An atom with a mass number of 61 amu has 33 neutrons in its nucleus.
What element is this ?

Answer:

b

Explanation:

Explanation:

these are quantities that have magnitude but no direction

Examples include : speed, mass, volume, density, everything, time

Energy released ,

[tex] = 3.789 \times {10}^{ - 12} [/tex] J

Answer:

v_f = 2.9m/s

Explanation:

v_f = v_I + a(t)

v_f = 0.6m/s + 1.2m/s²(1.9s)

v_f = 2.88m/s

v_f = 2.9m/s

Answer:

4.88 K.

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 5 moles

Pressure (P) = 1 atm

Volume (V) = 2 L

Gas constant (R) = 0.082 atm.L/Kmol

Temperature (T) =?

The temperature of the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

1 × 2 = 5 × 0.082 × T

2 = 0.41 × T

Divide both side by 0.41

T = 2 / 0.41

T = 4.88 K

Therefore, the temperature of the gas is 4.88 K.

Answer:

9 times

Explanation:

luminosity decreases to the square of the distance

Answer: hello your question is poorly written attached below is the well written question

answer: 4.9 * 10^-5 T

Explanation:

Current ( i ) = 39.30 A

Determine the magnitude of the B-field at Point P

Mag of B-field at Point P = μo*I / 2π*r --- ( 1 )

where : I = 39.30 , μo = 4π * 10^-7 , r = 8*10^(-2)

Input values into equation 1

( 10^-7 * 39.30 ) / ( 8*10^-2 )

∴ Mag of B-field = 4.9 * 10^-5 T

Answer:

C. Alleles

Explanation:

Our genes are responsible for our phenotype (physical traits). Alleles is the only genetic term in there.

Answer:

its two valence electrons

Explanation:

Calcium is a group 2 element with two valence electrons. Therefore, it is very reactive and gives up electrons in chemical reactions.

UR WELCOME!! :)

Answer:

Moment is the product of force and its perpendicular distance from a point along its line of action.

The net moments on clockwise and anticlockwise is zero at a point.

The net force applied on a spanner handle is equal to the moment of the rotation

Answer:

0.18 m/s²

Explanation:

The total time taken to cover a 10 km race is 27 min, 43.6 seconds.

At 25 min, Pre was at 7.71 km mark. Therefore the average speed = 7.71 km / 25 min = 7710 m / (25 * 60) s = 5.14 m/s

The distance remaining = 10 km - 7.71 km = 2.29 km = 2290 m

The remaining time = 27 min, 43.6 seconds - 25 min = 2 min 43.6 second = 163.6 seconds

She accelerates for 60 seconds, therefore the distance covered (S) during the acceleration (a) is:

S₁ = 5.14(60) + 0.5a(60)² = 308.4 + 1800a

She maintains the speed for the remaining distance (S₂). The remaining time = 163.6 seconds - 60 seconds = 103.6 seconds. The final speed after the acceleration = (5.14 + 60a) m/s

S₂ = (5.14 + 60a)* 103.6 = 532.5 + 6216a

S₁ + S₂ = 2290 m

(308.4 + 1800a) + (532.5 + 6216a) = 2290

8016a + 840.9 = 2290

8016a = 1449.1

a = 0.18 m/s²

Answer:

the baby would still be 30 kg on the moon because that’s mass, not weight.

However, it will feel like 5 kg, because the acceleration is 1/6 and the force is 1/6 and the actual weight (product of mass and acceleration will be 1/6).

The child has a mass of 30 kg on earth and has one-sixth gravity on the moon will have a weight of 50 kg on the moon.

Gravity is a fundamental interaction that causes all objects with mass or energy to attract one another. The weakest of the four fundamental interactions is gravity, by far. The Moon's gravity causes sublunar tides in the oceans, just as gravity on Earth imparts weight to physical objects.

Given:

The mass of the child (m) = 30 kg,

The gravity on the moon (G) = 1/6 of the earth.

To calculate the weight of the child on the moon, use the formula given below,

F = m × a

Here

[tex]F[/tex] is the weight, [tex]m[/tex] is the mass and [tex]a[/tex] is the gravity on earth, gravity on earth [tex]10\ m/s^2[/tex].

Substitute the values

[tex]F=30*10/6[/tex]

[tex]F = 50\ kg[/tex]

Therefore, the weight of a child on the moon is 50 kg.

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Answer:

[tex]k.e = \frac{1}{2} m {v}^{2} \\ {v}^{2} = {u}^{2} + 2gs \\ {v}^{2} = 0 + (2 \times 9.8 \times 15) \\ v = 17.1 \: m {s}^{ - 1} \\ k.e = \frac{1}{2} \times m \times {17.1}^{2} \\ = 147m \: joules \\ m \: is \: mass[/tex]

Kinetic energy of the rock just before hitting the water is 147m joules where m is the mass of the object.

The kinetic energy of an object is defined as the energy it possesses due to its motion. It is the work required to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains the kinetic energy as long as its momentum does not change.

It is directly proportional to the mass of the object and the square of its velocity:[tex]K.E. = 1/2 mv^2[/tex].

Where, m= mass of the object

v= velocity

So if the units of mass are in kilograms and the units of velocity are meters per second, then kinetic energy has units of kilogram-meter squared per second squared.

For above given information,

[tex]K.E= 1/2 mv^2\\v^2= u^2 + 2gs\\ \\v^2= 0+ (2* 9.8*15)\\v^2= 294 m/s\\So, K.E. = 1/2 m* 294 m/s\\K.E.= 147m joules[/tex]

Thus, Kinetic energy of the rock just before hitting the water is 147m joules where m is the mass of the object.

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Answer:

592000 J

Explanation:

We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:

1 Pa = 1 Kg/ms²

Therefore,

3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²

Next, we shall determine the workdone.

Workdone is given by the following equation:

Workdone (Wd) = pressure (P) × change in volume (ΔV)

Wd = PΔV

With the above formula, the work done can be obtained as follow:

Pressure (P) = 3.7×10⁵ Kg/ms²

Change in volume (ΔV) = 1.6 m³

Workdone (Wd) =?

Wd = PΔV

Wd = 3.7×10⁵ × 1.6

Wd = 592000 Kgm²/s²

Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:

1 Kgm²/s² = 1 J

Therefore,

592000 Kgm²/s² = 592000 J

Therefore, the Workdone is 592000 J.

Answer:

35%

Explanation:

win

The amount of work required to lift the 230 kg to a height of 3.5 m at  84% efficiency is 9391.67 J

This can be defined as the capacity to achieve a given task with little or no waste. Mathematically, it can be expressed as

Efficiency = (output / input) × 100

Work output = mgh

Work output = 230 × 9.8 × 3.5

Work output = 7889 J

Efficiency = (output / input) × 100

84% = 7889 / input

Cross multiply

84% × input = 7889

Divide both side by 84%

Input = 7889 / 84%

Input = 7889 / 0.84

Input = 9391.67 J

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Answer:

[tex]momentum = mass \times velocity \\ = 7.5 \times 1.6 \\ = 12 \: kg {ms}^{ - 1} [/tex]

Mass is 7.5 kg

Velocity is 1.6 m/s

we know that,

momentum = mass × velocity

or; momentum =7. 5 kg × 1. 6 m/s

or; momentum = 12 kgms^1

Answer:

(C) 250

Explanation:

Given the following data;

Mass = 10 kg

Distance = 50 m

Force applied = 10 N

Angle = 60°

To find the work done on the box, we would first of all find the horizontal component of the force applied.

[tex] Horizontal force, Fx = mgCosd[/tex]

Where;

Fx represents the horizontal force.

m is the mass of an object.

g is the acceleration due to gravity.

d is the angle of inclination (theta).

mg = weight = 10 N

Substituting into the equation, we have;

[tex] Fx = 10 * Cos60 [/tex]

[tex] Fx = 125 * 0.5 [/tex]

Fx = 5 Newton

Next, we find the work done.

Mathematically, workdone is given by the formula;

[tex] Workdone = force * distance[/tex]

Substituting into the formula, we have;

[tex] Workdone = 5 * 50 [/tex]

Workdone = 250 Nm

Answer:

new PE at top=original PE - energy lost

mgh=.2(9.8)20-30

Explanation:

mgh=.2(9.8)20-30 = 9.2

Answer:

120,000seconds

Explanation:

400 minutes x 5= 2000minutes (2000minutes x 60)seconds =120,000seconds.

btw :

stay safe! :3

Answer:

12000

Explanation:

5 x 400 = 2000, 2000 x 60 = 12000

Answer:

Explanation:

The formula for the magnitude of this resultant vector, which is in Q1, btw, is:

[tex]B_{mag}=\sqrt{6.18^2+4.43^2}[/tex] which gives us, to 3 sig fig's:

7.60 m. Since the resultant vector is in Q1, we don't need to add anything to the angle once we find it to 3 sig fig:

[tex]tan^{-1}(\frac{4.43}{6.18})[/tex] gives us an angle of 35.6 degrees

Answer:

Electricity is the set of physical phenomena associated with the presence and motion of matter that has a property of electric charge. Electricity is related to magnetism, both being part of the phenomenon of electromagnetism, as described by Maxwell's equations.

Answer:

Explanation:

You didn't fill in the proper masses which is why you never got an answer to this. But that's ok...I got you. I happen to know what they are! We will use the universal law of gravitation and the gravitational constant to solve this.

[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] and filling in:

[tex]F_g=\frac{(6.67*10^{-11})(5.98*10^{24})(7.36*10^{22})}{(3.84*10^8)^2}[/tex] The denominator is the radius of the earth plus the radius of the moon plus the distance between their surfaces, just FYI.

That gives us that

[tex]F_g=1.99*10^{20}N[/tex] Not sure what your choices entail, but I'd have to say, taking into consideration that maybe your problem didn't figure in the distance between the surfaces, you'd be at choice B.

Answer:

194 degrees farenheit! hot

Explanation:

(90°C × 9/5) + 32 = 194°F

Answer:

194

Explanation:

to get from celcius to Fahrenheit

multiply temperature by 2 and add 30

Answer:

 ΔT = [tex]\pi \ \frac{\Delta L}{\sqrt{Lg} }[/tex]

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

          w = [tex]\sqrt{\frac{g}{L} }[/tex]

angular velocity and period are related

          w = 2π / T

we substitute

          2π / T = \sqrt{\frac{g}{L} }

          T = [tex]2\pi \ \sqrt{\frac{L}{g} }[/tex]

In this exercise indicate that for a long Lo the period is To, then and increase the long

          L = L₀ + ΔL

we substitute

           T = [tex]2\pi \ \sqrt{\frac{L + \Delta L}{g} }[/tex]

            T = [tex]2\pi \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }[/tex]

in general the length increments are small ΔL/L «1, let's use a series expansion

           [tex]\sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...[/tex]  

we keep the linear term, let's substitute

           T = [tex]2\pi \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L} )[/tex]  

if we do

           T = T₀ + ΔT

           T₀ + ΔT = [tex]2\pi \sqrt{\frac{\Delta L}{g} } + \pi \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}[/tex]

            T₀ + ΔT = T₀ + [tex]\pi \sqrt{\frac{1}{Lg} } \ \Delta L[/tex]

            ΔT = [tex]\pi \ \frac{\Delta L}{\sqrt{Lg} }[/tex]

Answer:

h = 0.2 m

Explanation:

Given that,

A pendulum is moving 2.0 m/s at the bottom of its swing.

We need to find the height high it swing back. Let the height is h.

Using the conservation of energy such that,

[tex]\dfrac{1}{2}mv^2=mgh\\\\h=\dfrac{v^2}{2g}[/tex]

Put all the values,

[tex]h=\dfrac{(2)^2}{2\times 10}\\\\h=0.2\ m[/tex]

So, it will reach to a height of 0.2 m.

Answer:

W = 24 J

Explanation:

Given that,

Applied force, F = 60 N

Distance moved, d = 40 cm = 0.4 m

We need to find the work done by the carpenter. We know that,

Work done, W = Fd

Put all the values,

W = 60 N × 0.4 m

= 24 J

Hence, the required work done is equal to 24 J.

Answer:

36000

Explanation:

1/2mv²=0.5×500×12²=36000j

Answer:

Explanation:

First off, we need the mass in kg and it's in g so we have to convert it. Then we can do the problem. 200 g = .200 kg. Moving on.

The equation used to find the tension in the string is the same one we use to find the centripetal force, because the tension is what is upplying the centripetal force needed to keep the stone moving in a circular manner. The formula for that is

[tex]F_c=T=\frac{mv^2}{r}[/tex] where ma is the mass in kg, v is the velocity in m/s, and r is the radius of the circle about which the stone spins in meters.

Filling in:

[tex]T=\frac{(.200)(2)^2}{2}[/tex] which gives us

T = .4 N