As you get ready for bed, you roll up one of your socks into a tight ball and toss it into the laundry basket across the room. Then, you try to toss the other sock without rolling it up.. What effects whether or not your socks land in the basket?

Answer:

I dont know bro

Explanation:

Ask an expert

Answer:

Time=90s

Explanation:

Speed=distance /time

[tex]60 = \frac{5400}{t} where \: t \: is \: time \\60t = 5400 \\ t = \frac{5400}{60} \\ t =90 \\ hope \: this \: helps..good \: luck [/tex]

Answer:

1.25kg

Explanation:

Simply multiply volume and density together

Answer:

Tilted forward to keep the driest.

Explanation:

The rain is falling vertically so there is no wind. In these circumstances the umbrella should be tilted vertically forward.

The situation is the same as if you would stand still and the rain would come under an angle from the front.

The dragster's velocity v at time t with constant acceleration a is

[tex]v=at[/tex]

since it starts at rest.

After 2.1 s, it will attain a velocity of

[tex]v=\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)[/tex]

or 92.4 m/s.

Doubling the time would double the final velocity,

[tex]v=a(2t)=2at[/tex]

so the velocity would be twice the previous one, 184.8 m/s.

The dragster undergoes a displacement x after time t with acceleration a of

[tex]x=\dfrac12at^2[/tex]

if we take the starting line to be the origin.

After 2.1 s, it will have moved

[tex]x=\dfrac12\left(44.0\dfrac{\rm m}{\mathrm s^2}\right)(2.1\,\mathrm s)^2[/tex]

or 88 m.

Doubling the time has the effect of quadrupling the displacement, since

[tex]x=\dfrac12a(2t)^2=4\left(\dfrac12at^2\right)[/tex]

so after 4.2 s it will have moved 352 m.

Answer:

A) E = 278925.62 N/C with direction; radially out.

B) E = 43048.47 N/C with direction radially out.

C) E = -3214.29 N/C with direction radially in.

Explanation:

From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;

E = kQ/r²

where;

Q is the net charge within the distance r.

We are given the charge Q = 15-nC and

spherical shell of radius 10cm

A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²

E = 278925.62 N/C

This will be radially out ,since the net charge is positive.

B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²

E = 43048.47 N/C

This will be radially out ,since the net charge is positive.

C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Q = 15 nC - 22 nC

Q = -7 nC = -7 x 10^(-9) C

and;

E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²

E = -3214.29 N/C

This will be radially in, since the net charge is negative. You can indicate this with a negative answer.

A) When The distance r is = 2.2 cm = 0.022 m is between the surface and also the point charge, also that so only the point charge lies within this distance and also Q = 15 NC = 15 x 10^(-9) C

B) When The distance r = 5.6 cm = 0.056 m is between the surface and also the point charge, so only the point charge lies within this distance and also Q = 15 nC = 15 x 10^(-9) C

C) Then when The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

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Answer:

The correct answer is a

Explanation:

The cosine function is

      cos θ = ca / ​​H

done ca is the adjacent leg (x-axis) and H is the hypotenuse (vector module)

we clear

    H = ca / ​​cos θ

therefore, to find the magnitude of the vector, the cathete is divided into the cosine.

The correct answer is a

Complete Question

The complete question is shown on the first uploaded image

Answer:

The distance is [tex]r_2 = 4 \ AU[/tex]

Explanation:

From the question we are told that

    The size of Jupiter is  [tex]s_2 = 143,000 \ km[/tex]

    The  length of the International Space Station is [tex]r_1 = 380\ km[/tex]

    The  size of the International Space Station is  [tex]s_1 = 90 \ m =0.09 \ km[/tex]

The angular size where the same one night and this angular size is mathematically represented as

      [tex]\theta = \frac{s}{r}[/tex]

Since  [tex]\theta[/tex] is constant

        [tex]\frac{s_1}{r_1} = \frac{s_2}{r_2}[/tex]

substituting values

      [tex]\frac{0.09}{380} = \frac{143000}{r_2}[/tex]

=>   [tex]r_2 = 6.04 * 10^{9} \ km[/tex]

Now we are told to convert to AU and  1 AU  [tex]= 1.5 * 10^8 \ km[/tex]

  So

      [tex]r_2 = \frac{6.04 * 10^8}{1.5*10^{8}}[/tex]

      [tex]r_2 = 4 \ AU[/tex]

Answer:

The correct answer is option (c) a success because it did not detect a shift in the interference pattern.

Explanation:

In Michelson and Morley experiment  it was considered to be successful.

They both found out that the experiment that was carried out was not a failure  since it did not detect any shift in the interference pattern.

With this findings it was widely regarded as correct and precise.

Oooooo, that statement is not true.  Countries create quotas and tariffs to LIMIT the volume of trade with other countries, including their neighbors.

Answer:

False

Explanation:

I took the text :)

Answer:

Work-done in quarter an hour = 3.5 × 10⁶ J

Explanation:

Given:

Force (F) = 280 KN = 280,000 N

Velocity (V) = 50 km / h

Time (t) = 1 / 4 = 0.25 hour

Find:

Work-done in quarter an hour

Computation:

⇒ Displacement = Velocity (V) × Time

⇒ Displacement = 50 × 0.25

⇒ Displacement = 12.5 km

⇒ Work-done = Force (F) × Displacement

⇒ Work-done in quarter an hour = 280,000 × 12.5

⇒ Work-done in quarter an hour = 3,500,000

Work-done in quarter an hour = 3.5 × 10⁶ J

Answer:

At thestratosphere: it 20- 25km

Answer:

Vbg = 20.5 m/s

your velocity with respect to the ground Vbg = 20.5 m/s

Explanation:

Relative velocity with respect to the ground is;

Vbg = velocity of train with respect to the ground + your velocity with respect to the train

Vbg = Vtg + Vbt ......1

Given;

velocity of train with respect to the ground;

Vtg = 19 m/s

your velocity with respect to the train;

Vbt = 1.5 m/s

Substituting the given values into the equation 1;

Vbg = 19 m/s + 1.5 m/s

Vbg = 20.5 m/s

your velocity with respect to the ground Vbg = 20.5 m/s

Answer:

a) [tex]F = 210.803\,N[/tex], b) [tex]W_{F} = 779.971\,J[/tex], c) [tex]W_{f} = 235.683\,J[/tex], d) [tex]W_{N} = 0\,J[/tex]; [tex]W_{g} = 544.289\,J[/tex], e) [tex]W_{net} = 0\,J[/tex]

Explanation:

a) The net force exerted on the crate is:

[tex]\Sigma F = F - m\cdot g \cdot \sin \theta - \mu_{k}\cdot m\cdot g \cdot \cos \theta = 0[/tex]

The magnitud of the force that the work must apply to the crate is:

[tex]F = m\cdot g \cdot \sin \theta + \mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]

[tex]F = (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \sin 30^{\circ} + 0.25 \cdot (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]

[tex]F = 210.803\,N[/tex]

b) The work done on the crate due to the external force is:

[tex]W_{F} = (210.803\,N)\cdot (3.7\,m)[/tex]

[tex]W_{F} = 779.971\,J[/tex]

c) The work done on the crate due to the external force is:

[tex]W_{f} = (63.698\,N)\cdot (3.7\,m)[/tex]

[tex]W_{f} = 235.683\,J[/tex]

d) The work done on the crate due the normal force is zero, since such force is perpendicular to the motion direction.

[tex]W_{N} = 0\,J[/tex]

And, the work done by gravity is:

[tex]W_{g} = (147.105\,N)\cdot (3.7\,m)[/tex]

[tex]W_{g} = 544.289\,J[/tex]

e) Lastly, the total work done is:

[tex]W_{net} = W_{F} - W_{f} - W_{g} - W_{N}[/tex]

[tex]W_{net} = 779.971\,J - 235.683\,J - 0\,N - 544.289\,J[/tex]

[tex]W_{net} = 0\,J[/tex]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

[tex]W = F\times d[/tex]

[tex]KE = 0.5\times m\times v^2[/tex]

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

[tex]W = F \times d = 2.62 N \times 100 m[/tex]

[tex]W = 261.6 N\times m[/tex]

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

[tex]0.5 m\times v2^2 = 0.5 m\ v1^2 - W[/tex]

Now solve for v2

[tex]v2 = \sqrt{v1^2 - {\frac{2W}{M}}}[/tex]

[tex]= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}[/tex]

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

[tex]Ff = Us\times N[/tex]

Now solve for Us

[tex]= \frac{Ff}{N}[/tex]

[tex]= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}[/tex]

= 0.00494

Answer:

v = 19.2 m/s

Explanation:

In order to find the speed of the string you use the following formula:

[tex]f=\frac{v}{2L}[/tex]          (1)

f: frequency of the string = 80.0Hz

v: speed of the wave = ?

L: length of the string = 12.0cm = 0.12m

The length of the string coincides with the wavelength of the wave for the fundamental mode.

Then, you solve for v in the equation (1), and replace the values of the other parameters:

[tex]v=2Lf=2(0.12m)(80.0Hz)=19.2\frac{m}{s}[/tex]

The speed of the wave is 19.2 m/s

Complete Question

The complete question is shown on the first uploaded image

Answer:

The mass is    [tex]M =1.43 *10^{32} \ kg[/tex]

Explanation:

From the  question we are told that

       The mass of the stars are [tex]m_1 = m_2 =M[/tex]

        The orbital speed of each star is  [tex]v_s = 240 \ km/s =240000 \ m/s[/tex]

         The orbital period is [tex]T = 12.5 \ days = 12.5 * 2 4 * 60 *60 = 1080000\ s[/tex]

The centripetal force acting on these stars is mathematically represented as

      [tex]F_c = \frac{Mv^2}{r}[/tex]

The gravitational force acting on these stars is mathematically represented as

      [tex]F_g = \frac{GM^2 }{d^2}[/tex]

So  [tex]F_c = F_g[/tex]

=>        [tex]\frac{mv^2}{r} = \frac{Gm_1 * m_2 }{d^2}[/tex]

=>      [tex]\frac{v^2}{r} = \frac{GM}{(2r)^2}[/tex]

=>      [tex]\frac{v^2}{r} = \frac{GM}{4r^2}[/tex]

=>    [tex]M = \frac{v^2*4r}{G}[/tex]

The distance traveled by each sun in one cycle is mathematically represented as

     [tex]D = v * T[/tex]

      [tex]D = 240000 * 1080000[/tex]

      [tex]D = 2.592*10^{11} \ m[/tex]

Now this can also be represented as

      [tex]D = 2 \pi r[/tex]

Therefore

                  [tex]2 \pi r= 2.592*10^{11} \ m[/tex]

=>   [tex]r= \frac{2.592*10^{11}}{2 \pi }[/tex]

=>    [tex]r= 4.124 *10^{10} \ m[/tex]

So  

       [tex]M = \frac{v^2*4r}{G}[/tex]

=>    [tex]M = \frac{(240000)^2*4*(4.124*10^{10})}{6.67*10^{-11}}[/tex]

=>    [tex]M =1.43 *10^{32} \ kg[/tex]

Answer:

The initial angular speed is [tex]w_i = 48 \ rad/s[/tex]

Explanation:

From the question we are told that

   The angular displacement is  [tex]\theta = 220 \ rad[/tex]

    The angular speed is  [tex]w_f = 92 \ rad/s[/tex]

    The acceleration is  [tex]\alpha = 14 \ rad/s^2[/tex]

Generally the initial angular speed can be evaluated as

     [tex]w_f ^2 = w_i ^2 + 2 * \alpha * \theta[/tex]

=>  [tex]w_i ^2 = w_f ^2 - 2 * \alpha * \theta[/tex]

substituting values

=>     [tex]w_i ^2 = 92 ^2 - 2 * 14 * 220[/tex]

=>      [tex]w_i ^2 = 2304[/tex]

=>     [tex]w_i = 48 \ rad/s[/tex]

Answer:

  More than 48%

Explanation:

If the interest is computed monthly on the outstanding balance, it has an effective annual rate of ...

  (1 +4%)^12 -1 = 60.1% . . . .  more than 48%

The effective annual or yearly interest rate would be=30.56% which is Between 4% and 48%

The formula used to calculate annual Interest rate =

[tex](1+ \frac{i}{n} ) {}^{n} - 1[/tex]

where i= nominal interest rate = 4%

n= number of periods= 12 months

Annual Interest rate=

[tex](1 + \frac{4\%}{12} ) {}^{12} - 1[/tex]

= (1+0.333)^12 -1

= (1.333)^12-1

= 31.56 - 1

= 30.56%

Therefore, the effective annual or yearly interest rate would be= 30.56%

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Answer:

0.173 N.

Explanation:

We will calculate the mass and then use the following calculations on the surface of planet X that is :

                           [tex]W=mg[/tex]

We would use the following equation to get the value of g for planet X that is :

                   [tex]y_f-y_i=v_{yi}t+\frac{1}{2}gt^2[/tex]

Then, put the values in the above equation.

                          [tex]16=0+\frac{1}{2}\times g\times(2.90)^2[/tex]

                           [tex]\bf\mathit{g=3.80\;m/s^2}[/tex]

Now, we will measure the ball weight on planet X's surface:

                          [tex]m=\frac{100}{1000} \;\;\;\;\;\;\;\;\;\;[1kg=1000g][/tex]

Then, we have to put the value in the above equation.

                        [tex]W=0.1\times 1.73=0.173\:N[/tex]

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

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

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

[tex]y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m[/tex]

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

[tex]t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s[/tex]

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}[/tex]

Answer:

A.  xmax = 131.49 m

B.  t = 8.74 s

C.  ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

[tex]y=y_o+v_osin\theta-\frac{1}{2}gt^2[/tex]      (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

[tex]0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2[/tex] (2)

You use the quadratic formula to obtain the value of t:

[tex]t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s[/tex]

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

[tex]x_{max}=v_ocos\theta t[/tex]      

[tex]x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m[/tex]

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

[tex]y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}[/tex]     (3)

By replacing in the equation (3) the values of all parameters you obtain:

[tex]y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m[/tex]

The maximum height reached by the cannon ball is 220.33m

Answer:

About 2104m/s

Explanation:

[tex]F=ma \\\\F=\dfrac{2660kg}{2}\cdot 1614m/s=2,146,620N \\\\2,146,620N=1,269kg\cdot a \\\\a\approx 1691m/s \\\\v_f=v_o+at=413m/s+1691m/s=2104m/s[/tex]

Hope this helps!

_____________________________

Solution,

Work=50 Joule

Time=10 seconds

Power=?

Now,

Power=Work/time

= 50/10

= 5 Watt.

So the right answer is 5 W

Hope it helps..

Good luck on your assignment

__________________________

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

Answer:

Energy Stored = 36000 J = 36 KJ

Explanation:

The power of a battery is given by the formula:

P = IV

where,

P = Power delivered by the battery

I = Current Supplied to the battery

V = Potential Difference between terminals of battery = 12 volt

Now, we multiply both sides by the time period (t):

Pt = VIt

where,

Pt = (Power)(Time) = Energy Stored = E = ?

It = Battery Current Rating = 50 A.min

Converting this to A.sec;

It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec

Therefore,

E = (12 volt)(3000 A.sec)

E = 36000 J = 36 KJ

Answer:

I can help! What level of physics is it and what are your main topics?

Answer:

Explanation:

Time taken to accelerate to 28 m /s

= 28 / 2 = 14 s

a ) Total length of time in motion

= 14 + 41 + 5

= 60 s .

b )

Distance covered while accelerating

s = ut + 1/2 at²

= 0 + .5 x 2 x 14²

= 196 m .

Distance covered while moving in uniform motion

= 28 x 41

= 1148 m

distance covered while decelerating

v = u - at

0 = 28 - a x 5

a = 5.6 m / s²

v² = u² - 2 a s

0 = 28² - 2 x 5.6 x s

s = 28² / 2 x 5.6

= 70 m .

Total distance covered

= 196 + 1148 + 70

= 1414 m

total time taken = 60 s

average velocity

= 1414 / 60

= 23.56 m /s .

Answer:

ΔL = 1.011 mm

Explanation:

Let's begin by listing out the given information:

Length (L) = 600 mm = 0.6 m,

Diameter (D) = 40 mm = 0.04 m ⇒ Radius (r) = 20 mm = 0.2 m,

Area (cross sectional) = πr² = 3.14 x .02² = 0.001256 m²,

Modulus of Elasticity (E) = 85 GN/m²,

Compressive load (F) = 180 KN

Using the formula, Stress = Load ÷ Area

Mathematically,

σ = F ÷ A = 180 x 10³ ÷ 0.001256

σ = 143312.1 KN/m²

Modulus of elasticity = stress ÷ strain

E = σ ÷ ε

ε = ΔL/L

85 x 10⁹ = 143312.1 x 10³ ÷ (ΔL/L)

ΔL = 143312.1 x 10³ ÷ 85 X 10⁹ = 1686.02 * 10⁻⁶

ΔL = L x 1686.02 * 10⁻⁶

ΔL = 0.6 * 1686.02 * 10⁻⁶ = 1011.61 x 10⁻⁶

ΔL = 1.011 x 10⁻³ m

ΔL = 1.011 mm

∴The bar contracts by 1.011 mm