Two astronauts of equal mass are holding on to each other and moving at a speed of 30 m/s. They push off of each other and one of the astronauts moves in the same direction as the two were initially moving but at 60 m/s. The velocity of the second astronaut is

The package strike the ground relative to the point directly below it was released at 213.3m.

Given values,

Velocity of plane 45 m/s

Acceleration due to gravity 9.8 m/s²

horizontal component of plane velocity is

ux = 45 m/s

Vertical component is zero

In the same way

ay = - 9.8  m/s²

ax = .0 m/s^2

Motion in y direction = -110 m

From equation of motion time of flight is

d = u x t + 1/2 at²

where,

d is distance

u is velocity initial

a is acceleration

t is time

So, putting values

-110 = 0 + 1/2 (-9.8)t²

-110 = -4.9t²

t² = 110 / -4.9

t² = 22.45

t = √22.45

t = 4.74

Motion in x direction,

d = u x t + 1/2 at²

d = ( 45 x 4.74) + 1/2 (0) 4.74²

d = 213.3 + 0

d = 213.3m

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Answer:

As the parallel rays coming from the object converge at the principal focus, F of a concave mirror; after reflection through it. Therefore, when the object is at infinity the image will form at F. Properties of the image formed: Point-sized image, highly diminished in size.

At Infinity: At focus (F)

Position of object: Position of image

Between F and C: Beyond C

Beyond C: Between F and C

Explanation:

This is because when the object is at infinity, the incident rays are parallel to the principal axis. Hence the image is formed at the focus. And the image is real and inverted but in diminished size.

Answer: 12,339

Explanation:

Answer:

I assume they want answer A, but it's a silly question.

Explanation:

Well, if we assume that the girls are applying equal force, as they are the same size and strength (apparently also the same resolve etc.).

Equal force in opposite directions means the composite force is null. An object that no force is applied to continues moving in the same direction, so if the doll was moving somewhere, it would keep doing so, except all of this makes no sense the doll is not in gravityless vacuum what are we even talking about.

The answer A is the least ridiculous.

Answer:

C?

Explanation:

Yep,It's C all right.

Answer:

Yep,It's C all right.

Explanation:

We obtain the magnitude of the acceleration as   1.55 m/s^2 while its component parallel to the ground is 0.775  m/s^2.

The acceleration refers to change of speed with time. From the question;

It the follows that;

v^2 = u^2 + 2as

And;

u = 0m/s  owing to the fact that motion started from rest

v^2 = 2as

a = v^2/2s

a = (10.43 m/s)^2/2 * 35.0 m

a = 1.55 m/s^2

For the acceleration component which is parallel to the ground;

ax = a sin θ

ax =  1.55 m/s^2 cos 60

ax = 0.775  m/s^2

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given mass= 2 kg

height= 10m

g= 9.8.

we know that work down W = FD

= > W = (mg) (D)

= > W = (2* 9.8)(10)

= > W = 196 joules

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The linear distance and the tangential velocity of the end of the stick are 1.63 m and 10.9 m/s respectively.

Rotational motion is the motion of a body around a circular path, in a fixed orbit. The dynamics for rotational motion are entirely analogous to linear or translational dynamics. The equations for rotating objects are similar to the motion equations for linear motion.

Given, the length of the hockey stick, r = 1.1 m

The hockey stick completes one swing in time, t = 0.15 sec

The angle θ = 85° = 1.483 rad

The average angular velocity of the hockey stick,

ω = θ/t = 1.483rad/0.15s = 9.9 rad/s

The linear distance moved by the end of the hockey stick:

L = rθ = 1.1× 1.483 = 1.63 m

The tangential velocity of the end of the hockey stick:

v = rω = 1.1 × 9.9 = 10.9 m/s

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Decrease its voltage and decrease its current change to a circuit is most likely to decrease its electrical power.

The correct option is B.

When charged electrons (current) are forced through a conducting loop by the pressure of an electric circuit's power source, they may perform tasks like lighting a lamp. In a nutshell, voltage equal pressure and is expressed in volts.

Electric potential difference, often referred to as voltage, is measured in units called volts. According to official definitions, a volt is the difference in potential between two locations on a wire that's also conveying a current of one ampere and dissipating one watt of power.

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A hypothesis. It is the first step to the scientific method.

Answer:

Average speed is 5 m/s

Explanation:

The initial speed = 0 m/s (since the runner starts from rest)

Final speed = 10 m/s

time interval = 2 sec

average speed = ?

The speed is assumed to change at a constant rate

Average speed = (final speed + initial speed)/2

==> (10 + 0)/2 = 10/2 = 5 m/s

It sounds as though the two people are standing in front of the boat on opposite sides of it, so that they both make an angle of 30.0° with the axis of the boat, as in the attached free body diagram (ignoring the force of buoyancy and the weight of the boat).

By Newton's second law, the net vertical force is

∑ F = P₁ sin(60.0°) + P₂ sin(120.0°) - R = 0

where upward is positive and downward is negative, and the right side is 0 because the boat moves with constant velocity and thus zero acceleration.

We're told that P₁ = P₂ = 600 N, and we know sin(60°) = sin(120°), so the above reduces to

R = 2 P sin(60.0°) = 2 (600 N) sin(60.0°) ≈ 1040 N

We have that for the Question "Two people are pulling a boat through the water. Each exerts a force of 600 N directed at a 30.0 angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force exerted by the water on the boat." it can be said that the resistive force exerted by the water on the boat is

From the question we are told

Two people are pulling a boat through the water. Each exerts a force of 600 N directed at a 30.0 angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force exerted by the water on the boat.

Generally the equation for the Force  is mathematically given as

\(F=2*f cos\theta\\\\F=2*600*cos30\)

F=1039N

Therefore

the resistive force exerted by the water on the boat is

F=1039N

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Normal force is the force exerted when an object is on an surface. So an example could be a pile of books on top of a table.

For any freely falling item, the acceleration in the kinematic equations is -9.8 m/s/s, whether or not this is explicitly stated.

The supplied problem informs us that a ball was dropped and is supposed to be falling freely. Since the problem does not specify whether any other forces were present to influence the ball's motion, we must assume that the initial speed is v0=0 m/s.

Free-falling accelerates at a rate of 9.8 m/s2, which is referred to as the acceleration owing to gravity because it is being drawn towards the center of the earth. This indicates that a=9.8m/s2 is the value of the initial and final accelerations.

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Answer:

I don't understand your question ❓,the object.....of what experiment

Therefore, the charge on each capacitor is: Q1 = Q2 = 3 μC Q3 = 4.17 μC Q4 = 2 μC Q5 = 125 μC.

A two-terminal electrical device known as a capacitor is capable of storing energy in the shape of an electric charge. It is made up of two electrical wires that are spaced apart by a certain amount. Vacuum may be used to fill the area between the conductors, or a dielectric, an insulating substance, may be used instead.

We can examine the circuit using Kirchhoff's equations to determine the voltage across each capacitor and the charge on each capacitor.

First, by connecting the two 5 F capacitors in parallel, we can streamline the circuit and create a single identical capacitor with capacitance:

Ceq = C1 + C2 = 5 μF + 5 μF = 10 μF

Next, we can find the total resistance of the circuit by adding the resistances in series:

Rtotal = R1 + R2 = 30 Ω + 20 Ω = 50 Ω

We can now calculate the time constant of the circuit:

τ = Rtotal × Ceq = 50 Ω × 10 μF = 500 μs

Using the time constant, we can find the voltage across each capacitor at steady-state (when the circuit has reached equilibrium):

\(V1 = V × (1 - e^(-t/τ)) = 25 V × (1 - e^(-0/500 μs)) = 0 VV2 = V × (1 - e^(-t/τ)) = 25 V × (1 - e^(-0/500 μs)) = 0 VV3 = V × (1 - e^(-t/τ)) = 25 V × (1 - e^(-0/500 μs)) = 0 VV4 = V × (1 - e^(-t/τ)) = 25 V × (1 - e^(-0/500 μs)) = 0 VV5 = V × (1 - e^(-t/τ)) = 25 V × (1 - e^(-0/500 μs)) = 25 V\)

Therefore, the voltage across each capacitor is:

V1 = V2 = V3 = V4 = 0 V

V5 = 25 V

To find the charge on each capacitor, we can use the equation:

Q = C × V

Q1 = C1 × V1 = 3 μC

Q2 = C2 × V2 = 3 μC

Q3 = C3 × V3 = 4.17 μC

Q4 = C4 × V4 = 2 μC

Q5 = C5 × V5 = 125 μC

Therefore, the charge on each capacitor is:

Q1 = Q2 = 3 μC

Q3 = 4.17 μC

Q4 = 2 μC

Q5 = 125 μC

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a) The magnitude of the electric field at the center of the semicircle is given by E = Xo/(2πR).

An electric field is a physical field created by the presence of electric charge. It is an invisible force that can act on any other charged particles in its vicinity. Electric fields are generated by electric charges, or by time-varying magnetic fields.

The direction of the electric field is towards the center of the semicircle. This can be seen from the fact that the electric field is generated by the line of positive charge, thus pointing inwards.
b) The total force on the particle with negative charge -q placed at the center of the semicircle is given by F = -qE. Thus, the total force on the particle is given by F = -qXo/(2πR).
c) The magnitude of the electric field at the center of the semicircle is given by E = Xo/(2πR) cose. Here, cose is the cosine of the angle between the line of charge and the position of the particle. The direction of the electric field is still towards the center of the semicircle.
Bonus: If the semicircle has a total charge Q, the magnitude of the electric field at the center of the semicircle is given by E = Q/(2πR). The direction of the electric field is still towards the center of the semicircle.

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In order for the clown to land unharmed, the other clowns should hold the net at a point along the trajectory of the clown that is equal to the launch height of the clown.

The point can be calculated using the following equation:

Distance = velocity × time

Where velocity = 40 m/s and time = 2(velocity × sin(60o))/9.8 (where 9.8 is gravitational acceleration).

Therefore, the distance = 40 m/s × 2(40 m/s sin(60o))/9.8 = 80 m.

The other clowns should aim to place the net at a point that is 80 meters away from the cannon and at a height of 40 meters. This point should correspond to the apex of the trajectory, which is the highest point that the clown will reach when launched. The clowns should also make sure that the net is held at an angle of 30 degrees with respect to the horizontal, to ensure that the clown lands safely at the same level where it was launched.

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The temperature of the boiler is 150.9°C, the thermal energy absorbed each hour is 1.62 x 10^9 joules, and the amount of thermal energy lost per hour is 1.61 x 10^9 joules.

The Carnot cycle is a theoretical thermodynamic cycle that consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. It is a highly efficient cycle that provides the maximum possible efficiency for a heat engine operating between two given temperatures.

18. The efficiency of a Carnot engine is given by the equation:

efficiency = 1 - (T_C / T_H)

where T_C is the temperature of the cold reservoir and T_H is the temperature of the hot reservoir. We can rearrange this equation to solve for T_H:

T_H = T_C / (1 - efficiency)

Substituting the given efficiency and waste heat temperature values, we get:

T_H = 36.3°C / (1 - 0.761) = 150.9°C

Therefore, the temperature of the boiler is 150.9°C.

19. The efficiency of a Carnot engine is given by the equation:

efficiency = 1 - (T_C / T_H)

where T_C is the temperature of the cold reservoir and T_H is the temperature of the hot reservoir. We can rearrange this equation to solve for the thermal energy absorbed each hour:

thermal energy absorbed per hour = power output / efficiency

Substituting the given values, we get:

thermal energy absorbed per hour = 115 kW / (1 - (13°C / 569°C)) = 1.62 x 10^9 J/h

Therefore, the thermal energy absorbed each hour is 1.62 x 10^9 joules.

20. In a Carnot engine, all the input energy must be converted either into work output or waste heat. Therefore, the amount of thermal energy lost per hour is equal to the difference between the thermal energy absorbed and the power output:

thermal energy lost per hour = thermal energy absorbed per hour - power output

Substituting the given values, we get:

thermal energy lost per hour = 1.62 x 10^9 J/h - 115 kW x 3600 s/h = 1.61 x 10^9 J/h

So, the amount of thermal energy lost per hour is 1.61 x 10^9 joules.

Hence, The boiler has a temperature of 150.9°C, absorbs 1.62 × 109 joules of thermal energy every hour, and loses 1.61 x 109 joules of thermal energy every hour.

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The Answer Is - 0.58 Kilogram's

Hope This Help's

• Answer:

0.58 kg

• Explanation:

1 kg = 1000 g

580 g = 0.58 kg

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The greatest distance of the block from the center of rotation so that it does not slip is approximately 0.089 meters or 8.9 centimeters.

In order for the block to remain on the turntable without slipping, the frictional force between the block and the turntable must be equal to or less than the maximum friction that can be developed, which is 2 N in this case. The maximum frictional force is given by the product of the coefficient of friction and the normal force.The normal force is equal to the weight of the block, which is 5 N. Therefore, the maximum frictional force is 2 N.

To find the greatest distance of the block from the center of rotation without slipping, we need to consider the centrifugal force acting on the block due to the rotation of the turntable. The centrifugal force is given by the product of the mass of the block, the acceleration due to the rotation (which is equal to the square of the angular velocity times the radius), and the coefficient of friction.

Since the weight of the block is 5 N, the mass can be calculated as 5 N divided by the acceleration due to gravity (approximately 9.8 m/s^2), which gives us a mass of approximately 0.51 kg.

Given that the turntable is rotating at 0.6 revolution per second, the angular velocity can be calculated as 2π times the frequency, which gives us approximately 3.77 rad/s.

Let's assume the greatest distance of the block from the center of rotation is "r" meters.

The centrifugal force is then given by m * (ω^2) * r, and this force must be equal to or less than the maximum frictional force, which is 2 N.

Therefore, we can write the equation:

m * (ω^2) * r ≤ 2 N

Substituting the values we calculated:

0.51 kg * (3.77 rad/s)^2 * r ≤ 2 N

Simplifying the equation, we find:

r ≤ 0.089 m

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Answer:

The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.

Explanation:

The object is not under the influence of any external force, meaning that Principle of Momentum Conservation to calculation of the velocity of the third fragment:

\((m_{1}+m_{2}+m_{3})\cdot \vec {v}_{o} = m_{1}\cdot \vec v_{1} + m_{2}\cdot \vec v_{2} + m_{3}\cdot \vec v_{3}\) (1)

Where:

\(m_{1}\), \(m_{2}\), \(m_{3}\) - Masses of the first, second and third fragments, in kilograms.

\(\vec v_{o}\) - Initial velocity of the object, in meters per second.

\(\vec v_{1}\), \(\vec v_{2}\), \(\vec v_{3}\) - Velocities of the first, second and third fragments, in meters per second.

If we know that \(m_{1} = 0.5\,kg\), \(m_{2} = 1.3\,kg\), \(m_{3} = 1.2\,kg\), \(\vec v_{o} = (0,0)\,\left[\frac{m}{s} \right]\), \(\vec v_{1} = \left(-2.8, 0\right)\,\left[\frac{m}{s} \right]\) and \(\vec v_{2} = \left(0,-1.5\right)\,\left[\frac{m}{s} \right]\), the velocity of the third fragment is:

\((-1.4,0) + (0,-1.95) + 1.2\cdot \vec v_{3} = (0,0)\)

\(1.2\cdot \vec v_{3} = (1.4,1.95)\)

\(\vec v_{3} = (1.167, 1.625)\,\left[\frac{m}{s} \right]\)

The speed of the third fragment is the magnitude of the result found above:

\(v_{3} = 2\,\frac{m}{s}\)

And the direction of the third fragment is:

\(\theta_{3} = \tan^{-1} \left(\frac{1.625}{1.167}\right)\)

\(\theta_{3} \approx 54.316^{\circ}\)

The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.

Answer: you have to divide and multiple fam

Explanation:

Answer:

The frictional force between the wall and the ball

If A box of mass 210 kg is pulled from rest with a string of tension 1300n inclined at 35° to the horizontal. if the box moved with a speed of 10m/s and the frictional force between the box and surface is 100 n, Then the distance covered by the box is 10.89 meters.

To calculate the distance covered by the box, we need to analyze the forces acting on it and apply the work-energy principle.

Given:

Mass of the box, m = 210 kg

Tension in the string, T = 1300 N

The angle of inclination, θ = 35°

Frictional force, f = 100 N

Initial speed, u = 0 m/s

Final speed, v = 10 m/s

First, let's resolve the tension force into components parallel and perpendicular to the incline. The parallel component of the tension force can be calculated as:

T_parallel = T * cos(θ)

Next, let's calculate the net force acting on the box along the incline. The net force is given by:

Net force = T_parallel - f

Now, using Newton's second law, we can calculate the acceleration (a) of the box:

Net force = m * a

From the given information, we have the final velocity (v), initial velocity (u), and acceleration (a). We can use the following kinematic equation to calculate the distance covered (s):

v^2 = u^2 + 2as

Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Now, let's plug in the given values and calculate the distance covered:

T_parallel = 1300 N * cos(35°) ≈ 1067.35 N

Net force = 1067.35 N - 100 N = 967.35 N

a = (967.35 N) / (210 kg) ≈ 4.61 m/s^2

s = (10 m/s)^2 - (0 m/s)^2 / (2 * 4.61 m/s^2) ≈ 10.89 m

Therefore, the distance covered by the box is approximately 10.89 meters.

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Answer:

The word is "wind".

Explanation:

The sentence would be:

"The sun is also the reason wind exists. The wind is moving air."

Hope this helped !

The characteristics of the projectile launch allows to find the result for the velocity on the x axis is:

In projectile launching we have an application of kinematics to motion in two dimensions near the surface of the Earth.

In this case, on the vertical axis we have an acceleration directed towards the center of the planet and called the gravity acceleration.

In the horizontal axis or x axis there is not acceleration, except in special cases that we tone is how much the resistance of the air.

Consequently in the horizontal axis there is no acceleration and therefore the speed must be kept constant, the result for the horizontal speed is:

           vₓ = 8.0 m / s

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