A 8.0 kg block slides 2.0 m along a frictionless incline (measured along the incline L as shown). The incline makes an angle = 60° to the horizontal. Determine work done by normal force, and work done by gravitational force. O Work done by normal force is about 0 N, and work done by gravitational force is about 75.1J O Work done by normal force is about 78.4 N, and work done by gravitational force is about 156.8 J. O Work done by normal force is about 0 N, and work done by gravitational force is about 156.8 J.O Work done by normal force is about 0 N, and work done by gravitational force is about 135.8 J. O Work done by normal force is about 78.4 N, and work done by gravitational force is about 135.8 J.

The only object that could be able to float in water is the object A because it has a density that is less than that of water.

The term density has to do with the ratio of the mass to the volume of an object. We know that the volume is one of the properties of the material that we call the intrinsic properties. These are the properties that could be used in the identification of the material because they do not change.

The density of water is obtained as 1.00g/mL. The object that would float in water must have a density that is less than that of water. The only object that could be able to float in water is the object A because it has a density that is less than that of water.

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Answer:

si quiero ser tu novia❤❤

Answer:

2) a_y= -g  3) vₓ=constant v_y = v_{oy} - g t, 4)  vₓ = v₀ₓ - ax t

5)  changes the horizontal speed, should change range

7) changes the vertical speed change the maximum height

Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression

           v_y = v_{oy} - gt

at the point of maximum height, vy = 0 is equal to the maximum height

4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed

In the graph it would be directed to the left, therefore the velocity would be

           vₓ = v₀ₓ - ax t

5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

the equations of motion are

           x = v₀ₓ t

           y = v_{oy} t - ½ g t²

7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,

the equations of motion are the same.

Answer:

Should be -39.2 N

Explanation:

w=mg

w=4 x -9.8 m/s2

= -39.2 N

The value of the unknown resistance is approximately 44750 Ω.

In a balanced Wheatstone bridge, the ratio of the resistances in the two arms of the bridge is equal to the ratio of the resistances in the other two arms. That is:

R1/R2 = R3/R4

where R1, R2, R3, and R4 are the resistances in the four arms of the bridge.

In this problem, the resistances in three of the arms of the bridge are all exactly 500 Ω. Let the resistance in the fourth arm (the unknown resistance) be denoted by R. The voltage across the galvanometer is zero in a balanced Wheatstone bridge, so the current through the galvanometer is also zero.

Using Ohm's law, the current through the entire circuit is given by:

I = V/(R1 + R2 + R3 + R4)

where V is the voltage of the battery, and R1, R2, R3, and R4 are the resistances in the four arms of the bridge.

The voltage drop across the 500 Ω resistances is given by:

V_500 = (500/(500+500+R)) * V

The voltage drop across the unknown resistance R is also given by:

V_R = (R/(500+500+R)) * V

Since the voltage across the galvanometer is zero, the current through the unknown resistance R is equal to the current through the 80 Ω galvanometer. Using Ohm's law, we can write:

I_R = V_R/R = I_galvanometer = 0.08 μA

Substituting the expressions for the voltages and the current into the equation for the total current, we get:

V/(500+500+500+R) = 0.08 μA

Solving for R, we get:

R = (V/0.08 μA) - 1500 Ω

Substituting the given values, we get:

R = (3.7 V)/(0.08 μA) - 1500 Ω

R = 46250 Ω - 1500 Ω

R = 44750 Ω

Therefore, the value of the unknown resistance is approximately 44750 Ω.

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Answer:

Ionic bond

Explanation:

Hope it helps ya

The required, it experiences a downward force due to gravity and a force due to air resistance.

Projectile motion is the movement of an entity projected into space. After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory.

Here,

When throwing a ball vertically upward, there is a displacement of about 1.0 m from the initial position of the hand to the position where the ball leaves the hand. The mass of the ball is 0.80 kg and it reaches a maximum height of about 20 m above the initial position of the hand. While the ball is in the hand after it leaves, it experiences a downward force due to gravity and a force due to air resistance.

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Answer:

A. Gamma decay

Explanation:

A form of nuclear decay in which the atomic number is unchanged is a gamma decay.

The atom has undergone a gamma decay.

In a gamma decay, no changes occur to the mass and atomic number of the substance.

Explanation

the equation is base ond the Pythagorean theorem,it states that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)

so,

Step 1

if

We have the next information

d=190km

t=2.5 hour

Because the speed is requested in km per hour we don't need to change any unit so we only use the next formula

Answer:

Sunlight is the greatest source of UV radiation. Man-made ultraviolet sources include several types of UV lamps, arc welding, and mercury vapour lamps.

Explanation:

Answer:

The added mass will mean a longer period of oscillation.

Explanation:

The period of oscillation here is given by the formula;

T = 2π√(m/k)

Where m is mass and k is spring constant

From the equation of oscillation period above, it's obvious that when we increase the mass, the oscillation period will also increase.

Thus, the added mass will mean a longer period of oscillation.

The cost of boiling 500 cm³ of water using 1.4kW kettle is 0.945 p

We shall begin our calculation by obtaining the energy consumed when using 1.4 KW kettle. This is shown below:

E = Pt

E = 1.4 × 0.075

E = 0.105 KWh

Finally, we shall determine the cost of boiling the water using the 1.4 KW kettle. Details below

Cost = energy × Cost per KWh

Cost of boiling = 0.105 × 9

Cost of boiling = 0.945 p

Thus, we can conclude that the cost of boiling is 0.945 p

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The acceleration of the crate  after the frictional force is overcame is determined as 1 m/s².

Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity with time.

The  acceleration of the crate is calculated from the net force acting on the crate as shown below;

F(net) = applied force - frictional force

F(net) = 10 N - 5N

F(net) = 5 N

Apply Newton's second law of motion to calculate the acceleration of the crate.

F(net) = ma

where;

a = F(net) / m

a = 5/5

a = 1 m/s²

Thus, the acceleration of the crate  after the frictional force is overcame is determined as 1 m/s².

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The amount of work done by the gas is proportional to the pressure and the change in volume, as well as the efficiency of the process. If the pressure and volume are known, the work done by the gas can be calculated by multiplying these values by the efficiency of the process.

The amount of work done by a gas when it expands is proportional to the change in volume, pressure, and temperature. According to the first law of thermodynamics, the energy of a closed system is conserved, so the work done by the expanding gas is equal to the energy transferred from the gas to the environment in the form of work. Therefore, the work done by the gas is equal to the change in energy of the system. Assume that the process is 10% efficient. Then, only 10% of the energy available to the system is converted into work. This means that the remaining 90% of the energy is lost to the environment in the form of heat. As a result, the amount of work done by the gas expanding into the atmosphere is given by the formula

W = E x η, where W is the work done by the gas, E is the energy available to the system, and η is the efficiency of the process. The energy available to the system is determined by the difference between the internal energy of the gas before and after the expansion. The internal energy of a gas is determined by its temperature, pressure, and volume.

Assuming that the temperature and pressure are constant, the change in internal energy is proportional to the change in volume. Therefore, the energy available to the system is equal to the product of the pressure and the change in volume: E = P x ΔV, where P is the pressure of the gas and ΔV is the change in volume during the expansion. Substituting this equation into the formula for work, we get W = P x ΔV x η.

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Answer:

wait in comments.................

The candidate's speed (m/s), given that the candidate experiences a centripetal acceleration that is 2.8 times the acceleration due to gravity is 17.4 m/s

We understood that the centripetal acceleration is related to speed and radius according to the following formula:

a = v² / r

Cross multiply

v² = ar

Take the square root of both sides

v = √ar

Where

Withe the above formula, we can determin the speed of the candidate. Details below:

v = √ar

v = √(27.44 × 11.05)

v = 17.4 m/s

Thus, the speed of the candidate is 17.4 m/s

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To collect the asteroid an equal magnitude of the force in opposite direction is needed.

What is force ?

The motion or speed of an object depends upon the magnitude and the direction of the force.

If the direction of the force is against the direction of motion, then the object will slow down or stop. The two forces of the same magnitude in the opposite direction cancel each other and the collided objects stop each other.

Therefore, to collect the asteroid an equal magnitude of the force in opposite direction is needed.

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Explanation:

Object is under the effect of the acceleration of gravity

v = 1/2 a t^2       a = 9.81 m/s^2   t = 2.3

v = 25.9 m/s^2 = ~ 26 m/s   ( two significant digits)

The net force on A is  1.7312 × 10⁻⁷ N

It states that the force of attraction between two masses M and m is directly proportional to the product of their masses and inversely proportional to the square of their distance apart, R.

So, F = GMm/R² where G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Now let

So, the force of attraction between them is

F = Gm₁m₂/R₁² (positive since it is directed towards B in the positive x direction)

Substituting the values of the variables into the equation, we have

F = Gm₁m₂/R₁²

F = 6.67 × 10⁻¹¹ Nm²/kg² × 199.1 kg × 592.9 kg/(7.5 m)²

F = 787369.4213 × 10⁻¹¹ Nm²/56.25 m²

F = 13997.68 × 10⁻¹¹ N

F = 1.399768 × 10⁻⁷ N

F ≅ 1.3998 × 10⁻⁷ N

Also, let

So, the force of attraction between them is

F' = Gm₁m₃/R₂² (positive since it is directed towards C in the positive x direction)

Substituting the values of the variables into the equation, we have

F' = Gm₁m₂/R₁²

F' = 6.67 × 10⁻¹¹ Nm²/kg² × 199.1 kg × 330 kg/(11.5 m)²

F' = 438239.01 × 10⁻¹¹ Nm²/132.25 m²

F' = 3313.72 × 10⁻¹¹ N

F' = 0.331372 × 10⁻⁷ N

F' ≅  0.3314 × 10⁻⁷ N

So, the net force acting on object A is F" = F + F'

= 1.3998 × 10⁻⁷ N +  0.3314 × 10⁻⁷ N

= 1.7312 × 10⁻⁷ N

So, the net force on A is  1.7312 × 10⁻⁷ N

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The area of Mrs. Turner's apartment is approximately 0.05304 square feet.

To find the area of Mrs. Turner's apartment, we need to convert the measurements from the scale drawing to the actual measurements in feet, using the given scale factor of 1 inch/18 feet.

Length of apartment in feet = 6.25 inches × (1 foot/18 inches) = 0.34722 feet

Width of apartment in feet = 2.75 inches × (1 foot/18 inches) = 0.15278 feet

Now, we can calculate the area of the apartment in square feet by multiplying the length and width:

Area of apartment in square feet = Length × Width = 0.34722 feet × 0.15278 feet = 0.05304 square feet

Therefore, the area of Mrs. Turner's apartment is approximately 0.05304 square feet.

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The magnitude of the angular acceleration of the roller is approximately 108.8 rad/s².

To find the magnitude of the angular acceleration of the roller, we can use the rotational analog of Newton's second law: τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

First, let's calculate the moment of inertia of the roller. The moment of inertia of a solid cylinder rotating about its central axis is given by the formula: I = (1/2)mr², where m is the mass and r is the radius.

Given:

Mass of the roller (m) = 2.4 kg

Radius of the roller (r) = 3.8 cm = 0.038 m

Moment of inertia (I) = (1/2) * 2.4 kg * (0.038 m)² = 0.0021744 kg·m²

Next, we need to calculate the torque (τ) applied to the roller. Torque is given by the formula: τ = rFsin(θ), where r is the distance from the axis of rotation to the point of application of the force, F is the magnitude of the force, and θ is the angle between the force and the line connecting the axis of rotation and the point of application.

Given:

Force applied (F) = 16 N

Angle (θ) = 35°

Distance from the axis of rotation to the point of application (r) is equal to the radius of the roller, so r = 0.038 m.

Torque (τ) = (0.038 m) * (16 N) * sin(35°) = 0.2366 N·m

Now, we can use the equation τ = Iα and solve for the angular acceleration (α):

0.2366 N·m = (0.0021744 kg·m²) * α

α = 0.2366 N·m / 0.0021744 kg·m² ≈ 108.8 rad/s²

Therefore, the magnitude of the angular acceleration of the roller is approximately 108.8 rad/s².

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Conservation of momentum is a major law of physics which states that the momentum of a system is constant if no external forces are acting on the system. It is embodied in Newton’s First Law or The Law of Inertia.the mass of the second block is 1.1Kg.

principle of momentum conservation

M1u1 plus M2u2 equals M1v1 and M2V2.

As all collisions were elastic in nature and no energy loss through friction, heat, etc. was taken into account, theoretic calculations alone cannot guarantee that there was a complete transfer of energy.

Consider the scenario where a football with mass M2 is lying on the ground and a bowling ball with mass M1 is hurled at the football at a velocity of

The formula is: (2.8 kg * 1.1 m/s) + (m2 * 4.8 m/s) = (2.3 kg + m2). 2.3 m/s

The formula is 2.8 J + (4.8 m/s m2) = 4.8 J + (2.3 m/s m2).

4.8 m/s m2 = 2.8 J plus (2.3 m/s m2)

4.8 m2 = 2.8 + 2.3 m2

2.3 m2 on each side of the equation

2.5 m = 2.8 m = 2.8 / 2.5\sm = 1.1kg

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When a fullback moving south has a collision with an opponent running north, it is a type of head-on collision.

What are the types of collision?

1. Elastic collision: A collision in which the total kinetic energy of the system remains the same, and kinetic energy is exchanged between objects.

2. Inelastic collision: A collision in which the total kinetic energy of the system is not conserved, and some kinetic energy is converted into other forms, such as heat or sound.

3. Perfectly inelastic collision: A collision in which the total kinetic energy of the system is completely converted into other forms, such as heat or sound.

4. Partially inelastic collision: A collision in which the total kinetic energy of the system is partially conserved, and some kinetic energy is converted into other forms, such as heat or sound.

5. Plastic collisions: A collision in which the kinetic energy of the system is partially conserved, but some energy is also converted into permanent deformation.

The collision you described between the fullback moving south and the opponent running north is an example of a head-on collision. A head-on collision is defined as a type of collision where two objects collide with each other while moving in opposite directions. These types of collisions are often considered the most dangerous, due to the high energy that is released as the two objects collide.

Therefore, head-on collision is the answer.

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Answer:

2 * 10^5 pa

Explanation:

Pressure = Force / Area

Each thigh bone has a cross sectional area of 10cm²

Both thigh bones :

2 * 10cm² = 20cm²

To m² : 20 * (0.01)²

20 * 0.0001 m² = 0.002 m²

Force = mass * acceleration due to gravity(g)

g = 10m/s² ;

Force = 40 * 10 = 400N

Pressure = 400 N / 0.002 m²

Pressure = 200,000 N/m² = 2 * 10^5 pascal

The displacement of the runner after four seconds is 78.4 m.

We know that in this case, we are dealing with a case of an object that has a motion under gravity. We are told that the road runner an fell straight down off a cliff. The fact that we have been told that the runner just fell down the cliff means that the initial velocity of the runner would have to be taken in this context as zero since the runner was dropped from a height as shown.

Acceleration of the runner (g) = 9.8 m/s^2

Initial velocity of the runner (u) = 0 m/s

Time take (t) = 4 seconds

We then have;

h = ut + 1/2gt^2

If we then know that the initial velocity of the person is zero, then we have;

h = 1/2gt^2

h = 0.5 * 9.8 * (4)^2

h = 78.4 m

The height is 78.4 m.

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Answer:

1962 J

Explanation:

potential energy= mgh

m=10 kg

h=20 m

g=9.81 (always unless otherwise stated)

PE= 10(20)(9.81)=1962 J

Surface area of pyramid 1  539.2 \(m^{2}\) Pyramid 2: 370.9 \(m^{2}\) Pyramid 3: 446.4 \(m^{2}\)

To find the surface area of a pyramid, we need to know the dimensions of its base and its slant height. Assuming the base of each pyramid is a square, we can calculate the surface area using the formula:

Surface Area = Base Area + (0.5 × Perimeter of Base × Slant Height)

Let's calculate the surface area for each pyramid using the given dimensions:

1. Pyramid 1: Base = 12.2 m Slant Height = 16 m Base Area = (12.2 \(m^2\)= 148.84 \(m^2\)Perimeter of Base = 4 × 12.2 m = 48.8 m

Surface Area = 148.84 \(m^2\)+ (0.5 × 48.8 m × 16 m) = 148.84 m^2 + 390.4 \(m^2\) = 539.24 \(m^2\) (rounded to nearest tenth)

2. Pyramid 2: Base = 10.6 m Slant Height = 12.2 m Base Area = (10.6 \(m^2\))= 112.36\(m^2\) Perimeter of Base = 4 × 10.6 m = 42.4 m

Surface Area = 112.36\(m^2\)+ (0.5 × 42.4 m × 12.2 m) = 112.36 \(m^2\)+ 258.56 \(m^2\) = 370.92 \(m^2\)(rounded to nearest tenth)

3. Pyramid 3: Base = 12.2 m Slant Height = 12.2 m Base Area = (12.2 \(m^2\)= 148.84 \(m^2\)Perimeter of Base = 4 × 12.2 m = 48.8 m

Surface Area = 148.84 m^2 + (0.5 × 48.8 m × 12.2 m) = 148.84 \(m^2\)+ 297.52 \(m^2\) = 446.36 \(m^2\) (rounded to nearest tenth)

Therefore, the surface areas of the given pyramids are approximately: Pyramid 1: 539.2 \(m^2\)Pyramid 2: 370.9 \(m^2\) Pyramid 3: 446.4 \(m^2\).

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Answer:

\(12.79 m/s^2\)

Explanation:

1. Apply Newton's 2nd law:

\(F=ma\)

2. Use the definition of weight to find the mass of the box:

\(W=mg=m=\frac{W}{g}\)

3. Solve for acceleration in Newton's 2nd law and substitute for mass:

\(a=\frac{F}{\frac{W}{g} }\)

4. Plug in the given values from the question and assume \(g=9.81m/s^2\):

\(a=\frac{777N}{\frac{596N}{9.81m/s^2} } } =12.79m/s^2\)