A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to rotate around it. The box is 60 cmcm wide and 70 cmcm high. What minimum magnitude of the force F is needed to cause this movement

Answer:

Newton's first law of motion concerns any object that has no force applied to it.

Explanation:

three laws of motion and the law of universal gravitation.

Answer:

the mass of the object multiplied by the acceleration of the object

Explanation:

N2L states that F = ma (force equals mass times acceleration).

Answer:

Explanation:

mass per unit length ρ = .100 / 1.65 = .0606 . kg /m

length of wire L = 1.65 m

For fundamental frequency , the expression is as follows

n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]

L = 1.65 , T = 16 n and m = .0606

n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]

= 4.9 /s .

This is fundamental frequency .

other mode of vibration ( first three ) will be as follows

4.9 x 2 = 9.8 /s ,

4.9 x 3 = 14.7 /s .

Answer:

The rocket will appear larger than it actually is

Answer:

a) v = 1.1 m/s

b) A = 0.315 m

c) v = 1.1 m/s  A= 0.15 m

Explanation:

a)

        [tex]v = \lambda * f (1)[/tex]

       [tex]f = \frac{1}{T} = \frac{1}{5.20s} = 0.19 Hz (3)[/tex]

      [tex]v = \lambda * f = 5.70 m * 0.19 Hz = 1.10 m/s (4)[/tex]

b)

      [tex]A = \frac{0.630m}{2} = 0.315 m (5)[/tex]

c)  

       v = 1.10 m/s (6)

       [tex]A = \frac{0.30m}{2} = 0.15 m (7)[/tex]

Answer:

25N/s or 25kg*m/s^2

Explanation:

It is written wrong because the unit of momentum is kgm/s^2 or N/s

Answer:

[tex]5.76\times 10^{-18}\ \text{N}[/tex] perpendicular to the velocity and magnetic field

Explanation:

B = Magnetic field = [tex]4\times 10^{-5}\ \text{T}[/tex]

[tex]\theta[/tex] = Angle the magnetic field makes with the horizontal = [tex]30^{\circ}[/tex]

v = Velocity of electron = [tex]9\times 10^5\ \text{m/s}[/tex]

q = Charge of electron = [tex]1.6\times 10^{-19}\ \text{C}[/tex]

Magnetic force is given by

[tex]F=qvB\sin\theta\\\Rightarrow F=1.6\times 10^{-19}\times 9\times 10^5\times 4\times 10^{-5}\sin30^{\circ}\\\Rightarrow F=2.88\times 10^{-18}\ \text{N}[/tex]

The magnitude of the magnetic force is [tex]2.88\times 10^{-18}\ \text{N}[/tex] and the direction is perpendicular to the velocity and magnetic field.

Answer:

I think c I dont know sorry if I'm wrong

Answer: 5 km per hour

Explanation:

if in 10 km there is 2 hours, then 10 divided by 2 is 5.

Explanation:

it's true

Kiss me if I'm wrong. But dinosaurs still exist right?

Answer:

True

Definetely true

Answer:

The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".

Explanation:

The given values are:

Potential difference,

Δv = 400 v

Radius,

r = 0.5580 cm

As we know,

⇒  [tex]W=e \Delta v[/tex]

and,

⇒  [tex]\frac{1}{2}mv^2=e \Delta v[/tex]

then,

⇒  [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]

On substituting the values, we get

⇒     [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]

⇒     [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]

⇒     [tex]=1.185\times 10^7 \ m/sec[/tex]

Answer: i believe  its B Find the velocity of each block after they have moved apart sorry

Explanation: have a nice day buddy

Answer:

temperature

Explanation:

if you look at a hertzsprung-russel diagram. you can understand how I got that answer

Answer:

3.35

Explanation:

Got it on Acellus

The light of wavelength 485 nm passes through a single slit. The single between the first (m=1) and second (m=2) interference minima is 3.36°.

Diffraction is the phenomenon of bending of waves through obstacles.  

Given is the wavelength λ= 485 nm, silt width d =  8.32 *10⁻⁶ m, then the angle θ will be

d sinθ =mλ

for m=1, sin θ₁ = λ/d

for m=2, sin θ₂ = 2λ/d

Substitute the values into both expressions to find the angles,

sin θ₁ = 485 x 10⁻⁹ /  8.32 *10⁻⁶

θ₁ = 3.34°

and sin θ₂ = (2 x  485 x 10⁻⁹ )/  8.32 *10⁻⁶

θ₂ = 6.7°

The angle between m =1 and m=2 will be

θ₂ -θ₁ = 6.7° - 3.34° =3.36°

Thus, angle between the first (m=1) and second (m=2) interference minima is 3.36°.

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A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall, the maximum speed of the block when the spring is compressed to one-half of the maximum distance is 4.33 m/s

From the conservation of energy; the kinetic energy of the mass is equal to the work done on the spring.

i.e.

[tex]\mathbf{\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2_{max}}[/tex]

Given that:

∴

From the equation above, multiplying both sides with 2, we have:

[tex]\mathbf{mv^2 =kx^2_{max}}[/tex]

Making (k) the subject of the formula;

[tex]\mathbf{k = \dfrac{mv^2}{x^2_{max}}}[/tex]

[tex]\mathbf{k = \dfrac{4 \times 5^2}{0.68^2}}[/tex]

k = 216.26 N/m

However, when compressed to one-half of the maximum distance; the speed is computed as follows:

x = 0.68/2 = 0.34 m

∴

[tex]\mathbf{\dfrac{1}{2}mv_o^2 - \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2}[/tex]

[tex]\mathbf{m(v_o^2 -v^2) =kx^2}[/tex]

[tex]\mathbf{(v_o^2 -v^2) =\dfrac{kx^2}{m}}[/tex]

[tex]\mathbf{(5^2 -v^2) =\dfrac{216.26 \times 0.34^2}{4.0}}[/tex]

25 - v² = 6.25

25 -6.25 = v²

v² = 18.75

[tex]\mathbf{ v= \sqrt{18.75 }}[/tex]

v = 4.33 m/s

Therefore, we can conclude that the speed of the block when the spring is compressed to only one-half of the maximum distance is 4.33 m/s

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Answer:

c. Double Replacement

Explanation:

As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.

Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2

and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.

Answer:

I don't know how to do it the subject

The centripetal acceleration of the car when there is the radius and the speed is given so it should be considered as the option b. 36 [tex]m/s^2[/tex].

Since

The radius is 25 cm and the speed should be 30 m/s

Now the following formula should be used

Acceleration = speed^2/ radius

= 30^2 / 25

= 900 / 25

= 36 [tex]m/s^2[/tex]

Therefore, The centripetal acceleration of the car should be considered as the option b. 36 [tex]m/s^2[/tex].

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In a ocean full of storms

A new wave was born

Deep into that darkness flooding

Suddenly, I heard some pummeling

By the grave I saw the winds

And the sun just shined

Answer:

a friendly face that comes with waves,

the waves of all the memorial days,

and with these days we smile with pride,

as for the waves we used to ride,

given up the day has passed,

how it went away like an hour glass,

as if we knew the world was right,

just like the waves, oh so bright,

the time has come the days have passed,

the waves ashore the waves alast,

as if the friendly face was right,

the waves that rode, oh goodnight.

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A  = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]

Therefore, the length of the wire is 134.62 m.

Answer:

I'd say D

Explanation:

because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)

Answera black cube or dark colors cause dark colors suck in heat

Answer:

T = 2.5 s

Explanation:

Given that,

Number of complete orbits = 10

Time, t = 25 seconds

We need to find the period of the orbiting ball. Let it is T. We know that number of oscillations per unit time is called frequency and the reciprocal of frequency is called period of the ball.

So,

[tex]T=\dfrac{t}{n}\\\\T=\dfrac{25}{10}\\\\T=2.5\ s[/tex]

So, the period of the orbiting ball is equal to 2.5 seconds.

Answer:

  x = 0.455 L

Explanation:

For this exercise we must use the rotational equilibrium condition

        Σ τ = 0

it has two forces, the first is perpendicular to the rod, so its stub is

         τ₁ = F₁ L

the second force is applied with an angle, so we can use trigonometry to find its components

          sin θ = F_parallel / F₂

          cos θ = F_perpendicular / F₂

         F_parallel = F₂ sin θ

         F _perpendicular = F₂ cos θ

torque is

         τ₂ = F_perpendicular x + F_parallel 0

the parallel force is on the rod therefore its distance is zero

we apply the equilibrium equation

          τ₁  - τ₂ = 0

          F₁ L = F₂ cos θ  x

          x = [tex]\frac{L}{cos \theta} \ \frac{F_1}{F_2}[/tex]

let's calculate

          x = [tex]\frac{L}{cos \ 42.9} \ \frac{2.00}{6.00}[/tex]

          x = 0.455 L

The tension in the strings are 31.47 and 19.25 N respectively.

Mass of the block, m = 3 kg

From the figure, consider the vertical components,

T₁ sin45° + T₂ sin30° = mg

(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4

Also, consider the horizontal components,

T₁ cos45° = T₂ cos30°

T₁/√2 = T₂ x√3/2

T₁ = T₂ x √3/2 x √2

So,

T₁ = 0.612T₂

Applying in the first equation,

(T₁/√2) + (T₂/2) = 29.4

(0.612T₂/1.414) + 0.5T₂ = 29.4

0.434 T₂ + 0.5 T₂ = 29.4

0.934 T₂ = 29.4

Therefore, the tension,

T₂ = 29.4/0.934

T₂ = 31.47 N

So, the tension,

T₁ = 0.612 T₂

T₁ = 0.612 x 31.47

T₁ = 19.25 N

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Answer:

Explanation:

proton

Explanation:

As pressure increases, with temperature constant, density increases. Conversely when temperature increases, with pressure constant, density decreases. Air density will decrease by about 1% for a decrease of 10 hPa in pressure or 3 °C increase in temperature.

Answer:

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Explanation:

Answer: D) All of the above

Explanation: