A box of mass 7.0 kg is accelerated from rest across a floor at a rate of 2.0 m/s2 for 9.0 s .Find the net work done on the box. Express your answer to two significant figures and include the appropriate units.

Answer:

C

Explanation:

Current passing through the loop in either direction

Given :

Current, I = 3.75 A .

Magnetic Field, [tex]B = 2.61\times 10^{-4}\ T[/tex]

To Find :

The distance from the wire.

Solution :

We know,

[tex]B = K\dfrac{2i}{d}\\\\d = 10^{-7}\times \dfrac{2\times 3.75}{2.61\times 10^{-4}}\\\\d = 0.00287\ m \\\\d = 2.87\times 10^{-3}\ m[/tex]

Hence, this is the required solution.

Answer:

I thinks it's

deficit spending

Explanation:

cause When a government spends more than it collects in taxes, it is said to have a budget deficit.

Explanation:

There are still some questions beyond the Standard Model of physics, such as the strong CP problem, neutrino mass, matter–antimatter asymmetry, and the nature of dark matter and dark energy.

Answer:

0.7 m/[tex]s^{2}[/tex]

Explanation:

From Newton's law of universal gravitation,

F = [tex]\frac{GMm}{r^{2} }[/tex]

and from Newton's second law of motion,

F = mg

So that;

mg = [tex]\frac{GMm}{r^{2} }[/tex]

⇒ g = [tex]\frac{GM}{r^{2} }[/tex]

For the first planet,

7 = [tex]\frac{GM}{R^{2} }[/tex]

⇒ G = [tex]\frac{7R^{2} }{M}[/tex] .............. 1

For the second planet,

g = [tex]\frac{0.4GM}{(2R)^{2} }[/tex]

   = [tex]\frac{0.4GM}{4R^{2} }[/tex]

⇒ G = [tex]\frac{4gR^{2} }{0.4M}[/tex] ............. 2

Equating 1 and 2, we have;

[tex]\frac{7R^{2} }{M}[/tex] = [tex]\frac{4gR^{2} }{0.4M}[/tex]

g = [tex]\frac{7R^{2} *0.4M}{4R^{2}M }[/tex]

  = [tex]\frac{7*0.4}{4}[/tex]

  = [tex]\frac{2.8}{4}[/tex]

g = 0.7

Therefore, the acceleration due to gravity on the new planet is 0.7 m/[tex]s^{2}[/tex].

Answer:

The range of the projectile is 60 m

Explanation:

Horizontal Motion

When an object is thrown horizontally with a speed vo from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

[tex]v_x=v_o[/tex]

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

[tex]v_y=g.t[/tex]

The horizontal distance is calculated as a constant speed motion:

[tex]x = v_x.t[/tex]

Knowing the crossbow is fired horizontally at vo=vx=15 m/s and it takes t=4 s to hit the ground, thus the range of the projectile is:

x = 15*4 = 60

The range of the projectile is 60 m

Answer:

yes

Explanation:

Explanation:

The electromagnetic spectrum is the name given to the full range of frequencies and/or wavelengths that electromagnetic phenomena may have.

Human eyes respond to a small range of wavelengths in that spectrum. That response is called sight. Because humans can see that electromagnetic energy, it is called visible light.

9ycy8c8t 7f fixfuozofuxt8lsrupsurpaurae6pUeoUe6eoUeFipzuroz6d0, 7d0z6e0z7e0zurpz6e0z

Explanation:

force newton N - m·kg·s-2

pressure, stress pascal Pa N/m2 m-1·kg·s-2

energy, work, quantity of heat joule J N·m m2·kg·s-2

power, radiant flux watt W J/s m2·kg·s-3

volume cubic meter m3

Answer:

hsvshxansjusjsnwjwisks

Explanation:

When we stir a cup of tea we create a force in the center which pulls out all the particles towards it this is the basic reason for collection of tea leaves at the center of the cup rather than at the rim of the cup, it is similar to the the case of tornado where it takes all the particles present on it way to its ..

This question is incomplete, the complete question is;

A 8.45μC particle with a mass of 6.15 x 10⁻⁵ kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m.

How much time will it take for the particle to complete one orbit?

a. 92.7 s

b. 0.0927 s

c. 9.27 s

d. 927 s

Answer:

it will take 92.7 seconds for the particle to complete one orbit.

Option a) 92.7 s is the correct option

Explanation:

Given that;

mass m =  6.15 x 10⁻⁵ kg

q = 8.45μC = 8.45 × 10⁻⁶ C

B = 0.493

we know that

Time period T = 2πr / V

where r = mv/qB

so T = 2πm/qB

we substitute

T = (2 × 3.14 × 6.15 x 10⁻⁵) / ( 8.45 × 10⁻⁶ × 0.493)

T = 0.0003862 / 0.000004165

T = 92.7 sec

Therefore it will take 92.7 seconds for the particle to complete one orbit.

Option a) 92.7 s is the correct option

Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

       [tex]a_{c} = \omega^{2} * r (1)[/tex]

       ∴ ωp = ωw (2)

           ⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]

               [tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]

        [tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]

        [tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]

Answer:

Remote Sensing

Explanation:

Answer:

Select the second and the third options you listed.

Explanation:

Select the answer options:

"is equal to the force with which Dennis exerts on the  box."

and

"has an upper limit and Dennis has not yet exceeded the upper limit."

In fact, this upper limit of the static friction force is the product of the coefficient of static friction ([tex]\mu[/tex]) times the weight of the box.

Answer:

q = 5 C

Explanation:

The electric field is defined as the force experienced by a unit charge when it is brought into the field. Hence, the formula used to find the electrical field is given as follows:

E = F/q

where,

E = Electric Field Magnitude = 10 N/C

F = Force Experienced by the test charge = 50 N

q = Magnitude of the Charge = ?

Therefore,

10 N/C = 50 N/q

q = 50 N/(10 N/C)

Therefore,

q = 5 C

Answer:

268.22m/s

Explanation:

Given;

    10mile/min to m/s

We need to convert between the two units;

    1 mile  = 1609.34m

     60s  = 1min

Now;

    10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]

  = 268.22m/s

Answer:

Intensity at 12 meters will be 11.11 W/m^2

Explanation:

Recall that the intensity of sound is inversely proportional to the square of the distance from the source. Therefore, if at 4 m the intensity is 100 W/m^2

we have: 100 W/m^2 = k/16 and therefore, k = 1600 W

Then the intensity (I) at 12 m will be:

I = k/12^2 = 1600/144  W/m^2 = 11.11 W/m^2

Answer:

A

Explanation:

I'm pretty sure it's A. That's the one that makes the most sense and checks out

Answer:

Just here to confirm that it is A

Explanation:

Answer:

m = 4.9 10⁸ kg

Explanation:

The expression for the density is

           ρ = m / V

           m = ρ V

the volume of the atmosphere is the volume of the sphere of the outer layer of the atmosphere minus the volume of the plant

          V = V_atmosphere - V_planet

           V = 4/3 π R_atmosphere³ - 4/3 π R_venus³

           V = 4/3 π (R_atmosphere³ - R_venus³

)

the radius of the planet is R_venus = 6.06 10⁶ m.

The radius of the outermost layer of the atmosphere

          R_atmosphere = 50 10³ + R_ venus = 50 10³ + 6.06 10⁶

           R_atmosphere = 6.11 10⁶ m

let's find the volume

           V = 4/3 pi [(6,11 10⁶)³ - (6,06 10⁶)³]

            V = 23,265 10⁶ m³

let's calculate the mass

          m = 21  23,265 10⁶

          m = 4.89   10⁸ kg

with two significant figurars is

          m = 4.9 10⁸ kg

Answer:

T = 3990 N

Explanation:

The free body diagram for the elevator consists of a tension force pointing up, and its weight pointing down. So the elevator's net force is:

F = T - 2940N

ad at the same time, using Newton's second law, we have that this net force should equal the elevator's mass (300 kg) times its acceleration (a):

T - 2940N = 300kg (3.5m/s^2)

then

T = 2940 N + 1050 N

T = 3990 N

Answer:

A) r = 0.03 m

B) r = 0.0533 m

C) B_max = 0.00003 T

Explanation:

Formula for magnetic field inside the capacitor when it is parallel to the length element is;

B_in = (μ_o•I•r/(2πR²)

Formula for maximum magnetic field is;

B_max = (μ_o•I/(2πR)

Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)

A) Magnetic field inside the capacitor is gotten from our first equation above;

B_in = (μ_o•I•r/R²)

Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.

Thus;

B_in = 0.75B_max

(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))

μ_o•I and 2πR will cancel out to give;

r/R = 0.75

r = 0.75R

We are given R = 40 mm = 0.04 m

r = 0.75 × 0.04

r = 0.03 m

B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)

Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:

B_out = 0.75B_max

(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))

μ_o•I and 2π will cancel out to give;

1/r = 0.75/R

r = R/0.75

r = 0.04/0.75

r = 0.0533 m

C) B_max = μ_o•I/(2πR)

μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A

Thus;

B_max = (4π × 10^(-7) × 6)/(2π × 0.04)

B_max = 0.00003 T

Answer:

449.020m/s

Explanation:

We have been provided with the following information:

Mass of bullet = Mb = 4.2g

Mass of pendulum = 1.0 = Md

Vertical distance = h = 18cm

Gravitational force = g = 9.8

We have kinetic energy converted to potential energy for the entire system

1/2(Mb+Md)V² = (Mb + Md)gh

We have V as the speed of system during collision

Mbv = (Mb+Md)V

We divide through by Mb

v = (Mb+Md/Mb)√2gh

4.2x10^-3 = 0.0042

18cm = 0.18m

(0.0042+1.0/0.0042)√2x9.8x0.18

= 1.0042/0.0042√3.528

= 239.095x1.878

= 449.020m/s

This is the answer to the question

Thank you!

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, it should be noted that when objects of different sizes fall in absence of air resistance, the objects will get to the ground at the same time. But with the presence of air resistance, the heaviest object gets to the ground first; meaning it has the least air resistance while the lightest object will arrive at the ground last because it has the greatest air resistance and is slowed down the most by the air resistance.

Thus, the lightest object in the completed question is the answer.

Answer:

1.08 s

Explanation:

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

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

Answer:

Volume of Cuboid = Height*Width*Length

Explanation:

Volume of Cuboid = 10*5*7

= 350 cu² cm

Answer:

[tex]\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 5cm}\put(7.7,6.3){\sf 7cm}\put(11.3,7.45){\sf 10cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}[/tex]

It is a cuboid

where

As we know that in a cuboid

[tex]{\boxed{\sf Volume=lbh}}[/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf Volume =7×5×10 [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf Volume=35×10 [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf Volume=350cm^3 [/tex]

Answer:

0.87

Explanation:

To solve this, we use the principle of Archimedes. Archimedes Principal of flotation states that "the buoyant force of an object is equal to the total weight of the fluid it displaces."

In the attachment, I stated the mathemacal formula, of which

F(B) = The buoyant force

w(fl) = The weight of the salt water displaced

p(iron) = density of iron

p(salt) = density of the salt water = 1025 kg/m³

F' = weight of the iron in air

F = weight of the iron in salt water

p(man) = density of man = 7680 kg/m³

The rest are the easy calculations done by substituting the values

Given :

Potential difference, V = 300 V.

Magnetic Field, B = 0.001 T.

To Find :

The magnitude of the magnetic force on the electron.

Solution :

We know, for perpendicular orientation, force is given by :

[tex]F = qVB\\\\F = 1.6 \times 10^{-19} \times 300\times 0.001\ N\\\\F = 4.8\times 10^{-20}\ N[/tex]

Hence, this is the required solution.

Answer:

[tex]10.27m/s[/tex]

Explanation:

Given data

work W= -5.10 10^3 J

mass m= 71kg

final height of slide h2= 12.7m

initial height of slide h1=0m

initial velocity v1= 0m/s

final velocity v2=?

Step two:

required

Final velocity

The work-energy theorem is expressed as'

[tex]W=1/mv_2^2 +mgh_2-(1/mv_1^2+mgh_1)[/tex]

make V2 subject of formula we have final speed

[tex]v_2=\sqrt{\frac{2W}{m}+v_1^2-2g(h_1-h_2) } \\\\[/tex]

substitute our given data we have

[tex]v_2=\sqrt{\frac{2*(-5.1*10^3)}{71}+0^2-2*9.81(12.7) } \\\\v_2=\sqrt{143.66-249.174 } \\\\v_2=\sqrt{105.514 } \\\\v_2=10.27m/s[/tex]

The student going at  10.27m/s