An astronaut on the planet Zircon tosses a rock horizontally with a speed of 6.95m/s. The rock falls through a vertical distance of 1.40m and lands a horizontal distance of 8.75m from the astronaut. What is the acceleration of gravity on Zircon?

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Explanation: Edge 2020

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Answer:

It is C.

Explanation:

Got it right on edge.

Answer:

the solution for x is 0.2403 M-1s-¹

Explanation:

To solve for x, we simply need to plug in the given values and simplify:

x = (8.9 x 10³ M-2s-¹)(0.30 M)³

x = 8.9 x 10³ M-2s-¹ x (0.30 M)³

x = 8.9 x 10³ M-2s-¹ x 0.027 M³

x = 0.2403 M-1s-¹

Therefore, the solution for x is 0.2403 M-1s-¹.

Answer:

Newton's first law states that an object that is in rest or in motion will stay at rest or in motion until acted upon a unbalanced force. So when an object that is in motion wether it is slow or fast (it is in motion) and a unbalanced force makes it act in a different way than it is already and so which in this case it is making it stop

Explanation:

According to Newton's first law of motion, an object in motion will stay in motion unless it is restrained by an outside force. When an object slows down or stops, external forces on it, like friction or air resistance, are to blame for the object's loss of motion.

Newton's first law of motion states that an object in motion will remain in motion at a constant velocity and in a straight line unless acted upon by an external force. Similarly, an object at rest will remain at rest unless acted upon by an external force.

Here int he Question,

An object that is slowing down and coming to a stop, it is being acted upon by external forces that are causing it to lose its motion. These forces could be friction, air resistance, or any other opposing force that is acting on the object. As the object moves, these forces work to slow it down and eventually bring it to a stop.

For example, if a car is moving on a road, the friction between the tires and the road, as well as air resistance, will act to slow it down. The engine of the car provides the force necessary to overcome these opposing forces and maintain the car's motion. However, if the engine is turned off, the car will eventually slow down and come to a stop due to the opposing forces.

Therefore, Newton's first law of motion explains that an object in motion will continue to move unless acted upon by an external force. In the case of an object slowing down and coming to a stop, external forces such as friction or air resistance are acting upon the object, causing it to lose its motion.

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Answer:

Choice A. gamma rays

Explanation:

Answer:

homeostasis met be your's answer

Homeostasis is the state that leads to the maintenance of constancy in the properties and composition of the internal environment.

It is the capacity of the organism to present a characteristic and constant physical-chemical situation within certain limits.

Any system, including living beings, tend to adapt to changes and maintain a stable and constant internal environment through homeostasis.

Therefore, we can conclude that homeostasis is responsible for adapting our body to changes in the environment to help maintain the conditions of the internal environment.

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Answer:

Explanation:

put your question properly first!!!

Answer:

I believe the answer is They can warm up in winter and and keep cool in summer.

This is because thermostats are controlled by humans and will change the temperature in an enclosed area (usually houses). During the summer, people will usually keep their houses cool, and during the winter, people will usually keep their houses warm. Mammal's bodies can helps keep cool in the summer and can help keep warm in the winter. This is their similarity.

hope this helps

Answer:

The switching rate between the steady state and the normal state of a capacitor

Explanation:

This was a hard one! Could only find it in my textbook. Anyways this basically is the rate which the capacitors switches back from steady state and normal state from when it charges and discharges over time. This has many purposes as a special type of diode or any other transistor type device etc etc.

Answer:

uuuugggghhhhhhhhggggffffffffffffff

Answer:

The geocentric model could not fully explain these changes in the appearance of the inferior planets (the planets between the Earth and the Sun). Furthermore, Galileo's observations of Jupiter's moons made it clear that celestial bodies do move about centers other than the Earth.

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Answer: 0

Explanation: Trust

The magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.

The magnetic force on a charged particle moving through a magnetic field can be calculated using the formula:

F = q * v * B * sin(θ)

Where:

F is the magnetic force,

q is the charge of the particle (in this case, 4.88 x 10^(-6) C),

v is the velocity of the particle (in this case, 265 m/s),

B is the magnetic field strength (in this case, 0.0579 T),

θ is the angle between the velocity vector and the magnetic field vector (in this case, 90 degrees).

Plugging in the values:

F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * sin(90°)

Since sin(90°) is equal to 1, the equation simplifies to:

F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * 1

Calculating the value:

F = 6.47 x 10^(-4) N

Therefore, the magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.

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The depth of the barge below the waterline is 2.40 m.

To calculate the depth of the barge below the waterline, we need to consider the buoyancy force acting on the barge. The buoyancy force is equal to the weight of the water displaced by the barge.

First, we need to calculate the volume of water displaced by the barge.

Since the barge is flat-bottomed, we can assume that the shape of the displaced water is rectangular with a length of 20.0 m, a width of 10.0 m, and a depth of d (which is what we're trying to find).

Therefore, the volume of water displaced is V = 20.0 m x 10.0 m x d = 200.0 m³.

The weight of the displaced water can be calculated using its density and volume. In fresh water, the density of water is approximately 1000 kg/m³.

Therefore, the weight of the displaced water is W = 1000 kg/m³ x 200.0 m³ = 2.00 × 10⁵ kg.

Since the buoyancy force is equal to the weight of the displaced water, we have \(F_b\) = W = 2.00 × 10⁵ kg.

The weight of the barge is \(W_b\) = 4.80 × 10⁵ kg. According to Archimedes' principle, the buoyancy force acting on an object in a fluid is equal to the weight of the fluid displaced by the object, so we can write:

\(F_b\) = \(W_b\) - \(W_d\)

where \(W_d\) is the weight of the water displaced by the submerged part of the barge. Solving for \(W_d\), we get:

\(W_d\) = \(W_b\) - \(F_b\) = 4.80 × 10⁵ kg - 2.00 × 10⁵ kg = 2.80 × 10⁵ kg.

The volume of water displaced by the submerged part of the barge is equal to the volume of the rectangular prism with a length of 20.0 m, a width of 10.0 m, and a depth of d. Therefore, we can write:

\(V_d\) = 20.0 m x 10.0 m x d = 200.0 m³ x (d/10.0)

The weight of the displaced water is also equal to its density times its volume, so we have:

\(W_d\) = 1000 kg/m³ x \(V_d\)

Substituting \(V_d\) in terms of d and solving for d, we get:

d = (\(W_d\) / (1000 kg/m³ x 200.0 m²)) x 10.0 m = (2.80 × 10⁵ kg / (1000 kg/m³ x 200.0 m²)) x 10.0 m = 2.40 m

Therefore, the depth of the barge below the waterline is 2.40 m.

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Answer:

true

Explanation:

The  acceleration, tension of the cord between the 3.5 and 1 block, and the force excreted on the 2 block from the 1 block is 12.6 N.

To solve this problem, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

We can start by finding the acceleration of the system. Since the surface is frictionless, there are no horizontal forces acting on the blocks except for the tension in the string. Thus, we can write:

F - T = (m1 + m2 + m3) * a

where F is the force acting on the 3.5 kg block, T is the tension in the string between the 3.5 kg and 1 kg blocks, and a is the acceleration of the system. Substituting the values given in the problem, we get:

46 N - T = (1 kg + 2 kg + 3.5 kg) * a

46 N - T = 6.5 kg * a

Next, we need to find the tension in the string between the 3.5 kg and 1 kg blocks. This tension is the same throughout the string, so we can use the same equation we used to find the acceleration:

T - m1 * g = m1 * a

where g is the acceleration due to gravity (9.8 m/s^2). Substituting the values given in the problem, we get:

T - 1 kg * 9.8 m/s^2 = 1 kg * a

T = 1 kg * (a + 9.8 m/s^2)

Now, we can substitute this expression for T into the first equation we derived to get:

46 N - 1 kg * (a + 9.8 m/s^2) = 6.5 kg * a

Simplifying and solving for a, we get:

a = 2.84 m/s^2

Now that we know the acceleration of the system, we can find the force exerted by the 1 kg block on the 2 kg block. Since there are no horizontal forces acting on the 2 kg block, this force is equal in magnitude and opposite in direction to the force exerted by the 2 kg block on the 1 kg block. Thus, we can write:

F1-2 = -m2 * a

where F1-2 is the force exerted by the 1 kg block on the 2 kg block. Substituting the values given in the problem, we get:

F1-2 = -2 kg * 2.84 m/s^2

F1-2 = -5.68 N

Therefore, the force exerted by the 1 kg block on the 2 kg block is 5.68 N, directed to the left.

Finally, we can use the equation we derived earlier for the tension in the string to find the tension between the 3.5 kg and 1 kg blocks. Substituting the value we found for a, we get:

T = 1 kg * (2.84 m/s^2 + 9.8 m/s^2)

T = 12.6 N

Therefore, the tension in the string between the 3.5 kg and 1 kg blocks is 12.6 N.

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Answer:

Balanced forces are responsible for unchanging motion. Balanced forces are forces where the effect of one force is cancelled out by another. A tug of war, where each team is pulling equally on the rope, is an example of balanced forces. The forces exerted on the rope are equal in size and opposite in direction.

Explanation:

Answer:

Agua

Explanation:

Answer:

Explanation:

time to travel first leg

D = Vt₁

t₁ = D/V

time to travel second leg.

at constant deceleration, the average velocity will be half of the original.

2D = (V/2)t₂

t₂ = 4D/V

t = t₁ + t₂ = 5D/V

Vavg = d/t = (D + 2D)/(5D/V) = (3D/5D)V = 0.6V

With the concept of kinematics we find the average speed is 0.6V

given parameters

   * initial velocity V

   * the distances in each part of the movements x₁ = D and x₂ = 2D

to find

  The average speed of the entire trajectory

In kinematics we analyze the motion of bodies, the average velocity is defined by

        \(V_{avg} = \frac{\Delta x}{\Delta t}\)           1

where \(V_{avg}\) is the average velocity, Δx and Δt are the variation of displacement and time in the interval

For this exercise we have two types of movement in the first part a uniform movement and in the second part an accelerated movement  let's solve each onaccelerated movement,e separately

1 Part. uniform motion

we look for the time

          V = \(\frac{x_1}{t_1}\)

          t₁ = x₁ / V

          t₁ = \(\frac{D}{V}\)

2 Part. Accelerated movement, let's start by looking for acceleration, they tell us that the body stops at the end of the interval so its velocity is zero v_f=0, the initial velocity is v₀ = V

           \(v_f^2 = v_o^2 - 2 \ a \ x_2\)

           0 = v₀² - 2 a x₂

           a = \(\frac{v_o^2}{2x_2}\)

we substitute

           a = \(\frac{V^2}{D}\)

now we look for the time it takes to stop

          v = v₀ - a t₂

          0 = v₀ - a t₂

          t₂ = \(\frac{v_o}{a}\)

we substitute

          t₂ = \(\frac{V}{\frac{V^2}{4D} }\)

          t₂ = \(\frac{4D}{V}\)4D / V

for the entire movement, the total displacement is

          Δx = x₁ + x₂

          Δx = D + 2D

          Δx = 3D

the total time is  

          Δt = t₁ + t₂

          Δt = \(\frac{D}{V} + \frac{4D}{V}\)

          Δt = \(\frac{5D}{V}\)

we substitute in equation 1 of average velocity

         \(V_{avg} = \frac{3D}{\frac{5D}{V} }\)

         \(V_{avg} = \frac{3}{5} \ V\)

         \(V_{avg } = 0.6 V\)

With the use of the kinematics equations with constant acceleration we can find the average velocity of the body throughout the journey is 0.6V

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As the fish tank is 20 inches by 12 inches by 12 inches, its volume is 47194744.32 mm³.

The space that any three-dimensional solid occupies is known as its volume. These solids can take the form of a cube, cuboid, cone, cylinder, or sphere.

1 inch = 0.0254 meters.

20 inches = 20 × 0.0254 meters = 0.508 meters = 508 mm.

12 inches = 12 × 0.0254 meters = 0.3048 meters = 304.8 mm.

Hence, The volume of the fish tank  = length × width × height

=  508 mm ×  304.8 mm ×  304.8 mm.

= 47194744.32 mm³

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Answer:

7kg

Explanation:

Given parameters:

Force on the rock  = 28N

Acceleration  = 4m/s²

Unknown:

Mass of the rock  = ?

Solution:

According to newton's second law of motion;

           Force  = mass x acceleration

Insert the given parameters to find the mass;

         28  = mass x 4

             mass  = \(\frac{28}{4}\)   = 7kg

With a current shunt, the current is obtained by measuring voltage across the current shunt and calculating using Ohm's Law.

A current shunt is a device that is used to measure electric current. It is a small resistor placed in parallel with the load (or the element being measured), that creates a known small voltage drop proportional to the current flowing through it.

By measuring this voltage drop and using Ohm's law, the current flowing through the shunt (and the load) can be calculated. Current shunts are commonly used in high-current applications, such as in power plants, electrical distribution systems, and electric vehicles.

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With a current shunt, the current is obtained by measuring the voltage drop across the current shunt and calculating using Ohm’s Law. A current shunt is a device that creates a low-resistance path for electric current, to allow it to pass around another point in the circuit. In current measuring, shunts allow the measurement of high current values by placing a resistor of low, known resistance in parallel with a voltmeter.

Answer: 1. The movie one is virtual and the bathroom mirror is real

2. The image is distorted in a way

3. Behind the spoon

Explanation:

The movie one is virtual and the bathroom mirror is real and The image is distorted in a way and Behind the spoon.

In virtual desktops, the desktop environment is segregated from the physical device being used to access it. They are preloaded images of operating systems and apps. Over a network, users can remotely view their virtual desktops.

A virtual desktop can be accessed from any endpoint device, including a laptop, smartphone, or tablet. The user interacts with the client software that was installed on the endpoint device by the virtual desktop provider.

A virtual desktop mimics the appearance and feel of a real workstation. Because robust resources, like storage and back-end databases, are easily accessible, the user experience is frequently even better than that of a physical workstation.

Therefore, The movie one is virtual and the bathroom mirror is real and The image is distorted in a way and Behind the spoon.

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Answer:

Lets see this

Explanation:

So workout the mph of an average car then try and find out the speed

A hollow sphere made out of electrically insulating material is electrically neutral (no excess charge) then the correct option is  b) The excess charge has distributed itself evenly over the outside surface of the sphere.

The metal spherical is a conductor and hollow. Charges can therefore circulate freely across its surface as a result. A metal body serves as an equipotential body on the other side. Once a charge has been established and there is no longer a voltage differential imprinted across the body, the charges must be distributed uniformly across the exterior surface for the body to continue to be an equipotential body. There must not be any charge in the volume because otherwise an electrical field and consequent voltage differential would arise.

Additionally, the interior surface's charge distribution must be zero. The enclosed charge must be zero if the Gauss Theorem is applied to a gaussian sphere with a radius between the internal and external surfaces and the knowledge that field E is zero.

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Answer:

There are at least four different kinds of moonquakes: (1) deep moonquakes about 700 km below the surface, probably caused by tides; (2) vibrations from the impact of meteorites; (3) thermal quakes caused by the expansion of the frigid crust when first illuminated by the morning sun after two weeks of deep-freeze lunar ...

Explanation:

Answer:

The moonquakes are probably a result of the moon going contraction or tides

Answer:

g = 0.85 m\(s^{-2}\)

Explanation:

g = \(\frac{GM}{h^{2} }\)

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x \(10^{-11}\) N\(m^{2}\)\(kg^{-2}\)), M is the mass of the earth (5.972 x \(10^{24}\) kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

  = 3.40 x 6371 km

h = 21661.4 km

  = 21661400 m

Thus,

g = \(\frac{6.674*10^{-11}*5.972*10^{24} }{(21661400)^{2} }\)

  = \(\frac{3.9857 *10^{14} }{4.6922*10^{14} }\)

  = 0.84944

g = 0.85 m\(s^{-2}\)

The acceleration due to the Earth's gravitation is 0.85 m\(s^{-2}\).

Answer:

23. 63 degrees

Explanation:

Let us draw a free body diagram.

The net forces on ladder + painter must equal zero. Therefore, along the y-direction:

Along the x-direction

Now, we calculate the torque about the about O.

The torque is must be zero:

simplifying the above gives

Since

the above becomes

dividing both sides by cosine gives

further division gives

substitute the value of N and we get:

putting in m_p = 60 kg, m_l = 10 kg, l = 6.0 m and d = 1/3 * 6 = 2, and u = 0.5 gives,

taking the inverse tan of both sides gives

Therefore, the required angle against the wall is

Hence, the ladder should lean 23.63 degrees against the wall.

The net force on particle q₂, located between particles q₁ and q₃, is approximately 189000 N. The force exerted by particle q₁ on q₂ is positive and equals 252000 N, while the force exerted by particle q₃ on q₂ is negative and equals -63000 N.

To find the net force on particle q₂, we need to calculate the individual forces exerted on q₂ by particles q₁ and q₃ and then determine their sum.

The force between two charged particles can be calculated using Coulomb's law:

F = k * |q₁ * q₂| / r²

Where F is the force between the particles, k is the electrostatic constant (k ≈ 9.0 x \(10^9\) Nm²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them.

First, let's calculate the force exerted on q₂ by q₁:

F₁₂ = k * |q₁ * q₂| / r₁₂²

F₁₂ = (9.0 x \(10^9\) Nm²/C²) * |(8.0 μC) * (3.5 μC)| / (0.10 m)²

F₁₂ ≈ 252000 N

The force is positive because q₁ and q₂ have opposite charges.

Next, let's calculate the force exerted on q₂ by q₃:

F₂₃ = k * |q₂ * q₃| / r₂₃²

F₂₃ = (9.0 x \(10^9\)Nm²/C²) * |(3.5 μC) * (-2.5 μC)| / (0.15 m)²

F₂₃ ≈ -63000 N

The force is negative because q₂ and q₃ have the same charge.

Finally, we can find the net force on q₂ by summing the individual forces:

Net force = F₁₂ + F₂₃

Net force = 252000 N + (-63000 N)

Net force ≈ 189000 N

The net force on particle q₂ is approximately 189000 N.

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Answer:

I=0.016A

Explanation:

Explanation:

Given:

Diameter = 200 m

Radius, r = 200/2 = 100 m

Time taken, t = 30 seconds

Formula to be used:

Distance traveled, = circumference of circle = 2πr

Answer:

Putting all the values, we get

Distance traveled = 2πr

So, the distance traveled by Raju is 628.57 m.

Now, magnitude of the displacement,

At the end of 30 seconds, Raju will come to starting position or initial position, so displacement is zero.

Kepler's second law states that the radius vector drawn from the sun to the planet seeps out equal areas in equal intervals of time. The second law is referred to as the law of equal areas. Thus, the correct option is A

Answer:

44cm x 22cm

Explanation:

u= 10 cm

v= 1.1 cm

m=v/u= 1.1/10

m=11

hence the size of the image.

-0.5m/s^2 is the  acceleration of the car

Because we have the Initial velocity (20 m/s), the final velocity (0 m/s), and the distance traveled x=400m. we can use the kinematic equation

Vf^2= Vo^2 + 2ax,

vf will be zero.

0= Vo^2+ 2ax

Vo^2= -2ax

20^2= -2a(400)

1= -2a

a= -1/2= -0.5m/s^2

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