A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is at a radial distance of 1.8 m from the central axis. Which is true?A- Boy and girl have zero tangential and angular accelerations.B- The girl has a larger angular acceleration than the boy.C- The boy has a larger tangential acceleration than the girl.D- The boy has a larger angular acceleration than the girl.E- The girl has a larger tangential acceleration than the boy.

Answer:

\(Q=23,430J\)

Explanation:

Hello,

In this case, since we compute the required energy via:

\(Q=mC\Delta T\)

Whereas m is the mass which here is 70 g, C the specific heat which for water is 4.184 J/(g°C) and ΔT is the temperature difference which is:

\(\Delta T=100-20=80\°C\)

Therefore, the energy turns out:

\(Q=70g*4.184\frac{J}{g\°C}*80\°C\\ \\Q=23,430J\)

Best regards.

Explanation:

What impulse occurs when a cart that is originally at rest experiences an average force of

N for 2.5 s? *

(10 Points)

25 N

25 Nm

25 Ns

25 kg m/s

Determine (a) the value of k in light of the fact that the particle has a velocity of +9 m/s at x=-3 m and rests at the origin.

The speed of a particle—real or hypothetical—in a medium because it transmits a pulse is known as the particle velocity. The meter per second (m/s) is the metric unit for particle speed. Most of the time, this pressure wave is longitudinal, such as with noise, but it may also be transverse, similar to the vibration of the a tight string.

You could understand the phrase "A particle in repose prefers to stay at rest" to mean that what a particle that isn't moving relative to a certain context aware tends to remain stationary relative to those same things.

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Answer:

Explanation:

As warm air comes in contact with the glass, the air becomes cooler. When that happens the cool air can't hold as much water vapor which condenses on the glass.

The answer is C.

Option A) 3.9 km is the correct answer. the magnitude of Rhea's displacement vector is approximately 3.92 km.

In order to find out the magnitude of Rhea's displacement vector, we have to add up all of the displacement vectors.

Then we can use the Pythagorean theorem to calculate the magnitude of the resultant vector.

Since Rhea is first driving north for 2.4 km and then west for 3.1 km, we can represent her displacement vectors as follows: Δx = 0 km and Δy = 2.4 km for the first vector, and Δx = -3.1 km and Δy = 0 km for the second vector.

We can then add these vectors together by adding their components: Δx = 0 km + (-3.1 km) = -3.1 km and Δy = 2.4 km + 0 km = 2.4 km.

This gives us a resultant vector of -3.1 km east and 2.4 km north.

Using the Pythagorean theorem, we can find the magnitude of this vector: \(\sqrt{(\(-3.1 km)^{2} + (2.4 km)^{2} ) } = \sqrt{(9.61 + 5.76) km^{2} } = \sqrt{15.37 km^{2} } \approx 3.92 km.\)

Therefore, the magnitude of Rhea's displacement vector is approximately 3.92 km.

Therefore, option A) 3.9 km is the correct answer.

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Answer: B.75J

Explanation:

Answer:

So an object with mass is attracted to another object with mass, and the gravitational force is directly proportional to the masses of the two objects, and inversely proportional to the square of the distance between the two objects.

If distance  were to increase, than the gravitational force would decrease. If mass were to increase, so would the gravitational force.

Explanation:

Explanation:

Lets take gravitational force(F) and mass(m) and distance (r)

now for a body in contact with the surface of the earth, its mass is also considered(m‘),now the mass of the earth(m") is also considered,then the distance from the body to the center of the earth (r).ie it r because its practically the radius of the earth. is also considered

So by using dimensional analysis ....

we get F a m'•m"/r² ,where a is proportional to.

now since F is directly increaseproportional to m ie. F a m, then an increase in mass of the body increases it's gravitational force(and clearly that makes sense because the bigger you are the stronger you get pulled to the ground)

then we also see that F is inversely proportional to r ie.F a 1/r ,then an increase in the distance between the ground an the object decrease it's gravitational force ( meaning as any object on earth keeps on moving away from the ground the gravitational force between the object and the center of the earth is weak, when it reaches space then the force becomes virtually negligible!)

So to answer the second question, we clearly see that doubling the mass of the body increases the gravitational force between it and the earth

and doubling the distance on the other hand will decrease the attraction between the body and the earth

So a body forcefully projected into the air fights against gravity but its easier as it keeps on getting higher, If it has a greater mass like that of a trail or truck , it will not even probably stay in the air for long , unless its projected with a very high velocity

I hope this helps, and you can ask me any question concerning this via the comments platform.

Answer:

\(a=0.33\frac{m}{s^2}\)

Explanation:

Hello.

In this case, since the force is defined in terms of the mass and acceleration as follows:

\(F=ma\)

Given the force and the mass, we can compute the acceleration as shown below:

\(a=\frac{F}{m}=\frac{25N}{75kg}=\frac{25kg\frac{m}{s^2} }{75kg}\\ \\a=0.33\frac{m}{s^2}\)

Best regards.

Answer:

Solar energy? Solar rays? Solar Power?

Explanation:

not too sure what you mean, its kinda in the sentence. lmk if you need any help however

Answer:

Solar system is the energy output from the sun

A car is moving north at 5.2 m/s². SI unit in this value is m/s² (meter per second square), expressing Acceleration. Thus, Option D is the correct answer.

Here Acceleration of any object is given by the Rate of change in velocity in relation to time.

\(Acceleration = \frac{velocity}{time}\) ............(i)

The standard Indian (SI) unit of Acceleration is meter per second square(m/s²).

The velocity of any object is displacement per unit time.

\(velocity = \frac{Displacement}{Time}\)............(ii)

The standard Indian (SI) unit of Velocity is meter per second(m/s)

The Displacement of any object is the shortest distance covered by any object considering the direction of motion also.

The standard Indian (SI) unit of displacement is meter(m).

The standard Indian (SI) unit of Time is Second(s).

We can find standard Indian (SI) unit of Acceleration using formula as follows:

\(Acceleration= \frac{Velocity}{time*time}\)  (we got this formula from (i), (ii) )

The standard Indian (SI)  units of acceleration are \(=\frac{m}{s*s}\)

\(=m/s^{2}\)

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The ranking of the voltages, V₁, V₂, V₃, and V₄ in the order of decreasing voltage value is; V₂ > V₃ > V₁ > V₄

A voltage is the difference in potential or pressure that pushes or causes the flow of electric current in a circuit or conducting loop.

The current flowing through the circuit can be found using the following formula;

\(I=\dfrac{V}{R}\)

The formula for conductivity is presented as follows;

\(\sigma =\dfrac{l}{R\times A}\)

Where;

σ = The conductivity of the material

l = The length

R = The resistance to electricity

A = The cross-sectional area of the conductor

Therefore;

\(R=\dfrac{V}{I}\)

\(R=\dfrac{l}{A\times \sigma}\)

\(\dfrac{V}{I} =\dfrac{l}{A\times \sigma}\)

\(V=\dfrac{l \times I}{A\times \sigma}\)

Based on the dimensions of the cylinders, obtained from a similar question, we get;

\(V_1=\dfrac{3\cdot L \times I}{\pi \cdot D^2\times \sigma} = 3\times \dfrac{ L \times I}{\pi \cdot D^2\times \sigma}\)

\(V_2= \dfrac{2\cdot L \times I}{\pi \cdot \dfrac{D^2}{4} \times \sigma} = \dfrac{8\cdot L \times I}{\pi \cdot D^2\times \sigma} = 8\times \dfrac{ L \times I}{\pi \cdot D^2\times \sigma}\)

\(V_3=\dfrac{L \times I}{\pi \cdot \dfrac{D^2}{4} \times \sigma} =\dfrac{4\cdot L \times I}{\pi \cdot D^2\times \sigma}= 4\times \dfrac{L \times I}{\pi \cdot D^2\times \sigma}\)

\(V_4=\dfrac{L \times I}{\pi \cdot D^2\times \sigma} = 1\times \dfrac{ L \times I}{\pi \cdot D^2\times \sigma}\)

Therefore;

V₁ = 3·V₄

V₂ = 8·V₄

V₃ = 4·V₄

The rank of the voltages in the order of decreasing value is therefore;

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Answer:

The surface tension of the water is 6.278×10⁻² N/m

error = 13.65%

Explanation:

The surface tension of water is given by

\($ \gamma = \frac{F}{L} $\)

Where F is the force acting on water and L is the length over which is force is acted.

We are given the mass of 100 droplets of water

M = 3.78 g

n = 100

The mass of 1 droplet is given by

\(m = \frac{M}{n} \\\\m = \frac{3.78}{100}\\\\m = 0.0378 \: g \\\\m = 3.780\times10^{-5} \: kg\)

The force acting on a single droplet of water is given by

\(F = m \cdot g\)

Where m is the mass of water droplet and g is the acceleration due to gravity

\(F = 3.780\times10^{-5} \cdot 9.81\)

\(F = 3.708\times10^{-4} \: N\)

The circumferential length of the droplet is given by

\(L = \pi \cdot d\)

Where d is the diameter

\(L = \pi \cdot 1.88\times10^{-3}\\\\L = 5.906 \times10^{-3} \: m\)

Now we can find out the required surface tension of the water

\(\gamma = \frac{3.708\times10^{-4} }{5.906 \times10^{-3}} \\\\\gamma = 0.06278\: N/m\\\\\gamma = 6.278 \times10^{-2} \: N/m\\\\\)

Therefore, the surface tension of the water is 6.278×10⁻² N/m

The tabulated value of the surface tension of water at 20 °C is given by

\($ \gamma_t = 0.0727 \: N/m $\)

The percentage error between tabulated and calculated surface tension is given by

\($ error = \frac{\gamma_t - \gamma }{\gamma_t} $\)

\($ error = \frac{ 0.0727 - 0.06278}{0.0727} \times 100\% $\)

\($ error = 13.65 \%\)

Answer:

5% (w/v), by definition, means 5 grams of solute per 100 grams of solution.  Since the density is 1.2 g/ml, you can find the volume of 100 grams of solution.  100 g solution x 1 ml/1.2 g = 83.33 mls.  Hence...

83.33 mls x 5 g/100 mls = 4.17 g solute needed.

The mass of the object is approximately 0.457 kg.

The mass of the object attached to the trolley can be calculated using the principle of conservation of momentum. Since the two trolleys couple together and move as a single system after the collision, the total momentum before and after the collision should be the same. Given the mass of one trolley is 0.80 kg and the initial speed is 1.1 m/s, the momentum before the collision is 0.80 kg * 1.1 m/s = 0.88 kg·m/s. After the collision, the total mass is the sum of the two trolleys, and the final speed is 0.70 m/s.

Using the momentum equation, the mass of the object can be calculated as follows:

Total momentum before collision = Total momentum after collision

0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s

Solving for the mass of the object, we get:

0.88 kg·m/s = (0.80 kg + mass of the object) * 0.70 m/s

0.88 kg·m/s = 0.56 kg + 0.70 kg * mass of the object

0.88 kg·m/s - 0.56 kg = 0.70 kg * mass of the object

0.32 kg = 0.70 kg * mass of the object

Dividing both sides by 0.70 kg, we find:

mass of the object = 0.32 kg / 0.70 kg = 0.457 kg

The two trolleys collide and couple together, the total momentum before the collision is equal to the total momentum after the collision according to the principle of conservation of momentum.

The momentum of an object is defined as the product of its mass and velocity. In this case, the mass of one trolley is known (0.80 kg) and the initial speed is given (1.1 m/s), allowing us to calculate the momentum before the collision.

After the collision, the two trolleys move together at a new speed (0.70 m/s). By setting the initial momentum equal to the final momentum and solving for the unknown mass of the object, we can find its value.

In the calculation, we subtract the masses of the two trolleys from the total mass in order to isolate the mass of the object.

Dividing the difference in momentum by the product of the known mass and the new speed, we obtain the mass of the object. In this case, the mass of the object is approximately 0.457 kg.

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When a skydiver reaches terminal velocity, her acceleration is essentially zero. This is because the upward force of air resistance becomes equal to the downward force of gravity, and the net force acting on the skydiver becomes zero.

The skydiver's acceleration when they reach terminal velocity is essentially zero. So, the option "It is essentially zero" is the correct answer. This is due to the fact that when the terminal velocity is reached, the force of air resistance that the diver is subjected to increases in magnitude, preventing any further acceleration.The downward force exerted by gravity is balanced out by the upward force exerted by the air resistance on the skydiver as they fall at a constant speed. The acceleration of an object in free fall is typically 9.8 m/s² downward.

However, when a skydiver reaches their terminal velocity, this is no longer the case. With no net force, there is no acceleration according to Newton's second law of motion (F = ma).

In general, terminal velocity is defined as the highest velocity that a falling object can attain when the force of air resistance on the object is equal in magnitude and opposite in direction to the gravitational force acting on it. After this, the net force on the object becomes zero, and thus, the object no longer accelerates.

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contractility , excitability, extensibility, and ... Elasticity

the one which is NOT is abductability

Answer:

Explanation:

Formula (Force)

We have the values of :

Substitution

sound wave amplitude

Answer:

Single Displacement Reaction

Explanation:

Here, only one element is getting replaced during the reaction, that is, silver gets replaced by copper, hence, single displacement reaction.

Hope it helps :)

Answer:

b

Explanation:

when the wavelength increase it doesnt affect the frequency of a wave.

Answer: The wavelength increases by a factor of 3

Explanation:

Answer:

the proton speed of the proton was 1.6 × 10⁶ m/s

Explanation:

Given the data in the question;

Radius r = 6.42 cm = 0.0642 m

magnetic field B = 0.260 T

we know that; charge of proton q = 1.602 × 10⁻¹⁹ C

And mass of proton m = 1.672 × 10⁻²⁷ kg

we know that; Magnetic Force F = qvBsinθ

where q is the charge of proton, v is velocity, B is the magnetic field and θ is  angle ( 90° )

Also the Centripetal force experienced by the particle is;

F = mv² / r

where r is radius, m is mass of proton and v is velocity

hence;

qvBsinθ = mv² / r

we solve for v

rqvBsinθ = mv²

divide both sides by mv

rqvBsinθ / mv = mv² / mv

rqBsinθ / m = v

so we substitute

v = [ 0.0642 m × (1.602 × 10⁻¹⁹ C) × 0.260 T × sin(90°) ] /  1.67 × 10⁻²⁷ kg

v = 2.6740584 × 10⁻²¹ / 1.672 × 10⁻²⁷

v = 1.6 × 10⁶ m/s

Therefore, the proton speed of the proton was 1.6 × 10⁶ m/s

Answer:

The two forces acting on a boat or some other floating object are buoyancy and gravity

Answer: buoyant forceExplanation:two forces acting on a boat or some other floating object are buoyant force and gravityhi friend your answerI hope it will be helpful for you

mark as brainest answer

thank you

The concept of time in space can be different from time on Earth due to the effects of time dilation. Time dilation occurs due to differences in gravitational fields and relative motion.

If we consider an astronaut who is in a relatively low-gravity environment and not traveling at a significant fraction of the speed of light, the time experienced by the astronaut in space would be very similar to the time experienced on Earth. In this scenario, one hour in space would be approximately the same as one hour on Earth.

However, if we consider extreme cases, such as an astronaut near a black hole or traveling at near-light speeds, time dilation effects would become significant. In these situations, the time experienced by the astronaut would be different from the time on Earth.

It's important to note that the magnitude of time dilation effects depend on the specific conditions and relative velocities involved. For most common space travel scenarios, the difference in time between space and Earth is negligible, and one hour in space would correspond to one hour on Earth.

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Answer:

Answer:

A

Explanation:

Given data:

Force;

Time;

The impulse is given as,

Substituting all known values,

Therefore, the magnitude of the impulse of this force is 0.384 N.s.

three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650