What is the magnitude of vector X?​

Answer:

F = 1010 Lb

the tension on the cable is greater than its resistance, which is why the plan is not viable

Explanation:

For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.

          v = v₀ + a t

how the car comes out of rest v₀ = 0

          a = v / t

let's reduce to the english system

          v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s

let's calculate

          a = 66/10

          a = 6.6 ft / s²

now let's write Newton's second law

X axis

         Fₓ = ma

with trigonometry

         cos 20 = Fₓ / F

         Fₓ = F cos 20

we substitute

          F cos 20 = m a

          F = m a / cos20

          W = mg

          F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]

let's calculate

          F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20

          F = 1010 Lb

Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.

Answer:

The total distance at 7 s is:

[tex]x_{tot}=27\: m[/tex]

Explanation:

Distance due to the force

We can use second Newton's law to find the acceleration.

[tex]F=ma[/tex]

[tex]a=\frac{F}{m}=\frac{9}{5}=1.8\: m/s^{2}[/tex]

Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.

[tex]x_{1}=0.5at_{1}^{2}[/tex]

[tex]x_{1}=0.5(1.8)(3)^{2}[/tex]

[tex]x_{1}=8.1\: m[/tex]

In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.

First, we need to find the final velocity of the first interval

[tex]v=v_{i}+at_{1}=0+(1.8)3=5.4\: m/s[/tex]

So the second distance will be:

[tex]x_{2}=vt_{2}=5.4*4=21.6\: m[/tex]

Therefore, the total distance is:

[tex]x_{tot}=x_{1}+x_{2}=5.4+21.6=27\: m[/tex]

I hope it helps you!

The total distance at 7 s is: [tex]x_{total}=27\ m[/tex]

Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.

Distance due to the force

We can use second Newton's law to find the acceleration.

[tex]F=ma[/tex]

[tex]a=\dfrac{F}{m}=\dfrac{9}{5}=1.8\ \frac{m}{s^2}[/tex]

Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.

[tex]x_1=0.5at_1^2[/tex]

[tex]x_1=0.5(1.8)(3^2)[/tex]

[tex]x_1=8.1\ m[/tex]

In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.

First, we need to find the final velocity of the first interval

[tex]v=v_i+at_1=0+(1.8)(3)=5.4 \frac{m}{s}[/tex]

So the second distance will be:

[tex]x_2=vt_2=5.4\times 4=21.6\ m[/tex]

Therefore, the total distance is: [tex]x_{total}=27\ m[/tex]

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Answer:

Emissions per second = 0.36

Explanation:

Please find the attached question

Solution

Given  

Let X be the rate of background emission.

X = B/t  

Where B = 36

And t = 100

X = 36/100 = 0.36

Answer:

No. of Neutrons = 3

Explanation:

The atomic number of Lithium is given as 3 in the symbol while the mass number is given as 5.941 which is approximately equal to 6.

Mass Number = No. of Protons + No. of Neutrons = 6

Atomic Number = Number of Electrons = No. of Protons = 3

Therefore,

Mass Number - Atomic Number = (No. of Protons + No. of Neutrons) - No. of Protons

Mass Number - Atomic Number = No. of Neutrons

No. of Neutrons = 6 - 3

No. of Neutrons = 3

Answer:

3

Explanation:

Honestly, it is the only one in the picture and as an answer.

Answer:

Microeconomics is a part of economics and the study of decisions made by people and businesses regarding the allocation of resources, and prices at which they trade goods and services.

Microeconomics helps business planning i.e. helps the business community to plan their costs, production, etc. in anticipation of demand in order to maximize profits. Microeconomics is useful in explaining how market mechanism determines the price in a free market economy.

Explanation:

The centripetal force [tex]F_c[/tex] on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.

[tex]y:\:\:\:\:N - mg = 0[/tex]

[tex]x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N[/tex]

or

[tex]m \dfrac{v^2}{r} = \mu mg[/tex]

Solving for [tex]\mu[/tex],

[tex]\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28[/tex]

Answer:

dhfhffuththt9tr8tujtngigjtjrjrjrurur

Answer:

[tex]t=1.9 sec[/tex]

Explanation:

From the question we are told that:

Height [tex]h=28m[/tex]

Time [tex]t=3s[/tex]

Generally the Newton's equation for Initial velocity upward is mathematically given by

 [tex]s=ut+\frtac{1}{2}at^2[/tex]

 [tex]28=3u-\frac{1}{2}*9.8*3^2[/tex]

 [tex]u=24.03m/s[/tex]

Generally the velocity at  elevation and depression occurs  as ball arrives and passes through S=28

 [tex]v=\sqrt{24.03-2*9.8*28}[/tex]

 [tex]v=5.35m/s and -5.35m/s[/tex]

Generally the Newton's equation for time to reach initial velocity  is mathematically given by

 [tex]v=u+at[/tex]

 [tex]5.35=24.03-9.8t[/tex]

 [tex]t=\frac{28.03-5.35}{9.8}[/tex]

 [tex]t=1.9 sec[/tex]

Answer:

The energy of a photon is 2.94x10⁻¹⁹ J.

Explanation:

The energy of the photon is given by:

[tex] E = \frac{hc}{\lambda} [/tex]  

Where:

h: is Planck's constant = 6.62x10⁻³⁴ J.s

c: is the speed of light = 3.00x10⁸ m/s

λ: is the wavelength = 675 nm

Hence, the energy is:

[tex] E = \frac{hc}{\lambda} = \frac{6.62 \ccdot 10^{-34} J.s*3.00 \cdot 10^{8} m/s}{675 \cdot 10^{-9} m} = 2.94 \cdot 10^{-19} J [/tex]

Therefore, the energy of a photon is 2.94x10⁻¹⁹ J.

I hope it helps you!

Answer:

a1 =  v1^2 / R

a2 =  v2^2 / R

a2 = (v2 / v1)^2 = (3 / 2)^2 = 9/4

F2 = 9/4 * F1 = 9/4 * 146 = N     329 N      since F = m * a

Answer:

0.913

Explanation:

k.e=1/2mv square

k.e=1/2×0.078g×23.4256m/s square

k.e=0.913J

The kinetic energy when the lemming is 2.00 m above the ground is approximately 2.56 J (Joules).

To calculate the kinetic energy (KE) of the lemming when it is 2.00 m above the ground, we need to consider the change in its potential energy (PE) as it falls.

The potential energy at a height h is given by:

PE = m g h

Where:

m is the mass of the lemming (0.0780 kg)

g is the acceleration due to gravity (9.8 m/s²)

h is the height above the ground

Given:

Height of the cliff (h) = 5.36 m

Velocity of the lemming (v) = 4.84 m/s

Height above the ground (h') = 2.00 m

The lemming will lose potential energy as it falls from the cliff, which is converted into kinetic energy. Therefore, the kinetic energy when it is 2.00 m above the ground is equal to the difference between its total initial kinetic energy and the potential energy at that height.

Initial potential energy at the top of the cliff:

PE_initial = m g h

Potential energy when it is 2.00 m above the ground:

PE_final = m * g * h'

The change in potential energy is given by:

ΔPE = PE_final - PE_initial

The kinetic energy (KE) when it is 2.00 m above the ground:

KE = ΔPE = -ΔPE (due to energy conservation)

Let's calculate the potential energy at the top of the cliff and when it is 2.00 m above the ground:

PE_initial = m ×g × h

= 0.0780 kg × 9.8 m/s² × 5.36 m

PE_initial ≈ 4.09 J

PE_final = m ×g × h'

= 0.0780 kg ×9.8 m/s² ×2.00 m

PE_final ≈ 1.53 J

The change in potential energy (ΔPE) is:

ΔPE = PE_final - PE_initial = 1.53 J - 4.09 J

ΔPE ≈ -2.56 J

Since the change in potential energy is equal to the kinetic energy, the kinetic energy when the lemming is 2.00 m above the ground is approximately 2.56 J (Joules).

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Answer:

[tex]a=2.8*10^{-9}m/s[/tex]

Explanation:

From the question we are told that:

First Mass [tex]m=8.50kg[/tex]

2nd Mass [tex]m=14.5kg[/tex]

Distance

[tex]d_1=50=>0.50m\\\\d_2=>12cm=>0.12m[/tex]

Generally the Newtons equation for Gravitational force is mathematically given by

[tex]F_n=\frac{Gm_nm}{(r_n)^2}[/tex]

Therefore

Initial force on m

[tex]F_1=\frac{Gm_1m}{(r_1)^2}[/tex]

Final force on m

[tex]F_2=\frac{Gm_2m}{(r_2)^2}\\\\F=\frac{Gm_1m}{(r_1)^2}-\frac{Gm_2m}{(r_2)^2}[/tex]

Acceleration of m

[tex]a=\frac{F}{m}\\\\a=\frac{Gm_1}{r_1^2}-\frac{Gm_2}{r_2^2}[/tex]

[tex]a=6,67*10^{-11}{\frac{8.5}{0.12}}-\frac{14.5}{0.50}[/tex]

[tex]a=2.8*10^{-9}m/s[/tex]

Answer: 5.21 s

Explanation:

Given

Helicopter ascends vertically with [tex]u=5.4\ m/s[/tex]

Height of helicopter [tex]h=105\ m[/tex]

When the package leaves the helicopter, it will have the same vertical velocity

Using equation of motion

[tex]\Rightarrow h=ut+\dfrac{1}{2}at^2\\\\\Rightarrow 105=-5.4t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-5.4t-105=0\\\\\Rightarrow t=\dfrac{5.4\pm \sqrt{5.4^2+4\times 4.9\times 105}}{2\times 4.9}\\\\\Rightarrow t=\dfrac{5.4\pm 45.68}{9.8}\\\\\Rightarrow t=5.21\ s\quad \text{Neglect negative value}[/tex]

So, package will take 5.21 s to reach the ground

Answer:

the radio can tune wavelengths between 1.88 and 5.97 m

Explanation:

The signal that can be received is the one that is in resonance as the impedance of the LC circuit.

         X = X_c - X_L

         X = 1 / wC - w L

at the point of resonance the two impedance are equal so their sum is zero

         X_c = X_L

         1 / wC = w L

         w² = 1 / CL

         w = [tex]\sqrt{\frac{1}{CL} }[/tex]

let's look for the extreme values

C = 1  10⁻¹² F

         w = [tex]\sqrt{\frac{1}{ 1 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]

         w = [tex]\sqrt{1 \ 10^{18}}[/tex]

         w = 10⁹ rad / s

C = 10 10⁻¹² F

         w = [tex]\sqrt{\frac{1}{10 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]Ra 1/10 10-12 1 10-6

         w = [tex]\sqrt{0.1 \ 10^{18}}[/tex]Ra 0.1 1018

         w = 0.316 10⁹ rad / s

Now the angular velocity and the frequency are related

           w = 2π f

           f = w / 2π

the light velocity  is

           c = λ f

           λ = c / f

we substitute

          λ = c 2π/w

we calculate the two values

 C = 1 pF

          λ₁ = 3 10⁸ 2π / 10⁹

          λ₁= 18.849 10⁻¹ m

          λ₁ = 1.88 m

C = 10 pF

           λ₂ = 3 10⁸ 2π / 0.316 10⁹

           λ₂ = 59.65 10⁻¹ m

           λ₂ = 5.97 m

so the radio can tune wavelengths between 1.88 and 5.97 m

Answer:

time

Explanation:

Answer:

The mass decreases

Explanation:

Just smart

Answer: The voltage needed is 35.7 V

Explanation:

Assuming that the resistors are arranged in parallel combination.

For the resistors arranged in parallel combination:

[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]

We are given:

[tex]R_1=20\Omega\\R_2=20\Omega\\R_3=200\Omega[/tex]

Using above equation, we get:

[tex]\frac{1}{R}=\frac{1}{20}+\frac{1}{20}+\frac{1}{200}\\\\\frac{1}{R}=\frac{10+10+1}{200}\\\\R=\frac{200}{21}=9.52\Omega[/tex]

Calculating the voltage by using Ohm's law:

[tex]V=IR[/tex]         .....(1)

where,

V = voltage applied

I = Current = 3.75 A

R = Resistance = [tex]9.52\Omega[/tex]

Putting values in equation 1, we get:

[tex]V=3.75\times 9.52\\\\V=35.7V[/tex]

Hence, the voltage needed is 35.7 V

There are six different ways to arrange those resistors.

Each way leads to a different answer.

You didn't tell us how you want them connected.

The attached drawing shows them all.

Explanation:

Given: a = -3v^2

By definition, the acceleration is the time derivative of velocity v:

[tex]a = \frac{dv}{dt} = - 3 {v}^{2} [/tex]

Re-arranging the expression above, we get

[tex] \frac{dv}{ {v}^{2} } = - 3dt[/tex]

Integrating this expression, we get

[tex] \int \frac{dv}{ {v}^{2} } = \int {v}^{ - 2}dv = - 3\int dt[/tex]

[tex] - \frac{1}{v} = - 3t + k[/tex]

Since v = 10 when t = 0, that gives us k = -1/10. The expression for v can then be written as

[tex] - \frac{1}{v} = - 3t - \frac{1}{10} = - ( \frac{30 + 1}{10} )[/tex]

or

[tex]v = \frac{10}{30t +1 } [/tex]

We also know that

[tex]v = \frac{ds}{dt} [/tex]

or

[tex]ds = vdt = \frac{10 \: dt}{30t + 1} [/tex]

We can integrate this to get s:

[tex]s = \int v \: dt = \int ( \frac{10}{30t + 1}) \: dt = 10 \int \frac{dt}{30t + 1} [/tex]

Let u = 30t +1

du = 30dt

so

[tex] \int \frac{dt}{30t + 1} = \frac{1}{30} \int \frac{du}{u} = \frac{1}{30}\ln |u| + k[/tex]

[tex]= \frac{1}{30}\ln |30t + 1| + k[/tex]

So we can now write s as

[tex]s = \frac{1}{3}\ln |30t + 1| + k[/tex]

We know that when t = 0, s = 8 m, therefore k = 8 m.

[tex]s = \frac{1}{3}\ln |30t + 1| + 8[/tex]

Next, we need to find the position and velocity at t = 3 s. At t = 3 s,

[tex]v = \frac{10}{30(3) +1 } = \frac{10}{91}\frac{m}{s} = 0.11 \: \frac{m}{s} [/tex]

[tex]s = \frac{1}{3}\ln |30(3) + 1| + 8 = 9.5 \: m[/tex]

Note: velocity approaches zero as t --> [tex]\infty [/tex]

Answer:

The appropriate answer is "3761.69 N/C".

Explanation:

Given that:

Mass,

m = 4.99 g

or,

   = [tex]4.99\times 10^{-3} \ kg[/tex]

Charge,

q = 13 µC

or,

  = [tex]13\times 10^{-6} \ C[/tex]

As we know,

⇒ [tex]F=mg=Eq[/tex]

then,

⇒ [tex]E=\frac{mg}{q}[/tex]

By putting the values, we get

        [tex]=\frac{4.99\times 10^{-3}\times 9.8}{13\times 10^{-6}}[/tex]

        [tex]=3761.69 \ N/C[/tex]

Answer:

a) increases.

b) remains the same.

c) increases.

d) increases.

e) increases.

f) increases.

Explanation:

a)

       [tex]C = \epsilon_{0}*\frac{A}{d} (1)[/tex]

       where ε₀ = permitivitty of  free space

                   A = area of one of the plates

                   d=  plate separation

b)  

        [tex]C = \frac{Q}{V} (2)[/tex]

c)

d)

       [tex]U = \frac{1}{2}* C*V^{2} (3)[/tex]

e)

       [tex]V = E*d (4)[/tex]

f)

       [tex]u = \frac{U}{v} = \frac{\frac{1}{2}* C*V^{2}}{A*d} =\frac{1}{2}* \epsilon_{0}*\frac{A*V^{2} }{A*d*d} = \frac{1}{2} *\epsilon_{0}*E^{2} (5)[/tex]

Answer:

The angular velocity of the wheels is 276.314 radians per second.

Explanation:

The wheels are rolling, which is a combination of rotation and translation, whose center of rotation is the point of contact between the wheel and the ground and the geometrical center of the skateboard experiments pure translation. Then, the angular velocity can be found by using the following kinematic expression:

[tex]\omega = \frac{v}{R}[/tex] (1)

Where:

[tex]\omega[/tex] - Angular velocity, in radians per second.

[tex]v[/tex] - Velocity of the wheel at its center, in inches per second.

[tex]R[/tex] - Radius of the wheel, in inches.

If we know that [tex]v = 404.8\,\frac{in}{s}[/tex] and [tex]R = 1.465\,in[/tex], then the angular velocity of the wheels are:

[tex]\omega = \frac{404.8\,\frac{in}{s} }{1.465\,in}[/tex]

[tex]\omega = 276.314\,\frac{rad}{s}[/tex]

The angular velocity of the wheels is 276.314 radians per second.

Answer:2

Explanation:

Explanation:

[tex]which \: language \: is \: this[/tex]

[tex]pls \: write \: in \: english[/tex]

[tex]then \: only \: i \: can \: answer \: u[/tex]

[tex]otherwise \: sry[/tex]

Answer:

a. 3.09 m/s^2

b. 322.52 N

Explanation:

The computation of the acceleration of the object and the tension of the string is as follows:

The acceleration of the system is a

= (m1 - m2) × g ÷ (m1 + m2)

= (48 - 25) × 9.81 ÷ (48 + 25)

= 3.09 m/s^2

b. The tension in the string is T

= 2 × m1 × m2 × g÷   (m1+m2)

= (2  × 48  × 25  ×  9.81) ÷  (48 + 25)

= 322.52 N

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

Q =  ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q =  ∫∫0.2r⁴[-cosθ]drdΦ

Q =  ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q =  ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

Answer:

The centripetal force is 0.54 N.

Explanation:

mass, m = 0.56 kg

radius, r = 0.72 m

angular speed, w = 1.155 rad/s

The centripetal force is given by

[tex]F = m r w^2\\\\F =0.56\times 0.72\times 1.155\times 1.155\\\\F = 0.54 N[/tex]

I am guessing wind is caused by climate change in the atmosphere

Explanation:

wind is cause by climate change in the atmosphere that depends weather is is breezy really cold or rain and cold

Answer:

caused by the uneven heating of the Earth by the sun and the  own rotation.

Answer:

[tex]a=1440\ m/s^2[/tex]

Explanation:

Given that,

The angular velocity of a wave, [tex]\omega=12\ rad/s[/tex]

The maximum displacement of the wave, A = 10 cm (let)

The maximum acceleration of the wave is given by :

[tex]a=-A\omega^2[/tex]

Put all the values,

[tex]a=10\times 12^2\\\\a=1440\ m/s^2[/tex]

So, the maximum acceleration of the wave is equal to [tex]1440\ m/s^2[/tex].