequency is calculated by this expression:

[tex]\omega = 2\pi\cdot f[/tex] (4)

Where [tex]f[/tex] is the frequency, measured in hertz.

If we know that [tex]f = 3\,hz[/tex] and [tex]A = 0.13\,m[/tex], then the velocity when it passes the equilibrium point, which is the maximum and minimum velocities of the mass:

[tex]\omega = 2\pi\cdot (3\,hz)[/tex]

[tex]\omega \approx 18.850\,\frac{rad}{s}[/tex]

[tex]v = \pm \left(18.850\,\frac{rad}{s} \right)\cdot (0.13\,m)[/tex]

[tex]v = \pm 2.451\,\frac{m}{s}[/tex]

The velocity when it passes the equilibrium point is [tex]\pm 2.451[/tex] meters per second.

b) First, we need to determine the spring constant of the system ([tex]k[/tex]), measured in newtons per meter, in terms of the angular frequency ([tex]\omega[/tex]), measured in radians per second, and mass ([tex]m[/tex]), measured in kilograms. That is:

[tex]k = \omega^{2}\cdot m[/tex] (5)

If we know that [tex]\omega \approx 18.850\,\frac{rad}{s}[/tex] and [tex]m = 0.60\,kg[/tex], then the spring constant is:

[tex]k = \left(18.850\,\frac{rad}{s} \right)^{2}\cdot (0.60\,kg)[/tex]

[tex]k = 213.194\,\frac{N}{m}[/tex]

Lastly, we determine the velocity when the mass is 0.10 meters from equilibrium by the Principle of Energy Conservation:

[tex]U_{k} + K = K_{max}[/tex] (6)

[tex]\frac{1}{2}\cdot k\cdot x^{2} + \frac{1}{2}\cdot m\cdot v^{2} = \frac{1}{2}\cdot m\cdot v_{max}^{2}[/tex] (7)

Where:

[tex]U_{k}[/tex] - Current elastic potential energy, measured in joules.

[tex]K[/tex] - Current translational kinetic energy, measured in joules.

[tex]K_{max}[/tex] - Maximum translational kinetic energy, measured in joules.

[tex]v[/tex] - Current velocity of the system, measured in meters per second.

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v_{max}[/tex] - Maximum velocity of the system, measured in meters per second.

If we know that [tex]k = 213.194\,\frac{N}{m}[/tex], [tex]x = 0.10\,m[/tex], [tex]m = 0.60\,kg[/tex] and [tex]v_{max} = \pm 2.451\,\frac{m}{s}[/tex], then the velocity of the mass-spring system is:

[tex]\frac{k}{m} \cdot x^{2} + v^{2} = v_{max}^{2}[/tex]

[tex]v^{2} = v_{max}^{2}-\frac{k}{m}\cdot x^{2}[/tex]

[tex]v = \sqrt{v_{max}^{2}-\frac{k\cdot x^{2}}{m} }[/tex] (8)

[tex]v = \sqrt{\left(\pm 2.451\,\frac{m}{s} \right)^{2}-\frac{\left(213.194\,\frac{N}{m} \right)\cdot (0.10\,m)^{2}}{0.60\,kg} }[/tex]

[tex]v \approx \pm 1.567\,\frac{m}{s}[/tex]

The velocity when it is 0.10 meters from equilibrium is [tex]\pm 1.567[/tex] meters per second.

c) The total energy of the system ([tex]E[/tex]), measured in joules, can be determined by the following expression derived from the Principle of Energy Conservation:

[tex]E = \frac{1}{2}\cdot m\cdot v_{max}^{2}[/tex] (9)

If we know that [tex]m = 0.60\,kg[/tex] and [tex]v_{max} = \pm 2.451\,\frac{m}{s}[/tex], then the total energy of the system is:

[tex]E = \frac{1}{2}\cdot (0.60\,kg)\cdot \left(\pm 2.451\,\frac{m}{s}\right)^{2}[/tex]

[tex]E = 1.802\,J[/tex]

The total energy of the system is 1.802 joules.

d) Given that initial position of the mass-spring system is a maximum, then we conclude that the equation of motion has the following parameters: ([tex]A = 0.13\,m[/tex], [tex]\omega \approx 18.850\,\frac{rad}{s}[/tex] and [tex]\phi = 0.5\pi\,rad[/tex])

From (1) we obtain the resulting formula:

[tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex] (10)

The equation describing the motion of the mass, assuming that initial position is a maximum is [tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex].

Explanation:

The width of the central maximum is given by

W = 2 λ L / a

where W is the width of the central maximum

λ is the wavelength of the light used.

L is the distance between the aperture and screen

a is the width of the slit or aperture

So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.

Answer:

The torque on the child is now the same, τ.

Explanation:

       [tex]I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)[/tex]

       [tex]I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)[/tex]

       τ = 3/4*m*r² * (2α) = 3/2*m*r²

        same result than in (2), so the torque remains the same.

Answer:

√3 * Gm²/d²

Explanation:

m1 = m, m2= m, distance = d. hence:

F = Gm²/d²

Let the origin be A, the point x = d be B and the point above the first two is C.

The net force acting on the third mass (point C) [tex]F_{net}[/tex] = [tex]F_A+F_B[/tex]

Let j represent the vertical component and i the horizontal component. Hence:

[tex]F_B=-F_j\\\\F_A=-F(icos\frac{\pi}{6}+jsin\frac{\pi}{6} )=-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )\\\\F_{net} =F_A+F_B\\\\F_{net} =-F_j+{-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )}\\\\F_{net} =-\frac{F}{2} \sqrt{3}(i+j\sqrt{3} )\\\\The\ magnitude\ of\ the\ net\ force\ is:\\\\|F_{net}|=\frac{F}{2}\sqrt{3}(\sqrt{1^2+\sqrt{3}^2 })=\frac{F}{2} \sqrt{3}(\sqrt{4})\\\\|F_{net}|=\frac{F}{2} \sqrt{3}*2=F*\sqrt{3}\\\\|F_{net}|=\sqrt{3}*\frac{Gm^2}{d^2}[/tex]

Answer:

Explanation:

For equilateral triangle,

[tex]\theta = 30^o[/tex]

Force between masses,

[tex]F_1 = \frac{G*m*m}{d^2}\\\\F_! = \frac{Gm^2}{d^2}[/tex]

Therefore,

[tex]F_net = 2F_1cos\theta\\\\F_net = 2 * \frac{Gm^2}{d^2} * cos30\\\\F_net = 3^{0.5} * \frac{Gm^2}{d^2}[/tex]

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Answer:

Explained below.

Explanation:

Conductors can be defined as materials that permit electricity to flow through them easily.

Now, metals have a molecular structure that makes them good conductors because electrons in the atoms of these conductors tend to move freely from one atom to the other. So a majority of metals make good conductors because these metals tend to hold their electrons loosely. In short, it can help engineers make powerful batteries because then it means that they are capable of giving much more electrical energy since nowadays, advanced batteries make use of ion charges for the batteries.

Answer:

protons are in the nucleus .

Explanation:

there are 6 protons

Answer:

E_{pot} = 132.3 [J]

Explanation:

In order to solve this problem, we must use the definition of the potential energy which can be calculated by means of the following formula.

[tex]E_{pot}=W*h[/tex]

where:

Epot = potential energy [J]

W = weight = 5.4 [N]

h = elevation = 24.5 [m]

[tex]E_{pot}=5.4*24.5\\E_{pot}=132.3[J][/tex]

Answer:

3A. This phenomenon can be seen in the discrete emission of the molecules.

3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.

3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams

3D. This is due to the stimulated emission

3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus)

Explanation:

This problem asks for some experimental explanations of various quantum phenomena.

3A. This phenomenon can be seen in the discrete emission of the molecules.

In the classical explanation all states or energies are allowed, therefore when emitting energy (photons) there should be a continuum, this is not observed

In the correct quantum explanation only some states are allowed, therefore the emission must be discrete, which is observed in the emission or absorption of molecules and atoms

3B. The emotion of the atoms is observed, from states high in energy to a state of minimum energy that is stable indefinitely.

The incorrect classical explanation that if the minimum energy was zero the electrons cannot rotate around the nuclei and the atom collapses, this does not happen

3C. When an electron beam passes through an inhomogeneous magnetic field, it is divided into only two beams, which is evidence of the existence of two discrete states that we call spin, remember that a free electron beam has zero angular momentum.

3D. This is due to the stimulated emission that occurs when a photon passes through the emission zone, causing the atoms to have transitions and these emitted photons have the same initial photon location, the laser beam all photons are in phase and therefore it is coherent .

This is widely used for holographic and interference work

3E. The penetration of a potential barrier is observed in the radioactive emission of heavy atoms, where an alpha particle (Helium nucleus) leaves the atomic nucleus penetrating the barrier since its energy is lower than the nuclear barrier potential.

Answer:

Explanation:

Please provide some experimental demonstration or explanation of natural observations or characteristics of application which confirm(s) / display(s) the. Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature. Quantum mechanics arose gradually from theories to explain observations. By M Arndt · 2009 · Cited by 239 — Wave-particle duality of light: A paradigmatic example for this fact is the dual nature of light that manifests itself in Young's famous double-slit diffraction.

Answer:

[tex]\boxed{True}[/tex]

Explanation:

Objects are weightless in free fall because their is no other force pushing them rather than the gravitational force.

Hope it helps!<3

Answer:

Explanation:

Answer:

the normal force that is perpendicular to the surface has a component towards the center of the curve.

Explanation:

When curves with superelevation or inclination, the normal force that is perpendicular to the surface has a component towards the center of the curve. This component is what maintains the vehicle and gives it centripetal acceleration.

Consequently, the vehicle does not need the friction force since it does not have a tendency to slide and this is zero.

Answer:

Explanation:

The mass of the car can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{2250}{1.4} \\ = 1607.14285...[/tex]

We have the final answer as

Hope this helps you

Answer:

Explanation:

When the puck is sliding on the ice, there is no force being exerted on the puck to keep it moving forward. Instead, inertia keeps the puck moving forward. Friction between the puck and the ice gradually slows the puck down. You hit a hockey puck and it slides across the ice at nearly a constant speed

At constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.

According to Newton's second law of motion, the force applied to a an object is directly proportional to the product of mass and acceleration of the object.

F = ma

Acceleration is the change in the velocity of an object per change in time of motion.

Thus, we can conclude that at constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.

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Answer:

Please mark as brainliest!!

Explanation:

In coin card experiment smooth card is used so that the card can slide easily from glass.

Answer:

In coin card experiment smooth card is used so that the card can slide easily from glass.

Answer:

The apparent fit of the eastern coastline of South America and western coastline of Africa

Similarities of plants and animal fossils in South America and some parts of African continent which were separated by a vast ocean

Similarities in the sequence of rock layers of opposite sides of the Atlantic Ocean

Answer:

D

Explanation:

Nothing can enter or leave so it remains constant

Answer:

A

Explanation:

Weight = 55 * 9.8 = 539 N

Scale = 686 N

Since the scale reading is larger than the lady’s weight, the elevator must be accelerating as it moves upward.

Net upward force = 686 – 539 = 147 N

147 = 55 * a

a = 147 ÷ 55

The acceleration is approximately 2.67 m/s^2.