Physics question plz help ASAP

2. two dimensional motion

Because it has just 2 dimensions x and y

The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.

Answer:

25 you said ? thats incorecct

Explanation:

Answer:

44 N

Explanation:

Given that,

If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N, then we need to find the force the wall push on you.

It is based on Newton's third law of motion which states that for an action there is an equal and opposite reaction. If the you push the wall with a force of 44 N, the wall push on you is 44 N also as it is based on Newton's third law of motion.

Answer:

The correct answer would be - Low pitch.

Explanation:

As it is known that if frequency increases then pitch will be increase as well as pitch depends on frequency, Now for the question it is mentioned that the tube closed on one end frequency is:

f = v/2l

Where,

l = length of the tube

v = velocity of longitudinal wave of gas filled in the tube

Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases. As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.

Answer:

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Answer:

(a) The spring constant is 59.23 N/m

(b) The total energy involved in the motion is 0.06 J

Explanation:

Given;

mass, m = 240 g = 0.24 kg

frequency, f = 2.5 Hz

amplitude of the oscillation, A = 4.5 cm = 0.045 m

The  angular speed is calculated as;

ω = 2πf

ω = 2 x π x 2.5

ω = 15.71 rad/s

(a) The spring constant is calculated as;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m[/tex]

(b) The total energy involved in the motion;

E = ¹/₂kA²

E = (0.5) x (59.23) x (0.045)²

E = 0.06 J

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Answer:

100Hz

Explanation:

In a full wave rectifier, the fundamental frequency of the ripple is twice that of input frequency. Given the input frequency of 50 Hz, the fundamental frequency will be 2 × 50 = 100Hz

Answer:

HZ 100 is the right answer hope you like it

Complete question is;

One ball is projected in the upward direction with a certain velocity ‘v’ and other is thrown downwards with the same velocity at an angle θ.

The ratio of their potential energies at highest points of their journey, will be:

Answer:

u² : (u cos θ)²

Explanation:

Maximum potential energy for the first ball will be at a maximum height of;

H = u²/2g

Thus;

PE = mg(u²/2g)

For second ball at an angle θ, maximum PE will occur at a max height of (u cos θ)²/2g

PE = mg((u cos θ)²/2g)

The ratios of the potential energies are;

mg(u²/2g) : mg((u cos θ)²/2g)

mg will will cancel out since they are of same mass.

Thus;

(u²/2g) : (u cos θ)²/2g

Again 2g will cancel out to give;

u² : (u cos θ)²

Answer:

An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.

Explanation:

The Earth's axis is represented by the red line. The white circle represents axial precission, the slow "wobble" of the axis.

Answer:  

t = 1.27 x 10⁹ s  

Explanation:  

First, we will find the volume of the wire:

Volume = V = AL  

where,  

A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²  

L = Length of wire = 150 km = 150000 m  

Therefore,    

V = 47.12 m³

Now, we will find the number of electrons in the wire:  

No. of electrons = n = (Electrons per unit Volume)(V)  

n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)  

n = 3.97 x 10³⁰ electrons  

Now, we will use the formula of current to find out the time taken by each electron to cross the wire:

[tex]I =\frac{q}{t}[/tex]  

where,  

t = time = ?  

I = current = 500 A  

q = total charge = (n)(chareg on one electron)  

q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)  

q = 6.36 x 10¹¹ C  

[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]

Therefore,

t = 1.27 x 10⁹ s

Answer:

(a) 1.2 rad/s

(b) 1.8 rad

Explanation:

Applying,

(a) α = (ω-ω')/t................ Equation 1

Where α = angular acceleration, ω = final angular velocity, ω' = initial angular velocity, t = time.

From the question,

Given: α = 0.40 rad/s², t = 3 seconds, ω' = 0 rad/s (from rest)

Substitute these values into equation 1

0.40 = (ω-0)/3

ω = 0.4×3

ω = 1.2 rad/s

(b) Using,

∅ = ω't+αt²/2.................. Equation 2

Where ∅ = angle turned.

Substitutting the values above into equation 2

∅ = (0×3)+(0.4×3²)/2

∅ = 1.8 rad.

Answer:

3.55 T

Explanation:

Applying,

F = BILsin∅.............. Equation 1

Where F = Force, B = magnetic Field, I = current, L = Length of the wire, ∅ = Angle between the wire and the magnetic field

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 2.45 N, L = 3.00 cm = 0.03 m, I = 23.0 A, ∅ = 90° (Perpendicular)

Substitute these values into equation 2

B = 2.45/(0.03×23×sin90)

B = 2.45/0.69

B = 3.55 T

Answer:

solid cylinder

Explanation:

the object that arrives first is the object that has more speed, let's use the concepts of energy

starting point. Highest point

         Em₀ = U = m g h

final point. Lowest point

         Em_f = K = ½ mv² + ½ I w²

since the body has rotational and translational movement

how energy is conserved

         m g h = ½ mv² + ½ I w²

linear and angular velocity are related

          v = w r

          w = v / r

we substitute

          m g h = ½ mv² + ½ I (v/r) ²

          mg h = ½ v² (m + I /r²)

          v = [tex]\sqrt{2gh \ \frac{m}{m + \frac{I}{r^2} } }[/tex]

the tabulated moments of inertia for the bodies are

solid cylinder     I = ½ m r²

hollow cylinder  I = m r²

we look for the speed for each body

solid cylinder

          v₁ = [tex]\sqrt {2gh} \ \sqrt{\frac{m}{m + m/2} }[/tex]

          v₁ = [tex]\sqrt{2gh} \ \sqrt{2/3}[/tex]

let's call    v₀ = [tex]\sqrt{2gh}[/tex]

         v₁ = 0.816 v₀

hollow cylinder

          v₂ = [tex]\sqrt{2gh } \ \sqrt{\frac{m}{m+ m} }[/tex]

          v₂ = v₀ √½

          v₂ = 0.707 v₀

Therefore, the body that has the highest speed is the solid cylinder and since time is the inverse of speed, this is the body that spends less time to reach the bottom, that is, it is the first to arrive

Answer:

When studying nanotechnology, scientists must be aware that their ideas may not work out. Their work could be very time consuming and cost a lot of money. Finally, scientists do not yet know all of the effects of nanotechnology on human health.

Hope it helps u:)

Explanation:

a) zero, since the machine is mounted on the truck

b) 126 km/hr

c) 85 km/hr

d) 126 km/hr + 85 km/hr = 211 km/hr

Answer:

7.28×10⁻⁵ T

Explanation:

Applying,

F = BILsin∅............. Equation 1

Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.

make B the subject of the equation

B = F/ILsin∅.................. Equation 2

From the question,

Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°

Substitute these values into equation 2

B = 0.16/(68×34×sin72°)

B = 0.16/(68×34×0.95)

B = 0.16/2196.4

B = 7.28×10⁻⁵ T

Answer:

Renewable energy is a type of energy that can be renewed easily, such as sunlight. By using Solar panels to collect the suns energy, we are not depleting it, so this source is renewable.

Non-renewable  energy is something that cannot easily be replenished. An example would be oil because oil takes millions of years to form and cannot be renewed easily.

1433 km

Explanation:

Let g' = the gravitational field strength at an altitude h

[tex]g' = G\dfrac{M_E}{(R_E + h)^2}[/tex]

We also know that g at the earth's surface is

[tex]g = G\dfrac{M_E}{R_E^2}[/tex]

Since g' = (2/3)g, we can write

[tex]G\dfrac{M_E}{(R_E + h)^2} = \dfrac{2}{3}\left(G\dfrac{M_E}{R_E^2} \right)[/tex]

Simplifying the above expression by cancelling out common factors, we get

[tex](R_E + h)^2 = \dfrac{3}{2} R_E^2[/tex]

Taking the square root of both sides, this becomes

[tex]R_E + h = \left(\!\sqrt{\dfrac{3}{2}}\right) R_E[/tex]

Solving for h, we get

[tex]h = \left(\!\sqrt{\dfrac{3}{2}} - 1\right) R_E= 0.225(6.371×10^2\:\text{km})[/tex]

[tex]\:\:\:\:\:= 1433\:\text{km}[/tex]

Answer:

[tex]F_b= 0.720 N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=600N[/tex]

Average density [tex]\rho=1.20kg/m^3[/tex]

Mass

[tex]m=\frac{W}{g}[/tex]

[tex]m=\frac{600}{9.81}[/tex]

[tex]m=61.22kg[/tex]

Generally the equation for Volume is mathematically given by

[tex]V =\frac{ mass}{density}[/tex]

[tex]V= \frac{61.22}{1000}[/tex]

[tex]V=0.06122 m^3[/tex]

Therefore

Buoyant force [tex]F_b[/tex]

[tex]F_b=\rho*V*g[/tex]

[tex]F_b= rho_air*V*g[/tex]

[tex]F_b= 0.720 N[/tex]

Answer:

28 j

Explanation:

because when you add you get 28

Answer:

45

Explanation:

ft/sec2

Answer:

[tex]a=2.5\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of the car, u = 14 m/s

Finally, it comes to rest, v = 0

Time, t = 5.6 s

We need to find the average acceleration of the car during this time interval. We know that,

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-14}{5.6}\\\\a=-2.5\ m/s^2[/tex]

So, the acceleration of the car is [tex]2.5\ m/s^2[/tex] in the opposite direction of motion.

Answer:

oh my God I got really confused right now

Answer:

a) required area is 1.1318 m²

b) the maximum potential difference that can be applied across the compactor is 1931.1 V

Explanation:

Given the data in the question;

dielectric constant εr = 2.35

distance between plates ( thickness ) d = 0.0785 mm = 7.85 × 10⁻⁵ m

dielectric strength = 49.5 MV/m

a)

given that capacity capacitor C = 0.3 uF = 0.3 × 10⁻⁶ F

To find the Area, we use the following the expression.

C = ε₀εrA / d

we know that The permittivity of free space, ε₀ = 8.854 x 10⁻¹²  (F/m)

we substitute

0.3 × 10⁻⁶ = [ (8.854 x 10⁻¹²) × 2.35 × A  ] /  7.85 × 10⁻⁵

A = [ (0.3 × 10⁻⁶) × (7.85 × 10⁻⁵) ] / [ 2.35 × (8.854 x 10⁻¹²) ]

A = 2.355 × 10⁻¹¹ / 2.08069 × 10⁻¹¹

A = 1.1318 m²

Therefore, required area is 1.1318 m²

b)

the maximum potential difference that can be applied across the compactor.

We use the following expression;

⇒ 1/2 × dielectric strength × thickness d

we substitute

⇒ 1/2 × ( 49.5 × 10⁶ V/m ) × ( 7.85 × 10⁻⁵ m )

⇒ 1931.1 V

Therefore, the maximum potential difference that can be applied across the compactor is 1931.1 V

Answer:

Explanation:

The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.

From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.

Answer: [tex]283.2\times 10^{-9}\ nC[/tex]

Explanation:

Given

Cross-sectional area [tex]A=0.4\ cm^2[/tex]

Dielectric constant [tex]k=4[/tex]

Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]

Distance between capacitors [tex]d=5\ mm[/tex]

Maximum charge that can be stored before dielectric breakdown is given by

[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]

Answer:

The maximum charge is 7.08 x 10^-8 C.

Explanation:

Area, A = 0.4 cm^2

K = 4

Electric field, E = 2 x 10^8 V/m

separation, d = 5 mm = 0.005 m

Let the capacitance is C and the charge is q.

[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]