A man pushes a block of ice across a frozen pond at a constant velocity. While the coefficients of static and kinetic friction for ice are low, they are not zero. Consider this problem to involve friction. If necessary, use Fs for the force of static friction, and Fk as the force of kinetic friction.

Required:
Draw the Free Body Diagram for the block of ice.

The amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J (option A).

The amount of heat absorbed or released by a substance can be calculated using the following formula:

Q = mc∆T

Where;

Q = 6 × 0.385 × 15

Q = 90 × 0.385

Q = 34.65J

Therefore, the amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J.

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The kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.

To find the answer, we need to know about the concept of collision and kinetic energy.

                 [tex]TE=KE+PE\\T=PE=m_1gh=(0.4*9.8*20)=78.4 J[/tex]

                   [tex]TE=KE=78.4J[/tex]

                          [tex]U'=m_1gh'=mg\frac{h}{2} \\U'=39.2J[/tex]

                      [tex]TE=78.4 J\\KE'+PE'=78.4J\\KE'=78-U'=78.4-39.2=39.2J[/tex]

                            [tex]KE=\frac{1}{2} mv^2=78.4J\\v=\sqrt{\frac{2KE}{m} } =4m/s[/tex]

               [tex]\frac{1}{2}m_1u_1^2+ \frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]

                 [tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

                            [tex]m_1=m_2=m\\u_1=4m/s\\u_2=0\\v_1=?\\v_2=?[/tex]

                       [tex]\frac{1}{2}m*4^2=\frac{1}{2}m(v_1^2+v_2^2)\\(v_1^2+v_2^2)=16[/tex]  from resolving KE equation.

                        [tex]4m=m(v_2+v_1)\\4=v_2+v_1\\v_1=4-v_2[/tex] From resolving momentum conservation.

                            [tex]v_2=4m/s\\v_1=0[/tex]

Thus, we can conclude that, the kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.

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Answer:

the branch of science that is concerned with nature and properties of matter and energy.

Explanation:

a study of the basis of what does what in science.

The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

                      [tex]F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\[/tex]

                     [tex]Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N[/tex]

                [tex]F_V=(29*9.8)-(169.43*sin57)=142.10N[/tex]

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

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The hi.nge will apply a force of 142.10N on the beam in the vertical direction.

We must learn more about the tension in order to find the solution.

                           [tex]F_V=mg-Tsin\alpha[/tex]

                            [tex]Tlsin\alpha =mg\frac{l}{2}sin\beta \\T=\frac{mgsin90}{2sin57} =169.43N[/tex]

                     [tex]F_v=(29*9.8)-(169.43*sin57)=142.10N[/tex]

This leads us to the conclusion that the vertical component of the force the hi.nge exerts on the beam will be 142.10N.

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The primary means by which heat is transferred through fluids is convection currents (option D).

Convection is the transmission of heat in a fluid by the circulation of currents.

Heat can be transferred by different methods depending on the medium. Fluids like gases and liquids transfer heat through the process of convection.

Therefore, the primary means by which heat is transferred through fluids is convection currents.

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5.8 moles of nitrogen gas are needed to pressurize the air bag.

= (2.37×60)/(0.0821×298)

= 5.8 moles

Thus, we can conclude that 5.8 moles of nitrogen gas are needed to pressurize the air bag.

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The final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.

In an elastic collision, both momentum and energy are conserved. But only momentum is conserved in inelastic collision.

Given that two hockey pucks are moving towards each other. The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path.

The given parameters are;

The mathematical representation of the above question will be in two components.

Horizontal component

M1U1 - M2U2 = M1V1cosФ - M2V2cosФ

Substitute all the parameters

0.13 x 1.11 - 0.16 x 1.21 = 0.13 x V1 cosФ - 0.16 x 1.16cos42

0.1443 - 0.1936 = 0.13V1cosФ - 0.1379

0.13V1cosФ = 0.0886

V1cosФ = 0.0886/0.13

V1cosФ = 0.6815 ........ (1)

Vertical component

0 = M1V1sinФ - M2V2sinФ

M1V1sinФ = M2V2sinФ

Substitute all the parameters

0.13 x V1 sinФ = 0.16 x 1.16sin42

V1 sinФ = 0.1242/0.13

V1 sinФ = 0.9553 ......... (2)

Divide equation 2 by 1

V1 sinФ / V1 cosФ = 0.9553/  0.6815

Tan Ф = 1.40

Ф = [tex]Tan^{-1}[/tex](1.4)

Ф = 54.5°

Substitute Ф into equation 2

V1 sin54.5 = 0.9553

V1 = 0.9553 / 0.8141

V1 = 1.17 m/s

Therefore, the final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.

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Answer:

Explanation:

2.1 x 10^2 - 20J

The kinetic energy of the projectile at the highest point of its motion will be E/4.

When a projectile will be thrown obliquely near the surface of the earth, it travels a curved path under uniform acceleration directed toward the center of the Earth. The path of a particle is called a projectile while the motion of a projectile is projectile motion.

Given, the angle of projection  with horizontal, [tex]\theta = 60 ^o[/tex]

Consider that 'E' is the initial value of the kinetic energy of the projectile.

The equation for the initial kinetic energy is : [tex]E =\frac{1}{2}mu^2[/tex]

where m is the mass of the given projectile.

The component of the velocity of the projectile in the horizontal direction:

uₓ = u cosθ

uₓ = u cos 60°

uₓ = u/2

From the equation of motion: v = u +at

v = (u/2) + (0) t

v = u/2

The final kinetic energy of the projectile:

[tex]E_f = \frac{1}{2}mv^2[/tex]

[tex]E_f = \frac{1}{2}m(\frac{u}{2} )^2[/tex]

[tex]E_f = \frac{1}{4} (\frac{1}{2}mu^2 )[/tex]

[tex]E_f = \frac{E}{4}[/tex]

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The orbiting velocity of the satellite is 4.2km/s.

To find the answer, we need to know about the orbital velocity of a satellite.

=4.2km/s

Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.

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Answer: The mass of the object is 25kg.

The given question deals with Newton's second law of motion and its applications.

Explanation: Given force, F=500N

                                 acceleration, a=20 m/[tex]s^{2}[/tex]

  From Newton's 2nd law of motion , we have

                             F=ma where m=mass of the object

                         ⇒500=m×20

                         ⇒m=500/20=25

                 ∴ Mass of the object is 25 kg .

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The answer is A. Decreasing the distance between the particles by a factor of 3.

The Universal Law of Gravitation is :

F = Gm₁m₂ / r² (where 'r' is the distance between them)

Since force is inversely proportional to the square of the distance between them, distance has to be decreased by a factor of 3 to increase the force to 9 times the original force.

The position of the object is = -68cm

The magnification of the mirror= 0.3

The image distance = 20.5cm

The focal length= R/2 = 31.5/2= 15.75

The object distance= ?

Using the lens formula,1/f = 1/v-1/u

1/u = 1/v- 1/f

1/u = 1/20.5 - 1/15.75

1/u = 0.0489- 0.0635

1/u = -0.0146

u = -68cm

The magnification of the mirror is image size/object size

= 20.5cm/-68cm

= 0.3

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The buoyant force acting on the cylinder is, [tex]Fb = \rho Ahg[/tex]. Here A is the cross-sectional area of the cylinder, h is the height of the cylinder, ρ is the density of the cylinder, and g is the acceleration due to gravity.

This buoyant force is also equal to the volume of the fluid displaced. [tex]Fb = \sigma h(A-x)g[/tex]. Here σ is the density of the fluid.

Equate the above two equations and solve for x.

[tex]\rho Ahg = \sigma A(h-x)g\\\rho h = \sigma h - \sigma x\\x = \frac{(\sigma - \rho)h}{\sigma}[/tex]

So, the distance x depends on the density of the fluid, density of the cylinder and the height of the cylinder.

1. The density of the cylinder is same and distance x is independent of the diameter of the cylinder. Therefor, there will be no change in the distance x. Hence, the correct answer is No change.

2. Now the height is changing keeping the density same. As the distance x is directly proportional to the height, the distance x will increase.

3. The density of the added liquid is greater than of the water and it does not mix with the water. So, the liquid will settle down and there will be no change in the distance x.

4. The density of the added liquid is less than that of the water and it does not mix with the water. So, the liquid will not settle down and the distance x will change. The change in distance x can be determined as follow:

[tex]\rho Agh = \sigma' Axg + \sigma A(h-x)g\\\rho h=\sigma' x + \sigma h - \sigma x\\x=(\frac{\sigma - \rho}{\sigma - \sigma'})h[/tex]

Here, σ' is the density of the added liquid.

From the above relation it is clear, that on adding the liquid of the density less than that of water, the denominator term become small ando so the value of x will increase.  

5. On removing some of the water inside the glass, the height of the water column will decrease, but the value of x does not depend on the height of the water column. So, there will be no change in the distance x.

6. The density of the new cylinder is smaller than that of the earlier one. So, the numerator term will increase. Therefore, the value of x will increase.

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1. The speed with which the ball hits the ground is 17.1 m/s

2. The magnitude of the average force of air resistance exerted on it is 0.77 N

v² = u² + 2gh

v² = 2gh

Take the square root of both side

v = √(2 × 9.8 × 15)

v = 17.1 m/s

We'll begin by calculating the time to reach the ground. This is illustrated below:

h = ½gt²

15 = ½ × 9.8 × t²

15 = 4.9 × t²

Divide both side by 4.9

t² = 15 / 4.9

Take the square root of both side

t = √(15 / 4.9)

t = 1.75 s

Now we can determine the force. This can be obtained as illustrated below:

F = m(v –u) / t

F = 0.149(9 – 0) / 1.75

F = 0.77 N

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The original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.

The energy possessed by a body in motion is known as Kinetic Energy. The S. I unit is Joule.

Given that an 87-kg football player traveling 5.2 m/s is stopped in 1.0 s by a tackler. The given parameters are;

The original kinetic energy of the player can be calculated by using the formula K.E = 1/2m[tex]v^{2}[/tex]

Substitute all the parameters into the formula

K.E = 1/2 x 87 x [tex]5.2^{2}[/tex]

K.E = 1176.24

K.E = 1200 J

Power is the rate at which work is done.

Work done = energy

The average power is required to stop him can be calculated by using the formula P = E/t

Substitute all the parameters into the formula

P = 1200/1

P = 1200 W

Therefore, the original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.

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(a) The initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.

(b) The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.

The initial vertical velocity of the stone is calculated as follows;

Vi = Vsinθ

Vi = 22 x sin(50)

Vi = 16.85 m/s

Vxi = V cosθ

Vxi = 22 x cos(50)

Vxi = 14.14 m/s

Vf = Vi - gt

where;

Vfx = Vxi = 14.14 m/s

Thus, the initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.

The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.

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If you speed through a construction zone while workers are present, your fines could be as much as one thousand dollars.

This is referred to as the amount which is instructed by a court or authority to be paid as a result of it being a penalty for various types of offences. each crime has its specific fine which helps to serve as deterrent for unlawful behavior in the community.

it is always best not to speed when within a construction zone which has workers present in the area. This is ideal as it helps to prevent incidences of accidents or death.

it is therefore the reason why a fine of 1000 dollars is to be paid so that people can think of such high amount before performing certain types of activities when driving and makes it the most appropriate choice.

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From the calculation, the normal force is 6161.2 N.

The normal force is given by the expression;

N - mg = ma

Then;

N = mg + ma

m = 84.4 kg

g = 9.8 m/s^2

a = 63.2 m/s2

Now we have;

N = m(g + a)

N = 84.4 (9.8 + 63.2)

N = 6161.2 N

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Answer:

 no

Explanation:

The angle subtended by the radius of the target at Emily's distance can be found using the tangent relation.

 tan(α) = opposite/adjacent = (1/4 ft)/(150 ft) = 1/600

The angle is found using the inverse relation -

α = arctan(1/600) ≈ 0.095°

If Emily's aim is off by 0.2°, she will miss the target by several inches.

Emily's projectile will miss her aiming point by ... (150 ft)tan(0.2°) ≈ 0.524 ft ≈ 6.28 in

a) Her feet are in the air for 0.73+0.41 = 1.14 seconds

b) Her highest height above the board is 0.82 m

c) Her velocity when her feet hit the water is 7.16 m/s

t = Time taken

u = Initial velocity = 4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

a) Her feet are in the air for 0.73+0.41 = 1.14 seconds

s = ut + [tex]\frac{1}{2}[/tex]at²

2.62 = 0t + [tex]\frac{1}{2}[/tex] ₓ 9.8  ₓ [tex]t^{2}[/tex]

t = 0.73 s

b) Her highest height above the board is 0.82 m

The total height she would fall is 0.82+1.8 = 2.62 m

v = u + at

0 = 4 ₋ 9.8 ₓ t

t = 0.41 s

s = ut +[tex]\frac{1}{2}[/tex] at²

s = 4 ₓ 0.41 ₊ [tex]\frac{1}{2}[/tex] ₓ ₋9.8 ₓ 0.41 [tex]t^{2}[/tex]

c) Her velocity when her feet hit the water is 7.16 m/s

[tex]v = u + at \\v = 0 + 9.8[/tex] ₓ [tex]0.73[/tex]

v = 7.16 m/s

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The horizontal distance XQ traveled by the bomb is 250 m.

Let the XQ be the horizontal distance traveled by the bomb.

h = vt + ¹/₂gt

Let the vertical velocity = 0

h = 0 + ¹/₂gt²

h = ¹/₂gt²

2h = gt²

t² = 2h/g

t = √(2h/g)

t = √(2 x 100 / 9.8)

t = 4.5 s

XQ = vx(t)

where;

vx is horizontal speed, = 200 km/hr = 55.56 m/s

XQ = 55.56 x 4.5

XQ = 250 m

Thus, the horizontal distance XQ traveled by the bomb is 250 m.

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The highest point of a roller coaster is almost always the first hill. In the majority of roller coasters, the hills get smaller as the train travels down the track.

To find the answer, we have to know more about the mechanical energy of a system.

                  Mechanical energy = U = mgh

Where m represents the car mass, g represents gravity, and h represents height

Finally, by applying the principle of energy conservation, we may determine that, the initial hill must be the highest.

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The force F, necessary to support an object with a mass of 991 kg placed on piston 2. is 20.61J.

It should be noted that by Pascal Law, the pressure on piston 1 will have the same value as the pressure on piston 2.

This will be:

(991 × 10) /(π × 0.0946/2)²

= 9910/0.022

= 450454.6 Pa

F1 = A1 × 450454.6 = 3.14 × (0.0187/2)² × 450454.6

= 123.64

F = 123.64/6

F = 20.61

Therefore, the force F, necessary to support an object with a mass of 991 kg placed on piston 2. is 20.61J.

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The tension in the cable is equal to 323.5 N.

The tension, T in the cable is determined by taking moments about the pivot  marked X.

The angles of the boom and the cable with the horizontal are first calculated.

Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°

The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80

Taking moments about the pivot:

175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1

Tension = 241.68/0.747

Tension = 323.5 N

In conclusion, the tension in the cable helps to suspend the crate.

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5.48 V is the potential drop across the cable for a copper transmission cable of 50.0 km long and 10.0 cm in diameter carries a current of 105 A

Ohm's Law states that the potential drop is determined by the equation: V = IR, where I is the current and R is the wire resistance.
R=PL/A
Under the assumption that all physical parameters and temperatures remain constant, Ohm's law asserts that the voltage across a conductor is directly proportional to the current flowing through it.

Only when the given temperature and the other physical variables remain constant does Ohm's law apply. Increasing the current causes the temperature to rise in some components. The filament of a light bulb serves as an illustration of this, where the temperature increases as the current increases. Ohm's law cannot be applied in this situation. The filament of the lightbulb defies Ohm's Law.

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The escape velocity from the surface of the planet X is 2,249.2 m/s.

[tex]v = \sqrt{\frac{2GM}{r} } \\\\v^2 = \frac{2GM}{r}\\\\v^2r = 2GM\\\\G = \frac{v^2r}{2M}[/tex]

where;

[tex]\frac{v_x^2 \ r_x}{2M_x} = \frac{v_y^2 \ r_y}{2M_y} \\\\\frac{v_x^2 \ r_x}{M_x} = \frac{v_y^2 \ r_y}{M_y} \\\\v_x^2 = \frac{v_y^2 \ r_yM_x}{M_yr_x}\\\\v_x^2 = \frac{(1695)^2 (r_y)(1.59M_y)}{M_y(0.903r_y)} \\\\v_x^2 = 5,058,814.78\\\\v_x = \sqrt{5,058,814.78} \ \ = 2,249.2 \ m/s[/tex]

Thus, the escape velocity from the surface of the planet X is 2,249.2 m/s.

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The horizontal force applied is  160 N while the velocity is  2.03 m/s.

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

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