A small sphere of mass m is launched horizontally over a body of water from a height h above the water and with a launch speed v0. Determine expressions for the following in terms of m, v0, h, and g. Air resistance is negligibly small.

(a) W is the amount work done by the force of gravity on the projectile during its flight.

W =
−mgh


For a conservative force, how does the work done by the force compare to the change in potential energy associated with the force?

Answer:

C) Both 1 and 2

Explanation:

The scan tool may include a bi-directional control that allows the technician to control the output of the alternator for testing purposes.

The Diagnostic Trouble Codes (DTCs), are used by automobile manufacturers to diagnose problems related to the vehicle.

The scan tool can also be used to monitor the output voltage of the vehicle to verify if the correct amount of voltage is supplied by the alternator.

Both Technicians A and B are correct because the steps they both take are necessary for the diagnosis of the vehicle.

Answer:

Solid

Explanation:

The plasma is the liquid part of blood, it is 90% and accounts for 55% of blood volume. It is what red blood cells, white blood cells, and platelets move around in. These cells remain solid within the plasma. I hoped this helped!

Although blood cells are contained within a special liquid called plasma, the cells themselves are suspended in plasma.

Blood is a fluid tissue that is made up of plasma, red blood cells, white blood cells, and platelets. Plasma is the liquid part of blood that makes up about 55% of blood volume.

Red blood cells make up about 45% of blood volume and are responsible for carrying oxygen to the tissues. White blood cells make up about 1% of blood volume and are responsible for fighting infection. Platelets are responsible for clotting blood.

The cells are suspended in the plasma because they are too small to sink to the bottom of the blood. The plasma also helps to transport the cells throughout the body.

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Answer:

The height will be 917431.2 m.

Explanation:

Power of bulb = 75 W

Time kept on 1 hr = 60 x 60 = 3600 sec

Energy of bulb = power x time

E = 75 x 3600 = 270000 J

From conservation of energy, kinetic energy of the bulb is equal to the potential energy of the bulb due to its height of fall.

Potential energy = m x g x h

Where g = acceleration due to gravity 9.81 m/s2

m = mass = 30 g = 0.03 kg

PE = 0.03 x 9.81 x h = 0.2943h

Equating withe the energy of bulb (still obeying energy conservation)

270000 = 0.2943h

h = 270000/0.2943 = 917431.2 m

The incandescent bulb would have to be dropped from a height of approximately 918,367.35 meters to have its kinetic energy equal to the energy consumed in one hour of use.

To determine from what height you should drop a 75-watt incandescent bulb so that upon reaching the floor, its kinetic energy will equal the energy used to power it for 1.0 hour, we use the formula for gravitational potential energy (PE) and kinetic energy (KE), where PE = mgh and KE = ½ mv². The power consumption of the bulb indicates how much energy it uses per second, so we first calculate the energy consumed in one hour (E = power × time). Since the bulb is 75 watts, and there are 3600 seconds in an hour, it uses 75 W × 3600 s = 270,000 joules in one hour. This is the energy we want the bulb to have as kinetic energy when it hits the ground. To find the height, we set this equal to the potential energy at the start (mgh) and solve for h:

E = mgh

270000 J = (0.03 kg)(9.8 m/s²)(h)

h = 270000 J / (0.03 kg × 9.8 m/s²)

h = 918367.35 m

Thus, the bulb would have to be dropped from a height of approximately 918367.35 meters to have the same energy when it impacts the ground as it uses in one hour of being lit.

Answer:

The car was 12.8m/s fast when the driver applied the brakes.

Explanation:

The equations of motion of the car in the horizontal and vertical axes are:

[tex]x: f_k=ma\\\\y: N-mg=0[/tex]

Since the kinetic friction is defined as [tex]f_k=\mu_kN[/tex] and [tex]N=mg[/tex] we have:

[tex]\mu_kmg=ma\\\\a=\mu_kg[/tex]

Next, from the kinematics equation of speed in terms of distance, we have:

[tex]v^2=v_0^2-2ax\\\\v^2=v_0^2-2\mu_kgx[/tex]

Since the car came to a stop, the final velocity [tex]v[/tex] is zero, and we get:

[tex]0=v_0^2-2\mu_kgx\\\\v_0=\sqrt{2\mu_kgx}[/tex]

Finally, plugging in the known values, we obtain:

[tex]v_0=\sqrt{2(0.300)(9.81m/s^2)(28.0m)}\\\\v_0=12.8m/s[/tex]

It means that the car was 12.8m/s fast when the driver applied the brakes.

The car was moving at the speed of 12.8 m/s, when the driver applied the brakes.

From kinamatic equation,

[tex]\bold {v^2 = v_0^2-2ax}\\[/tex]

Since, car stops the final veocity will be zero. and

Acceleration [tex]\bold {a = \mu_kg}[/tex]

So,

[tex]v_0 = \sqrt {2\mu k_gx}[/tex]

Where,

Vo - initial speed =?

[tex]\mu[/tex] - friction constant = 0.3

g - gravitational acceleration = 9.8 m/s²

x - distance = 25 m

Put the values,

[tex]v_0 = \sqrt {2\times 0.3 \times 9.81 \times 0.25 m}\\\\v_0 = 12.8\ m/s[/tex]

Therefore, the car was moving at the speed of 12.8 m/s, when the driver applied the brakes.

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Answer:

B.

Explanation:

The distance between two crests of adjacent waves is called wavelength.

Final answer:

Wavelength is (option B) the distance between consecutive crests of adjacent waves and is parallel to the direction of wave propagation, typically represented by the Greek letter lambda (λ).  

Explanation:

The correct description of wavelength in the context of waves is the distance between crests of adjacent waves. Wavelength is the distance between two consecutive points of similar position along the wave such as from one crest to the next crest, or one trough to the next trough. It's important to note that wavelength is parallel to the direction in which the wave is traveling, so it measures the length of one complete cycle of the wave.

Options such as the number of waves passing a point in a given time (frequency) and the wave's height (amplitude) are different wave properties. Thus, the correct answer is option B: the distance between crests of adjacent waves.

Answer:

hall mobility = 139.8 x 10 ∧3

carrier density = 83.1 x 10 ∧14

Explanation:

The pictures attached below shows the full explanation

Answer:

1.10^6 kg of mass per second

Explanation:

All energy lost by the sun comes from nuclear fusion.

Sun loses energy at 2.5*10^{19}J per hour, that is 9*10^{22}J/s

To find the mass lost by the sun in liberation of energy you use the famous Einstein's equation:

[tex]E=mc^2\\\\m=\frac{E}{c^2}=\frac{9*10^{22}}{(3*10^{8}m/s)^2}=1*10^6kg[/tex]

hence, the sun liberates 1.10^6 kg of mass per second

Answer:

The maximum vertical displacement is 2.07 meters.

Explanation:

We can solve this problem using energy. Since there is a frictional force acting on the block, we need to consider the work done by this force. So, the initial potential energy stored in the spring is transferred to the block and it starts to move upwards. Let's name the point at which the block leaves the ramp "1" and the highest point of its trajectory in the air "2". Then, we can say that:

[tex]E_0=E_1\\\\U_e_0=K_1+U_g_1+W_f_1[/tex]

Where [tex]U_e_0[/tex] is the elastic potential energy stored in the spring, [tex]K_1[/tex] is the kinetic energy of the block at point 1, [tex]U_g_1[/tex] is the gravitational potential energy of the block at point 1, and [tex]W_f_1[/tex] is the work done by friction at point 1.

Now, rearranging the equation we obtain:

[tex]\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgh_1+\mu Ns_1[/tex]

Where [tex]k[/tex] is the spring constant, [tex]x[/tex] is the compression of the spring, [tex]m[/tex] is the mass of the block, [tex]v_1[/tex] is the speed at point 1, [tex]g[/tex] is the acceleration due to gravity, [tex]h_1[/tex] is the vertical height of the block at point 1, [tex]\mu[/tex] is the coefficient of kinetic friction, [tex]N[/tex] is the magnitude of the normal force and [tex]s_1[/tex] is the displacement of the block along the ramp to point 1.

Since the force is in an inclined plane, the normal force is equal to:

[tex]N=mg\cos\theta[/tex]

Where [tex]\theta[/tex] is the angle of the ramp.

We can find the height [tex]h_1[/tex] using trigonometry:

[tex]h_1=s_1\sin\theta[/tex]

Then, our equation becomes:

[tex]\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgs_1\sin\theta+\mu mgs_1\cos\theta\\\\\implies v_1=\sqrt{\frac{2(\frac{1}{2}kx^{2}-mgs_1\sin\theta-\mu mgs_1\cos\theta)}{m}}=\sqrt{\frac{kx^{2}}{m}-2gs_1(\sin\theta+\mu \cos\theta)}[/tex]

Plugging in the known values, we get:

[tex]v_1=\sqrt{\frac{(15190N/m)(0.25m)^{2}}{10kg}-2(9.8m/s^{2})(1.12m)(\sin36.87\°+(0.300) \cos36.87\°)}\\\\v_1=8.75m/s[/tex]

Now, we can obtain the height from point 1 to point 2 using the kinematics equations. We care about the vertical axis, so first we calculate the vertical component of the velocity at point 1:

[tex]v_1_y=v_1\sin\theta=(8.75m/s)\sin36.87\°=5.25m/s[/tex]

Now, we have:

[tex]y=\frac{v_1_y^{2}}{2g}\\\\y=\frac{(5.25m/s)^{2}}{2(9.8m/s^{2})}\\\\y=1.40m[/tex]

Finally, the maximum vertical displacement [tex]h_2[/tex] is equal to the height [tex]h_1[/tex] plus the vertical displacement [tex]y[/tex]:

[tex]h_2=h_1+y=s_1\sin\theta +y\\\\h_2=(1.12m)\sin36.87\°+1.40m\\\\h_2=2.07m[/tex]

It means that the maximum vertical displacement of the block after it becomes airborne is 2.07 meters.

Answer:

All of the above!

Explanation:

All of the answers are true! I hope I helped!

Correct question:

A child bounces a 57 g superball on the sidewalk. The velocity change of the superball is from 24 m/s downward to 11 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Answer in units of N.

Answer:

The magnitude of the average force exerted on the superball by the sidewalk is 592.8 N

Explanation:

Given;

mass of the superball, m = 57 g = 0.057 kg

initial velocity of the superball, u = 24 m/s

final velocity of the superball, v = 11  m/s

contact time with the sidewalk, t =  1 / 800 s

To determine the magnitude of the average force exerted on the superball by the sidewalk, we apply Newton's second law of motion;

F = ma

[tex]But, a = \frac{v-u}{t} \\\\Thus, F = m(\frac{v-u}{t} )\\\\F = 0.057(\frac{11-24}{1/800} )\\\\F = 0.057(\frac{-13}{1/800})\\\\F = -0.057(\frac{800*13}{1})\\\\ F = -592.8 \ N\\\\Magnitude \ of \ the \ force \ is \ 592.8 \ N[/tex]

Therefore, the magnitude of the average force exerted on the superball by the sidewalk is 592.8 N

Here is the complete part of the question

A block is released from rest at height d= 40 cm and slides down a frictionless ramp and onto a first plateau, which has length d and where the coefficient of kinetic friction is 0.50. If the block is still moving, it then slides down a second frictionless ramp through height d/2 and  onto a lower plateau, which has length d/2 and where the coefficient of kinetic friction is again 0.50. If the block is still moving, it then slides up a frictionless ramp until it (momentarily) stops. Where does the block stop?

Answer:

0.3 m

Explanation:

Given that :

the height h = 40 cm = 0.40 m

Coefficient of kinetic friction is [tex]\mu[/tex] =0.50

Using the Law of conservation of energy = [tex]\frac{1}{2} m \mu_1^2 = mgd[/tex]

As the blocks slides down a frictionless ramp and onto a first rough plateau region.So kinetic energy is decreased to :

[tex]\frac{1}{2}mu^2_2 = mgd - \mu_2_k mgd \\ \\ u^2_2 = 2gd - 2 \mu_kgd[/tex]

If the block is still moving, it then slides down a second frictionless ramp through an height h = d/2

Then , we can say that the gained kinetic energy is :

[tex]\frac{1}{2} mu_2^2 = mg (\frac{d}{2})+\frac{1}{2}mu_2^2 \\ \\ \frac{1}{2} mu_2^2 = mg (\frac{d}{2})+ mgd - \mu_k mgd \\ \\ \frac{1}{2} mu_2^2 = 2g(\frac{d}{2})+2gd - 2 \mu_k gd[/tex]

Futhermore , it moves on the horizontal surface where the coefficient of friction causes some of the kinetic energy to disappear

So, the final value of kinetic energy at the end just before climbing is :

[tex]\frac{1}{2}mv^2 = \frac{1}{2}m \mu_2 ^2 - \mu_k mg (\frac{d}{2}) \\ \\ \frac{1}{2}mv^2 = mg \frac{d}{2} + mgd - \mu_k mgd - \mu_k mg (\frac{d}{2}) \\ \\ v^2 = gd + 2gd - 2 \mu_kgd -\mu_kgd[/tex]

[tex]= 3gd - 3 \mu_k gd \\ \\ = 3[g- \mu_kg ]d[/tex]

Let represent H to be the height above the lower plateau when it momentarily stops; From the law of conservation of energy :

[tex]\frac{1}{2}mv^2 = mgH \\ \\ \frac{3}{2}[g-\mu_kg]d = gH \\ \\ H = \frac{3}{2}[1-\mu_k]d \\ \\ = \frac{3}{2}[1-(0.50)](0.40 \ m) \\ \\ =0.3 m[/tex]

Answer:

We conclude that ball 2 has a larger mass.

Explanation:

This is because the internal forces at collision are an action reaction pair. The force on ball 1 due to ball 2 equals the force on ball 2 due to ball 1. Since these forces are equal in magnitude, let's denote it by F. So

For each ball, F = ma and a = F/m. Since F is the same for both balls, if follows that the acceleration of each ball is inversely proportional to its mass. That is, a ∝ 1/m.

Since ball 1 has a higher acceleration, it has a lower mass. Also ball 2 has a small acceleration, so it has a higher mass.

Answer:

See explanation

Explanation:

Solution:-

- Denote the following:

  mass of ball 1 = m1

  initial velocity of ball 1 = vi,1

  Final velocity of ball 1 = vf,1

  mass of ball 2 = m2

  initial velocity of ball 2 = vi,2

  Final velocity of ball 2 = vf,2

- Since there are no external forces acting on the system containing the two balls then we can say that the linear momentum of the system is conserved.

- The conservation of linear momentum says that the momentum before or after the collision remains conserved:

                                    Pi = Pf

Where,                        p = m*v

- The conservation of momentum for the given system:

                    m1*vi,1 + m2*vi,2 = m1*vi,1 + m2*vi,2  

- Since no information is given about the initial conditions. We will consider the internal force (F) that acts when two balls come in contact. Then apply Newton's second law of motion on both balls individually:

                                F = m*a

Where,             m: Mass of object

                        a: Acceleration of the object

- For ball 1:

                               F1 = m1*a1

- For ball 2:

                              F2 = m2*a2

Given that a1 > a2, and the the internal force that acts on both balls during collision obeys Newton's 3rd Law i.e for every action there is an equal but opposite reaction: So , F1 = F2

                             F = m1*a1

                             F = m2*a2

                             m1*a1 = m2*a2

- Since, a1 > a2 for above equation to hold true then m1 < m2. That is the mass of ball 1 is quite less compared to ball 2.

Answer:

The speed at the top of the hill is 4.38 m/s

Explanation:

Here we have total Kinetic = KE (translational) +KE (rotational)

=0.5 m v² + 0.5m·r²v²/r² = m·v²

Therefore at height 0.6 m we have

0.6 mg = mv²

When v = 5.01 m/s maximum height is

m·g·h=m·5.01²

or h = 2.56 m

Therefore at 0.6 m we have 2.56 - 0.6 more height energy to climb

which gives

1.96·m·g = m v₂²

or v₂² = 19.22

v₂ = 4.38 m/s.

Answer:

6.45 x 10-7 meters

Explanation:

Taking speed of light as 300000000 m/s

We know that speed of light is product of wavelength and frequency and expressed as

s=fw

Where s represent speed, f is frequency and w is wavelength

Substituting 4.65*10^14 Hz for frequency then the wavelngth will be given by

300000000=4.65*10^14 *w

w=300000000÷4.65*10^14

w=6.451612903225*10^-7 m

Rounded off, the wavelength is 6.45 x 10-7 meters

Answer:

0.01 rad

Explanation:

For a double-slit experiment, we have that dsinθ = nλ. The separation between the central maximum and an adjacent maximum is when n = 1

dsinθ = λ

The angular separation θ = sin⁻¹(λ/d)

now d = 100λ

θ = sin⁻¹(λ/100λ) = sin⁻¹(0.01) = 0.573°

θ = 0.573° × π/180 = 0.01 rad

Answer:

Explained.

Explanation:

Let me explain the two cases in by one.

(1) When Person wearing a jacket, heat from his or her body is going to get trapped inside his or her jacket to keep this person warm when a thermometer is placed inside, it would indicate a higher temperature on the scale.

(2) When Jacket is Hanging in a closet, its temperature would be roughly close to room temperature and when we would place a thermometer it would not indicate any change on its scale.

Final answer:

Technician A is correct, as there is slippage between input and output speeds in fluid couplings without a mechanical lock-up. Technician B is incorrect, as the impeller and turbine rpms do not match in a torque converter coupling.

Explanation:

Technician A is correct. In any type of fluid coupling without a mechanical lock-up, there is always some slippage between the input and output speeds. This means that the output speed will be less than the input speed, resulting in a difference between them.

Technician B is incorrect. In a torque converter coupling, the impeller and turbine do not have matching rpm (revolutions per minute). The impeller rpm is higher than the turbine rpm, which allows the torque converter to efficiently transfer power from the engine to the transmission.

Answer:

Chemical

Explanation:

Answer:

The right answer is "chemical"

Explanation:

Batteries store energy in the chemical bonds in the metals inside them.

Answer:

Voltage can be calculated in various ways but I am a easier solution to it:

Explanation:

V=I×R instead of this formula I use:

P=I×V

600=400×V

600=400V

Divide both sides by 400

V=3÷2

V=1.5 J/C

1volt=1coulomb or joule

Answer:1.5V

Explanation: power =IV

Where I= current

V= voltage

Therefore Voltage=power/current

V= 600/400= 1.5

Answer:

See explanation

Explanation:

Depth: Earthquakes can happen anywhere from at the surface to 700 kilometers below. In general, deeper earthquakes are less damaging because their energy dissipates before it reaches the surface. So it is highly likely that the earthquake 1 was shallower than earthquake 2.

Distance from the epicenter: The epicenter is the point at the surface right above where the earthquake originates and is usually the place where the earthquake's intensity is the greatest. Plausible that earthquake 1 might have been closer to its epicenter as compared to earthquake 2.

Local geologic conditions: The nature of the ground at the surface of an earthquake can have a profound influence on the level of damage. Loose, sandy, soggy soil, like in Mexico City, can liquefy if the shaking is strong and long enough, for example. That doesn't bode well for any structures on the surface. Plausible that the area around the earthquake 1 might have poorer geologic conditions as compared to area of earthquake 2.

Answer:[tex]I_2>I_1[/tex]

Explanation:

Given

Shape of the object is dumbbell shaped

Moment of Inertia w.r.t an axis passing through center and perpendicular to it

[tex]I_1=m(\frac{r}{2})^2+m(\frac{r}{2})^2[/tex]

[tex]I_1=\frac{mr^2}{2}[/tex]

For the axis which passes through one of the masses

[tex]I_2=mr^2[/tex]

so [tex]I_2>I_1[/tex]

Sound waves spread out and bend around corners, while light waves behave like rays and do not bend around corners.

The difference between sound waves and light waves when passing through an opening is that light waves behave like rays and do not bend around corners, while sound waves spread out and bend around corners due to their longer wavelengths.

Light waves have very short wavelengths, which allows them to act like rays and create sharp shadows when passing through openings. On the other hand, sound waves have longer wavelengths, on the order of the size of the opening, and can bend around corners.

For example, when light shines through an open door into a dark room, we expect to see a sharp shadow of the doorway on the floor and no light bending around corners. However, when sound passes through a door, we expect to hear it everywhere in the room, indicating that sound waves spread out and bend around corners.

Answer:

See explanation

Explanation:

If the magnetic field increases up through the loop a current will flow such that the magnetic field from this current will oppose the change in magnetic flux. (As per Lenz's  Law)

The magnetic field is increasing up so the field from the current must be directed down.

From the application of right hand rule by pointing the thumb downwards and curl our fingers the current flows clockwise then the magnetic field from this current will be directed downward inside the loop of wire.

According to Lenz's law, the induced current in a horizontal loop of wire with an increasing magnetic field passing upward through it will flow clockwise when viewed from above, as it acts to oppose the change in the magnetic field.

When a horizontal loop of wire has an increasing magnetic field passing upward through it, Lenz's law helps us determine the direction of the induced current in the loop. According to Lenz's law, the induced current will flow in such a way that it creates a magnetic field that opposes the change in magnetic flux. Since the magnetic field through the loop is increasing upwards, the induced current must produce a magnetic field that points downwards.

Applying the right-hand rule, if you orient your right hand with the thumb pointing downwards (opposing the increasing magnetic field), your fingers would curl in a clockwise direction when viewed from above. Therefore, the induced current in the loop will flow in a clockwise direction, as viewed from above, when the magnetic field is increasing through the plane of the loop.

Answer:

60V

Explanation:

240V/12 loops = 20 volts per loop.

20 * 3 = 60 volts  

Answer:

60V

Explanation:

Kinetic energy is formed when water is heated.

Explanation:

There are three types of energy transformations taking place

(i) conversion of kinetic energy to rotational energy

(ii) conversion of rotation energy to electrical energy

(iii) conversion of electrical energy to heat energy

As seen in the image attached, there is running water that is falling on the blades of a small turbine and makes it rotate. The turbine is connected to the heating system with wires.

Now, the energy transformations that take place in heating water are described as follows:

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Answer:

-(0.330m/s² ) kˆ

Explanation:

given data:

Mass of particle 'm'= 1.81 x [tex]10^{-3}[/tex] kg

Velocity 'v'= (3.00 x[tex]10^{4}[/tex] m/s)j

Charge of particle 'q'= 1.22 x [tex]10^{-8}[/tex] C

Uniform magnetic field 'B' = (1.63iˆ + 0.980jˆ )T

In order to calculate particle's acceleration, we'll use Newton's second law of motion i.e F=ma

Also,the force a magnetic field exerts on a charge q moving with velocity v is called the magnetic Lorentz force. It is given by:

F = qv × B

F= ma = qV x B

a= [tex]\frac{q(v*B)}{m}[/tex] --->eq(1)

Lets determine the value of (v x B) first

v x B= (3.00 x[tex]10^{4}[/tex] m/s)j x (1.63iˆ + 0.980jˆ )

v x B= 4.89 x [tex]10^{4}[/tex]

Plugging all the required values in eq(1)

a= [1.22 x [tex]10^{-8}[/tex] x (4.89 x [tex]10^{4}[/tex]kˆ)] / 1.81 x [tex]10^{-3}[/tex]

a=  -(0.330m/s² ) kˆ

-ve sign is representing the opposite direction

Answer:

Drinking 20oz of sports drinks decrease the urine volume compared to 20oz of water

Explanation:

Sport drinks typically used by athletes contain water,carbs and electrolytes which helps to replenish the lost minerals from the body during exercises.

Electrolytes an important ingredients of sports drinks is typically sodium and potassium they help the body to retain water,hence less volume of urine is released from the body.