Hi Friends!

please help me with these questions!

SUBJECT: Chemistry, Physics,Biology

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.

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:

Minimum area of rectangle = 24 centimeter²

Explanation:

Given:

Length of rectangle = 6 centimeter

Width of rectangle = 4 centimeter

Find:

Minimum area of rectangle

Computation :

Minimum area of rectangle = Length of rectangle x Width of rectangle

Minimum area of rectangle = 6 x 4

Minimum area of rectangle = 24 centimeter²

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:

workdone= 1/2mv^2

1/2×1500×30^2

675000J

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:

45

Explanation:

ft/sec2

Answer:

a)  N₁ = 14083 turns,  b)   I₁ = 2.58 A

Explanation:

The relationship that describes the relationship between the primary and secondary of the transformer is

         [tex]\frac{V_2}{N_2} = \frac{V_1}{N_1}[/tex]

a) They indicate that the secondary has N2 = 130 turns, the turns of the primary are

         N₁ = [tex]N_2 \frac{V_1}{V_2}[/tex]

         N₁ = [tex]130 \ \frac{13000}{120}[/tex]

         N₁ = 14083 turns

b) since there are no losses, the power of the neighboring transformer is

          P = V I

          P = 120 280

          P = 33600 W

this is the same power of the substation

          P = V₁  I₁

          I₁ = P / V₁

          I₁ = 33600/13000

          I₁ = 2.58 A

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.

Here are some healthy ways to gain weight when you're underweight:

Eat more frequently. When you're underweight, you may feel full faster. Eat five to six smaller meals during the day rather than two or three large meals.

Choose nutrient-rich foods. As part of an overall healthy diet, choose whole-grain breads, pastas and cereals; fruits and vegetables; dairy products; lean protein sources; and nuts and seeds.

Try smoothies and shakes. Don't fill up on diet soda, coffee and other drinks with few calories and little nutritional value. Instead, drink smoothies or healthy shakes made with milk and fresh or frozen fruit, and sprinkle in some ground flaxseed. In some cases, a liquid meal replacement may be recommended.

Watch when you drink. Some people find that drinking fluids before meals blunts their appetite. In that case, it may be better to sip higher calorie beverages along with a meal or snack. For others, drinking 30 minutes after a meal, not with it, may work.

Make every bite count. Snack on nuts, peanut butter, cheese, dried fruits and avocados. Have a bedtime snack, such as a peanut butter and jelly sandwich, or a wrap sandwich with avocado, sliced vegetables, and lean meat or cheese.

Top it off. Add extras to your dishes for more calories — such as cheese in casseroles and scrambled eggs, and fat-free dried milk in soups and stews.

Have an occasional treat. Even when you're underweight, be mindful of excess sugar and fat. An occasional slice of pie with ice cream is OK. But most treats should be healthy and provide nutrients in addition to calories. Bran muffins, yogurt and granola bars are good choices.

Exercise. Exercise, especially strength training, can help you gain weight by building up your muscles. Exercise may also stimulate your appetite.

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:

28 j

Explanation:

because when you add you get 28

Answer:

sgsbssbduebubbeeifirjeirneejrbb8m!keoejr

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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:

(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:

oh my God I got really confused right now

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

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:

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:

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

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:

Number of oscillation = 106 oscillations

Explanation:

Given the following data;

To find how many complete oscillations the pendulum makes in 5.00 min;

First of all, we would determine the period of oscillation of the pendulum using the following formula;

[tex] T = 2 \pi \sqrt{\frac{l}{g}} [/tex]

Where;

Substituting into the formula, we have;

[tex] T = 2 * 3.142 \sqrt{\frac{2}{9.8}} [/tex]

[tex] T = 6.284 \sqrt{0.2041} [/tex]

[tex] T = 6.284 * 0.4518 [/tex]

Period, T = 2.84 seconds

Next, we would determine the number of complete oscillation in 5 minutes;

We would have to convert the time in minutes to seconds.

Conversion:

1 minutes = 60 seconds

5 minutes = X seconds

Cross-multiplying, we have;

X = 5 * 60 = 300 seconds

Mathematically, the number of oscillation of a pendulum is given by the formula;

[tex] Number \; of \; oscillation = \frac {Time}{Period} [/tex]

Substituting into the formula, we have;

[tex] Number \; of \; oscillation = \frac {300}{2.84} [/tex]

Number of oscillation = 105.63 ≈ 106 oscillations

Number of oscillation = 106 oscillations

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:

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

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]