In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a calculation in a very artificial situation. Suppose a man standing on frictionless ice fires a rifle horizontally. The mass of the man together with the rifle is 70 kg, and the mass of the bullet is 10 g. If the bullet leaves the muzzle at a speed of 500 m/s, what is the final speed of the man?

Answer: T = 692.82 and 346.4 N

Explanation:

Given that;

w = 600 N

∅ = 30°

ΣFy = ma

a = 0 m/s²

ΣF = T(cos30°) - W = 0

T(cos30°) = W

we Divide both sides by cos30°

T = W / cos30o

T= 600N / cos30°

T = 692.82

and ∑fx

F = T sin∅

F = 692.82 × (sin30°)

F = 346.4 N

The equilibrium condition allows finding the result for the force of the ball against the wall is:

Newton's second law gives a relationship between force, mass and acceleration of bodies. In the case where the acceleration is zero, it is called the equilibrium condition.

            ∑ F = 0

A free-body diagram is a diagram of the forces without the details of the bodies. In the attached we have a free-body diagram of the system.

Let's use trigonometry to break down stress.

          sin 30 = [tex]\frac{T_x}{T}[/tex]  

          cos 30 = [tex]\frac{T_y}{T}[/tex]  

          T_y = T cos 30

          Tₓ = T sin 30

Let's write the equilibrium condition for the system.

y-axis.

         T_y -W = 0

          T cos 30 = W

          [tex]T = \frac{W}{cos 30}[/tex]  

x-axis.

        R - Tₓ = 0

        R = T sin 30

We substitute

        [tex]R = \frac{W}{cos 30} \ sin 30 \\R = W \ tan 30[/tex]

Let's calculate.

        R = 600 tan 30

        R = 346.4 N

This force is directed from the wall towards the ball, by Newton's third law the force of the ball is of equal magnitude and opposite direction, that is, directed towards the wall.

In conclusion with the equilibrium condition we can find the result for the force of the ball against the wall is:

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Answer:

energy

Explanation:

im sure

Answer:

A

Explanation:

Hopefully this helps.

Answer:

The value is  [tex]n = 18.5 \ oscillations[/tex]

Explanation:

From the question we are told that

   The mass is  [tex]m = 350 \ g = 0.350 \ kg[/tex]

   The length is  [tex]L = 45 \ cm = 0.45 \ m[/tex]

    The angle is  [tex]\theta = 4.5^o[/tex]

    The damping constant is  [tex]b = 0.010 \ kg/s[/tex]

    The time taken is [tex]t = 25 \ s[/tex]

Generally the angular frequency of this damped oscillation is mathematically evaluated as

     [tex]w = \sqrt{ \frac{ g}{L} + \frac{b^2}{4m^2} }[/tex]  

=>   [tex]w = \sqrt{ \frac{9.80 }{ 0.45} + \frac{0.010 ^2}{4* 0.350^2} }[/tex]  

=>   [tex]w = 4.667 \ s^{-1}[/tex]

Generally the period of the oscillation is mathematically represented as

      [tex]T = \frac{2 \pi }{w}[/tex]  

=>  [tex]T = \frac{2 * 3.142 }{ 4.667 }[/tex]

=>  [tex]T = 1.35 \ s[/tex]

Generally the number of oscillation is mathematically represented as

        [tex]n = \frac{t}{T}[/tex]

=>     [tex]n = \frac{25}{1.35}[/tex]

=>     [tex]n = 18.5 \ oscillations[/tex]

Answer:

m = 1.35 kg

Explanation:

       [tex]U_{i} + K_{i} = U_{f} + K_{f} (1)[/tex]

        [tex]U = \frac{1}{2} * k * \Delta x^{2} (2)[/tex]

        where k is the spring constant = 60.0 N/m, and Δx is the distance from

        the equilibrium  position.

        [tex]\frac{1}{2} * k *\Delta x_{i} ^{2} =\frac{1}{2} * k *\Delta x_{f} ^{2} +\frac{1}{2} * m *v ^{2} (3)[/tex]

       m =1.35 kg

Answer: True

Explanation: I took the quiz and got it right! have a great day.

True, tornadoes are accompanied by spinning downdrafts and updrafts that form a funnel cloud.

What is a Tornado?

A tornado can be defined as a spinning air column, usually violet resulting from a thunderstorm in the cloud and extends to the ground.

These tornadoes are accompanied by spinning downdrafts and updrafts that form a funnel cloud.

The major constituent of tornadoes are:

Thus, we can conclude that the given statement about tornadoes is true.

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Answers:

Question 1: 3 sig figs

Question 2: 4 sig figs

Question 3: 5 sig figs

Question 4: 8 sig figs

The rule for sig figs:

Any number after a zero in a decimal is one sig fig.

Example: 0.00359, it has three sig figs.

Also, remember if there are zeros after a number greater than zero, it is also a sig fig as well.

Hope this helps you!

Answer:

1) 3 sf  2) 3 sf  3) 5 sf  4) 6 sf

Explanation:

00s at the start dont count towards the sig figs and 00s at the end dont either but 00s in the middle do

your welcome :)

Answer:

3.49m/s

Explanation:

We have mess of kit = 4kg

Then pieces are 2kg

3m/s due north

5m/s, 30⁰ due south

MiVi = m1v1 + m2v2

We break velocities to have x and y components

MiVi = m1v1(cosθ+sin θ1j) + m2v2(v2cos θ2i+sin2 θj)

θ1 = 90⁰north

θ2 = 30⁰ north east

Vi = 2/4x3m/s(cos90i + sin90j)+2/4x5m/s(cos30i + sin30j)

= 2.16i + 2.75j

Vi = √Vx²+Vy²

Vi = √ (2.16)²+(2.75)²

Vi = 3.49m/s

This is the original speed of the kit

Answer:

im just so focused on the fact that im going to mars :O

Answer:

2 m/s 2

Explanation: Just is....

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, to determine the bicyclist that traveled the fastest at the end of the race, the speed of the bicyclists at the end of the race will determine this (not the bicyclist that came first nor there overall speed). The speed of the bicyclist at the end of the race can be determined by using the formula below

s = d ÷ t

Where s is the speed of each bicyclist at the end of the race

d is the specific distance covered by the bicyclist at the end of the race

t is the time taken for the bicyclist to complete that distance

It should be noted that to get an accurate result, the distance covered at the end of the race must be the same for all the bicyclists.

Answer:

960.4N

Explanation:

Given parameter:

Mass of the body  = 98kg

Unknown:

Force exerted by gravity  = ?

Solution:

Force exerted by gravity on the body is the weight

  Weight  = force x acceleration due to gravity

Acceleration due to gravity = 9.8m/s²

  Weight  = 98kg x 9.8m/s²

  Weight  = 960.4N

Answer:

Tension= 475N

Force= 225N

Explanation:

The question is not complete, here is the complete question

Also, see attached a free body diagram for your reference

"A 20.0-kg uniform plank is supported by the floor at one end by a vertical rope at the other as shown in the figure. A 50.0-kg mass person stands on the plank a distance three-fourths of the length plank from the end on the floor.

a. What is the tension in the rope?

b. What is the magnitude of the force that the floor exerts on the plank?"

given data

mass of man=50kg

mass of plank=20kg

length of plank=10m

let us make the lenght of the rope be d

The torque about the floor

That is taking moment about the floor

[tex]N*0+T*d=20*10*d/2 + 50*10*3d/4\\\\T=100+375=475N\\\\[/tex]

Force will be also zero  

[tex]N+T=20*10+50*10\\\\N+T=700 \\\\N=700-475=225 newtons\\\\N+T=20*10+50*10\\\\N+T=700\\\\N=700-475=225newtons[/tex]

Answer:

P = 2893.95 [kPa]

Explanation:

The manometric pressure can be calculated by means of the following equation.

[tex]P=Ro*g*h[/tex]

where:

Ro = density of the water = 1000 [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = deep = 295 [m]

Now replacing:

[tex]P = 1000*9.81*295\\P = 2893950 [Pa]\\P = 2893.95 [kPa][/tex]

Answer:

[tex]v=\sqrt{26}~m/s[/tex]

Explanation:

Parametric Equation of the Velocity

Given the position of the particle at any time t as

[tex]r(t) = (x(t),y(t))[/tex]

The instantaneous velocity is the first derivative of the position:

[tex]v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))[/tex]

The speed can be calculated as the magnitude of the velocity:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]

We are given the coordinates of the position of a particle as:

[tex]x=5t-3t^2[/tex]

[tex]y=5t[/tex]

The coordinates of the velocity are:

[tex]v_x(t)=(5t-3t^2)'=5-6t[/tex]

[tex]v_y(t)=(5t)'=5[/tex]

Evaluating at t=1 s:

[tex]v_x(1)=5-6(1)=-1[/tex]

[tex]v_y(1)=5[/tex]

The velocity is:

[tex]v=\sqrt{(-1)^2+5^2}[/tex]

[tex]v=\sqrt{1+25}[/tex]

[tex]\mathbf{v=\sqrt{26}~m/s}[/tex]

Answer:

[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]

Explanation:

Given that,

A radio wave transmits 38.5 W/m² of power per unit area.

A flat surface of area A is perpendicular to the direction of propagation of the wave.

We need to find the radiation pressure on it. It is given by the formula as follows :

[tex]P=\dfrac{2I}{c}[/tex]

Where

c is speed of light

Putting all the values, we get :

[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]

So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].

When a substance undergoes fusion, it melts.

Fusion is a physical process. The process of fusion of substances takes place during the conversion of a solid substance into a liquid.

For this process to take place, heat is being applied to the solid substance, and the particles of the solid loose, then move freely thereby reducing the intermolecular forces in the solid substance before changing into liquid.

Learn more about physical changes here:

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Answer:

1. A. 8.22. m/s

Explanation:

Answer: The Answer is attached to the image below

Answer:

0.1875 m/s leftward

Explanation:

Taking rightwards as positive

We are given:

Ball 1:

Mass (m1) = 1.5 kg

velocity (u1) = 2 m/s

Ball 2:

Mass (m2) = 2.5 kg

velocity (u2) = -1.5 m/s          [negative because it is in the opposite direction]

Speed and Direction of Ball 2:

We are told that the balls stick together after the collision

We can say that the balls have the same velocity since they are sticking together

So, Final velocity of Ball 1 (v1) = Final velocity of Ball 2 (v2) = V m/s

According to the law of conservation of momentum

m1u1 + m2u2 = m1v1 + m2v2

replacing the variables

1.5(2) + (2.5)(-1.5) = V (1.5 + 2.5)                     [v1 = v2 = V]

3 + (-3.75) = 4V

-0.75 = 4V

V = -0.75/4                                                    [dividing both sides by 4]

V = -0.1875 m/s

Hence, the balls will move at a velocity of 0.1875 m/s in the Leftward direction

Answer:

A) I = 0.09947 W , β = 109 db , B) β = 116 db , β = 116 db , c) Δβ = 7 dB,

D)  P = 50.27 W

Explanation:

A) The intensity of a spherical sound wave is

          I = P / A

where A is the area of ​​the sphere where the sound is distributed

          A = 4π R²

we substitute

          I = P / 4πR²

let's calculate

          I = 500 / (4π 20²)

          I = 0.09947 W

to express this quantity in decibels we use relate

         β = 10 log (I / I₀)

The detectivity threshold is I₀ = 1 10⁻¹² W / m²

         β = 10 lob (0.09947 / 10⁻¹²)

         β = 10 (10.9976)

         β = 109 db

B) intensity at r = 10m

          I = 500 / (4π 10²)

          I = 0.3979 W / m²

          β = 10 log (0.3979 / 10⁻¹²)

          β = 10 (11.5997)

          β = 116 db

C) the change in intensity in decibles is

          Δβ = β₁ - β₂

          Δβ = 116 - 109

          Δβ = 7 dB

D) let's find the intensity for 100 db

          I = I₀ 10 (β / 10)

          I = 10⁻¹² 10 (100/10)

          I = 10⁻² W / m²

Thus

           P = I A

           P = I 4π R²

           P = 10⁻² 4π 20²

           P = 50.27 W

Answer:

Explained below

Explanation:

In skying down a hill, usually the skiers start at an elevated position and this means that they possess a large quantity of potential energy since they are in vertical position.

Now, as the skiers start to descend down the hill, they will lose potential energy while they gain kinetic energy since they are in motion. This is because there is reduction in height which results in a loss of potential energy and there is an increase in their speed which results in an increase in kinetic energy.

Now, immediately the skiers reaches the bottom of the hill, it means they are now at zero level height which means potential energy is now zero and it implies they have completely depleted the potential they had at the beginning at the top of the hill.

In contrast, at this zero level height, their speed and kinetic energy would have reached a maximum and this kinetic energy state will be maintained until they encounter a section of unpacked snow where they have to skid to a stop under force of friction. This friction force will carry out work on the skiers which will make their total mechanical energy to decrease. This means that as the force of friction keeps acting over an increasing distance, the quantity of work will therefore increase while the mechanical energy of the skiers will gradually be dissipated.

Eventually, the skiers will run out of energy and comes to a rest position and therefore they wouldn't be able to go as high as they first were before pushing off from the gate.

Answer:

2.00 A

Explanation:

Total resistance = 10 + 20 + 30 = 60 ohms

Potential difference (V) = 120volts

Current (I) =?

from Ohms law V = IR

==> I = V/R = 120/60 = 2 A

Note: for resistors in series equal amount of electric current flows through the circuit

Answer:

The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.

Explanation:

We can determine the final velocity of the skier at the bottom of the ramp by Principle of Energy Conservation and Work-Energy Theorem, whose model is:

[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+W_{disp}[/tex] (1)

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.

[tex]W_{disp}[/tex] - Work dissipated by friction, measured in joules.

By definitions of gravitational potential and translational kinetic energy and work, we expand and simplify the model:

[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2}) =\mu_{k}\cdot N\cdot \Delta s[/tex] (2)

Where:

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

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the skier, measured in meters.

[tex]N[/tex] - Normal force from the incline on the skier, measured in newtons.

[tex]\Delta s[/tex] - Distance covered by the skier, measured in meters.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.

The normal force exerted on the skier and the covered distance are, respectively:

[tex]N = m\cdot g\cdot \cos \theta[/tex] (3)

[tex]\Delta s = \frac{z_{1}-z_{2}}{\sin \theta}[/tex] (4)

Where [tex]\theta[/tex] is the angle of the incline above the horizontal, measured in sexagesimal degrees.

By applying (3) and (4) in (2), we get that:

[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot m\cdot g \cdot \cos \theta \cdot \left(\frac{z_{1}-z_{2}}{\sin \theta} \right)[/tex]

[tex]g\cdot (z_{1}-z_{2}) +\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})= \mu_{k}\cdot g \cdot \left(\frac{z_{1}-z_{2}}{\tan \theta} \right)[/tex] (5)

Then, we clear the velocity of the skier at the bottom of the ramp is: ([tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.1[/tex], [tex]\theta = 40^{\circ}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}-z_{2} = 40\,m[/tex])

[tex]\left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = \frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex]

[tex]2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = v_{1}^{2}-v_{2}^{2}[/tex]

[tex]v_{2} = \sqrt{v_{1}^{2}-2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2})}[/tex] (6)

[tex]v_{2} = \sqrt{\left(0\,\frac{m}{s} \right)^{2}-2\cdot \left(\frac{0.1}{\tan 40^{\circ}} -1\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m)}[/tex]

[tex]v_{2} \approx 26.288\,\frac{m}{s}[/tex]

The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.

7.53 m

We are given:

Initial Horizontal Velocity of the Ball = 4.6 m/s

Initial Vertical Velocity of the Ball = 0 m/s

Height from which ball is kicked = 13.4 m

Time taken by the ball to reach the ground:

The ball has an initial vertical velocity of 0 m/s

it also has a downward acceleration of 10 m/s² due to gravity

Solving for the time taken:

s = ut + 1/2(at²)                 [second equation of motion]

replacing the values

13.4 = (0)(t) + 1/2 (10)(t²)

13.4 = 5t²

t² = 13.4/5                  [dividing both sides by 5]

t² = 2.68

t = 1.637 seconds     [taking the square root of both sides]

Horizontal distance covered by the ball:

Since there are no horizontal opposing forces on the ball,

the ball will more horizontally at a velocity of 4.6 m/s until it hits the ground

We calculated that the ball will hit the ground in 1.637 seconds

Distance covered:

s = ut + 1/2 (at²)                            [seconds equation of motion]

s = ut                                            [since a = 0m/s² in the horizontal plane]

replacing the values

s = 4.6 * 1.637

s = 7.53 m

Hence, the ball landed 7.53 m from the cliff

Explanation:

Using the formula;

2x = vt

x is the distance up from the ocean floor the submarine is

v is the speed of sound in water

t is the time

Given

t = 2.3s

v = 1490m/s

Required

how high (approx) up from the ocean floor is the submarine x

From the formula;

x = vt/2

x = 1490(2.3)/2

x = 745(2.3)

x = 1,713.5m

Hence the submarine is 1713.5m high up from the ocean floor

Answer:

the thickness of the glass divided by thickness of water is going to be 1.333 divided by 1.52, which is 0.877. So, the height of this glass, in order to have the same number of wavelengths as in water, the height of the glass will be 0.877 times the height of the water, and so it will be smaller.

Answer:

Potential energy is the energy that is stored in an object. ... When you park your car at the top of a hill, your car has potential energy because the gravity is pulling your car to move downward; if your car's parking brake fails, your vehicle may roll down the hill because of the force of gravity.

Answer:

a)  P = 4.03 10⁵ Pa, b)  P = 4.03 10⁵ Pa

Explanation:

a) The pressure as a function of the depth of a fluid is

          P =[tex]P_{atm}[/tex] + ρ g y

where Patm is the atmospheric pressure, the sea density of about 1025 kg / m³

let's calculate

         P = 1.01825 10⁵ + 1025 9.8 30

         P = 4.03 10⁵ Pa

b) When the hood is submerged, the water exerts a perpendicular force on the entire surface, in the equilibrium position, the air is compressed by this force until the pressure it exerts is equal to the external pressure (open at the lower), therefore the air pressure is

        P = 4.03 105 Pa

"[tex]3.958\times 10^5 \ Pa[/tex]" would be the pressure outside the bell .as well as the air pressure inside it. A further explanation is below.

According to the question,

Depth,

Pressure,

             [tex]= 1.018\times 10^{5} \ Pa[/tex]

Now,

→ The water pressure outside the bell will be:

= [tex]P_o +p\times g\times d[/tex]

By putting the values, we get

= [tex]1.018\times 10^5+1000\times 9.8\times 30[/tex]

= [tex]3.958\times 10^5 \ Pa[/tex]

And inside the air pressure will be same as the water pressure.

Thus the response above is right.

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