A straight wire in a magnetic field experiences a force of 0.011 N when the current in the wire is 2.3 A. The current in the wire is changed, and the wire experiences a force of 0.051 N as a result. What is the new current

Explanation:

The artificial satellite experiences a centripetal force [tex]F_c[/tex] as it moves around the earth and it is defined as

[tex]F_c = m\dfrac{v^2}{r} = m\left(\dfrac{2\pi r}{T}\right)^2\left(\dfrac{1}{r}\right) = \dfrac{4\pi^2mr}{T^2}[/tex]

where m is the mass of the satellite, r is its orbital radius and T is its orbital period. But we need to find the radius first.

Recall that the satellite is orbiting at a height where its acceleration due to gravity is 6.44 m/s^2. Since we know that the weight mg of the satellite is equal to the gravitational force [tex]F_G[/tex] between the earth and the satellite, we can write

[tex]mg = F_G = G\dfrac{mM}{r^2}[/tex]

[tex]\Rightarrow g = G\dfrac{M}{r^2}[/tex]

where M is the mass of the earth (=[tex]5.972×10^{24}\:\text{kg}[/tex]) and G is the universal gravitational constant (=[tex]6.674×10^{-11}\:\text{N-m}^2\text{/kg}[/tex]). Plugging in the values, we find that the radius of the satellite's orbit is

[tex]r = \sqrt{\dfrac{GM}{g}} = \sqrt{\dfrac{(6.674×10^{-11}\:\text{N-m}^2\text{/kg})(5.972×10^{24}\:\text{kg})}{6.44\:\text{m/s}^2}}[/tex]

[tex]\:\:\:\:\:= 7.87×10^6\:\text{m}[/tex]

Now that we have the value for the radius, we can now calculate the orbital period T. Recall that the centripetal force is equal to the weight of the satellite at its orbital radius. Therefore,

[tex]F_c = mg \Rightarrow \dfrac{4\pi^2mr}{T^2} = mg[/tex]

or

[tex]4\pi^2r = gT^2[/tex]

Solving for T, we get

[tex]T^2 = \dfrac{4\pi^2r}{g} \Rightarrow T = \sqrt{\dfrac{4\pi^2r}{g}}[/tex]

We can further simplify the above expression into

[tex]T = 2\pi\sqrt{\dfrac{r}{g}}[/tex]

Plugging in the values for r and g, we get

[tex]T = 2\pi\sqrt{\dfrac{(7.87×10^6\:\text{m})}{(6.44\:\text{m/s}^2)}}[/tex]

[tex]\:\:\:\:\:= 6945\:\text{s} = 1.93\:\text{hrs}[/tex]

________

= 310 - 273

= 37°C

Actually,310 Kelvin is same with 37°C, and as you see, there is no 37°C

So, The Nearest Number To 37°C is 36,9°C

Answer:

36.9

Explanation:

Plato

Answer:

she will eventually slow down and come to a stop

Answer:

L=mvr

α=0

v=ωr

L=mωr

2

2

1

mv

2

=k

L=Iω

k=

2

1

Iω

2

ω

1

=2ω

k

1

=1/2k

∴I

1

=

8

1

I

L

1

=

8

1

I×2ω

=

4

1

Iω

=L/4

Physical motion that occurs when an item rotates or spins on an axis is known as “rotatory motion,” sometimes known as “rotational motion” or “circular motion.”

Angular displacement per unit of time is measured by angular frequency, also referred to as radial or circular frequency (). Therefore, it has degrees (or radians) per second as its units. 1 Hz = 6.28 rad/sec, thus. 1 radian equals 57.3° because 2 radians equal 360°.

The frequency of rotation is the rate at which an item rotates around an axis, sometimes referred to as rotational speed or rate of rotation.

However, the linear velocity also depends on how far the particles are from the axis of rotation. Because every particle distances are different, so too will the linear velocities for each particle.

Therefore, All the particles in a rigid body's rotational motion move in the same direction in a given interval. Therefore, all the particles' angular velocities will be the same.

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In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).

Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:

[tex]x=\frac{m_{steam}}{m_{total}}[/tex]

Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:

[tex]x=\frac{2kg}{10kg} =0.20[/tex]

Now, at this temperature and pressure, the volume of a saturated vapor is  2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:

[tex]v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg[/tex]

This means that the volume of the container will be:

[tex]V=10kg*0.4727m^3/kg\\\\V=4.73m^3[/tex]

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Answer:

C

Explanation:

Definition of drug: a medicine or other substance which has a physiological effect when ingested or otherwise introduced into the body

Answer:

Why is community action important? ... Involving communities in the design and delivery of services can help to achieve a number of objectives, including: Building community and social capacity – helping the community to share knowledge, skills and ideas. Community resilience – helping the community to support itself

Answer:

F12 = G M1 M2 / R12^2

F12' = G M1 M2 / R12'^2

F12' / F12 = R12'^2 / R12^2 = (1/3)^2

F12' = 1/9 F12

The new force is 1/9 the of the old force

Answer:

Explanation:

The work of the brakes will equal the initial kinetic energy of the car

Fd = ½mv²

F = mv²/2d

F = 2457(18²) / (2(62))

F = 6,419.903...

F = 6.4 kN

Answer:

The box arrives first.

Explanation:

Hope this helps!! :))

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According to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

Friction may be defined as the resistance that is offered by the surfaces that are in contact when they move past each other. It is a type of force that opposes the motion of a solid object over another.

There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. According to the context of this question, the sphere possesses less friction as compared to the box. This is because the box has an irregular surface that possesses high friction over the inclined surface.

Therefore, according to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

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Hello!

To solve, we can begin by using the kinematic equation:

[tex]d = v_it + \frac{1}{2}at^2[/tex]

Where:

vi = initial velocity (m/s)

t = time (s)

a = acceleration (in this case, due to gravity. g = 9.8 m/s²)

Since the object falls from rest, the initial velocity is 0 m/s.

[tex]d = \frac{1}{2}at^2[/tex]

Plug in the given values:

[tex]d = \frac{1}{2}(9.8)(3.5^2) = \boxed{60.025 m}[/tex]

Answer: increases

Explanation:

Temperature is a measure of the average velocity of the molecular particles. The faster they go, the higher the temperature.

Answer:

ma*XsinB

option 2 is correct

Answer:

Approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex], assuming that no external force (e.g., gravitational pull) was acting on this rocket.

Explanation:

Assume that no external force is acting on this rocket. The system of the rocket and the fuel on the rocket would be isolated (an isolated system.) The momentum within this system would be conserved.

Let [tex]v_{0}\; \rm m\cdot s^{-1}[/tex] be the initial velocity of the rocket.

The velocity of the exhaust gas would be [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] since the gas is ejected away from the rocket.

Let [tex]\Delta v\; \rm m\cdot s^{-1}[/tex] denote the increase in the velocity of the rocket. The velocity of the rocket after ejecting the gas would be [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex].

The momentum [tex]p[/tex] of an object of velocity [tex]v[/tex] and mass [tex]m[/tex] is [tex]p = m \cdot v[/tex].

The combined mass of the rocket and the fuel was [tex]10000\; \rm kg[/tex]. The initial momentum of this rocket-fuel system would be:

[tex]\begin{aligned}p_{0} &= m \cdot v\\ &= 10000\; {\rm kg} \times v_{0}\; {\rm m \cdot s^{-1}} \\ &= (10000\; v_{0})\; \rm {kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].

The momentum of the [tex]5.0\; \rm kg[/tex] of fuel ejected at [tex](v_{0} - 5000)\; \rm m\cdot s^{-1}[/tex] would be:

[tex]\begin{aligned} & 5.0 \; {\rm kg} \times (v_{0} - 5000)\; {\rm m\cdot s^{-1}}\\ =\; & (5.0\, v_{0} - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

After ejecting the [tex]5.0\; \rm kg[/tex] of the fuel, the mass of the rocket would be [tex]10000\; \rm kg - 5.0\; \rm kg = 9995\; \rm kg[/tex]. At a velocity of [tex](v + \Delta v)\; \rm m\cdot s^{-1}[/tex], the momentum of the rocket would be:

[tex]\begin{aligned} & 9995 \; {\rm kg} \times (v_{0} + \Delta v)\; {\rm m\cdot s^{-1}}\\ =\; & (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

Take the sum of these two quantities to find the momentum of the rocket-fuel system after the fuel was ejected:

[tex]\begin{aligned}p_{1} &= (5.0\, v_{0} - 25000)\; {\rm kg \cdot m\cdot s^{-1}\\ &\quad\quad + (9995\, v_{0} + 9995\, \Delta v)\; {\rm kg \cdot m \cdot s^{-1}} \\ &= (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}\end{aligned}[/tex].

The momentum of the rocket-fuel system would be conserved. Thus [tex]p_{0} = p_{1}[/tex].

[tex](10000\, v_{0})\; {\rm kg \cdot m\cdot s^{-1}} = (10000\, v_{0} + 9995\, \Delta v - 25000)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Solve this equation for [tex]\Delta v[/tex], the increase in the velocity of the rocket.

[tex]10000\, v_{0} = 10000\, v_{0} + 9995\, \Delta v - 25000[/tex].

[tex]9995\, \Delta v = 25000[/tex].

[tex]\begin{aligned}\Delta v &= \frac{25000}{9995} \approx 2.5\end{aligned}[/tex].

Thus, the velocity of the rocket would increase by approximately [tex]2.5\; \rm m\cdot s^{-1}[/tex] after ejecting the [tex]5.0\; \rm kg[/tex] of fuel.

Answer:

I feel like its the second one but I'm not completely sure..

Explanation:

Explanation:

The orbiting period of a satellite at a height h from earth' surface is

T=2πr32gR2

where r=R+h.

Then, T=2π(R+h)R(R+hg)−−−−−−−−√

Here, R=6400km,h=1600km=R/4

T=2πR+R4−−−−−−√R(R+R4g)−−−−−−−−−⎷=2π(1.25)32Rg−−√

Putting the given values,

T=2×3.14×(6.4×106m9.8ms−2)−−−−−−−−−−−−√(1.25)32=7092s=1.97h

Now, a satellite will appear stationary in the sky over a point on the earth's equator if its period of revolution around the earthh is equal to the period of revolution of the earth up around its own axis whichh is 24h. Let us find the height h of such a satellite above the earth's suface in terms of the earth,'s radius.

Let it be nR.Then

T=2π(R+nR)R(R+nRg)−−−−−−−−−−√

=2π(Rg)−−−−−√(1+n)32

=2×3.14(6.4×106m/s9.8m/s2)−−−−−−−−−−−−−−−⎷(1+n)32

(5075s)(1+n)32=(1.41h)(1+n)32

For T=24h, we have (24h)=(1.41h)(1+n)32

or (1+n)32=241.41=17

or 1+n(17)23=6.61

or n=5.61

The height of the geostationary satellite above the earth's surface is nR=5.61×6400km=6.59×104km.

The work done by the man against friction is 4,192.86 J.

The given parameters;

The work done by the man against friction is calculated as follows;

[tex]W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J[/tex]

Thus, the work done by the man against friction is 4,192.86 J.

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Answer:  You need to use Newton's law for the equation --->

Explanation: G  ×  M  ×  m / separation. Thats how youll get your answer !!

Answer:

0

Explanation:

0 is the answer

The volume of a material is the total amount of matter that it can contain. The volume of the given coin has been determined to be 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex]. Since the gold doubloon do not absorb water, then its volume remains constant at the ocean floor.

The volume of the gold doubloon can be determined by;

volume = [tex]\pi r^{2}[/tex] + h

where r is the radius of the coin and h is its thickness.

Such that; diameter = 6.1 cm (61 mm) and h = 2.0 mm

r = [tex]\frac{diameter}{2}[/tex]

 = [tex]\frac{61}{2}[/tex]

r = 30.5 mm

Thus,

volume of the coin = [tex]\frac{22}{7}[/tex] x [tex](30.5)^{2}[/tex] x 2

                               = 5847.2857

Therefore, the volume of the gold doubloon is 5847.3 [tex]mm^{3}[/tex]. This can also be expressed as 5.85 x [tex]10^{-6}[/tex] [tex]m^{3}[/tex].

Since the gold doubloon is  not miscible with water, thus its volume at a depth of 770 m at the ocean floor is the same as its initial volume.

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Answer:

all are harmful to the body

The moral condition of the world today appears to be worse than it was in the antediluvian era.

The biblical account of the flood records that the world delved into apostasy in the days of Noah so much so that God regretted the fact that he created man. Some of the evils of the antediluvian world include; sodomy, drunkenness, lewdness and several forms of immorality.

We can see that these vices that led to the destruction of the world due to moral bankruptcy in the antediluvian era is still very much prevalent in our world today. The moral condition of the world today appears to be worse than it was in the antediluvian era.

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Answer:

All organisms begin their lives as single cells.Overtime,these organisms grow and take on the characteristics of their species...All organisms grow,and different parts of organisms may grow at different rates.Organisims made out of only one cell

may change little during their lives, but they do grow

Explanation:

brainlest me please

L=1m

=2mm²=(2/1000²)=2(10^-6)m

y=4x10¹¹N/m²

∆l=2mm=2/10.00=0.002m

=(4x10¹¹x2x10^-6x2x10^-3)÷1m

=16x10¹¹-⁶-³

=16x10¹¹-⁹

=16x10²

=1600N

where a=cross sectional area=2x10^-6m

C=2mm= 2x10^-3m

L=1m

hope it helps

Answer:

c) cornea being wavy or not spherical

Answer:

d = -33.1 m and Vf = -25.5 m/s

Explanation:

a = -9.8 m

t = 2.6 s

Vᵢ = 0 m/s

d = ?

Vf = ?

Now,

d = Vᵢ × t + 0.5 × a × t²

d = (0 m/s) × (2.60 s) + 0.5 × (-9.8 m/s²) × (2.60 s)²

d = -33.1 m (- indicates direction)

Vf = Vᵢ + a × t

Vf = 0 + (-9.8 m/s²) × (2.60 s)

Vf = -25.5 m/s (- indicates direction)

Thus, d = -33.1 m and Vf = -25.5 m/s

-TheUnknownScientist 72

Answer:

0.067 births per chipmunk per year

Explanation:

I took the test