The image of the object formed by the lens is real, enlarged and inverted. What is the kind of lens ?​

Answer:

Force is zero.

Explanation:

According to the Newton's second law, when an object is moving with an acceleration the force acting on the object is directly proportional to the rate of change of momentum of the object.

F = m a

if the object is moving with uniform velocity, the acceleration is zero, and thus, the force is also zero.  

Answer: Near the moon’s surface, a thrust over 11,250 N but under 13,500 N would make it travel at a constant vertical velocity.

Explanation: .Newton’s first law of motion states that an object in motion continues to move in a straight line at a constant velocity unless acted upon by an unbalanced force. In accordance with this law, the lunar lander moves in a downward direction toward the surface of the moon under the influence of force due to gravity. A thrust somewhere between 11,250 and 13,500 balances this gravitational force out.

Answer:

13.94 rpm

Explanation:

Given that,

The diameter of the pulley, d = 14 foot

Radius, r = 7 foot

The linear velocity of the pulley, v = 14 mph = 20.53 ft/s

We need to find the angular velocity in rpm.

We know that, the relation between the linear velocity and the angular velocity is as follows :

[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{20.53}{14}\\\\\omega=1.46\ rad/s[/tex]

or

[tex]\omega=13.94\ rpm[/tex]

So, the angular velocity of the pulley is 13.94 rpm.

Answer:

frequency

Explanation:

The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.

So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is  

[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]

where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.

Answer: hello your question is incomplete below is the missing part

A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.

answer:

- q

Explanation:

Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero

given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q

Answer:

Explanation:

Answer:

B. Is its acceleration constant

Explanation:

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. ... An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction.

Answer:

In 1 second the amount of charge flowing through the conductor is 2.5 Q.

Explanation:

Answer:

Explanation:

From the given information:

From the rotational axis, the distance of the force of gravity is:

d_g = 25+5.0 cm

d_g = 30.0 cm

d_g = 30.0 × 10⁻² m

However, the relative distance of FB  cos 75.9° from the axis is computed as:

d_B = 5.0 cm

d_B = 5.0 × 10⁻² m

The net torque rotational equilibrium = zero (0)

i.e.

[tex]\tau_g -\tau_B = 0 \\ \\ F_gd_g -F_gcos 75.9^0 d_B = 0 \\ \\ F_B = \dfrac{F_g d_g}{F_g cos 65.6} \\ \\ F_B = \dfrac{(3.2)(9.8)(30*10^{-2})}{(5.0*10^{-2} * cos 75.9)} \\ \\ \mathbf{F_B = 772.4 N}[/tex]

= 772.4 N

Thus, the force exerted = 1772.4 N

Answer:

30J

Explanation:

Given data

The total quantity of heat recieved= 50J

Quantity of heat used to do work= 20J

Hence the net change is

ΔU= Total Heat - Net work

ΔU= 50-20

ΔU= 30J

Hence the change in the internal energy is 30J

Answer:

E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules

Answer:

The answer is D. 2.25 × 1017 J

Explanation:

got it right on edge 2021

Explanation:

Given:

L = 0.02 H

C = [tex]2\:\mu \text{F}[/tex]

f = 200 Hz

The general form of the impedance Z is given by

[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]

Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to

[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]

[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]

Answer: 2.4 millimeters = 0.0024 meters

Explanation: A millimeter is 1/1000 of a meter. By diving 2.4 by 1000, you get 0.0024.

Answer:

Natae Si Jordan Kaya Sya Napaihe

Explanation:

haha

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

To find another angle of projection to give the same range;

[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]

Check:

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

Answer:

b)

Explanation:

I see six (6) loops.

I attached a drawing to show where I get six loops from.

Explanation:

[tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]

Here,

We have to find the equivalent resistance of the circuit.

Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,

[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]

Reciprocating both sides,

[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]

Henceforth, Option A is correct.

[tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

[tex] \longrightarrow [/tex] V = IR

[tex] \longrightarrow [/tex] 3 = I × 3.33

[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I

[tex] \longrightarrow [/tex] 0.90 Ampere = I

Henceforth, Option B is correct.

[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]

Answer:

7.5 kg

Explanation:

We are given that

[tex]m_1=10 kg[/tex]

Length of plank, l=3 m

Distance of fulcrum from one end of the plank=1 m

[tex]m_2=20 kg[/tex]

We have to find the mass must be on the other end if the plank remains balanced.

Let m be the mass must be on the other end if the plank remains balanced.

In balance condition

[tex]20\times 1=10\times (1.5-1)+m\times (1.5+0.5)[/tex]

[tex]20=10(0.5)+2m[/tex]

[tex]20=5+2m[/tex]

[tex]2m=20-5=15[/tex]

[tex]\implies m=\frac{15}{2}[/tex]

[tex]m=7.5 kg[/tex]

Hence, mass 7.5 kg   must be on the other end if the plank remains balanced.

Answer:

The mass at the other end is 7.5 kg.

Explanation:

Let the mass is m.

Take the moments about the fulcrum.

20 x 1 = 10 x 0.5 + m x 2

20 = 5 + 2 m

2 m = 15

m = 7.5 kg

Answer:

978.6084 Newton

Explanation:

Given the following data;

To find the weight in Newtown;

Conversion:

1 lb = 4.448220 N

220 lb = 220 * 4.448220 = 978.6084 Newton

220 lb = 978.6084 Newton

Therefore, the weight of the astronaut in Newton is 978.6084.

Weight can be defined as the force acting on a body or an object as a result of gravity.

Mathematically, the weight of an object is given by the formula;

Weight = mg

Where;

Note:

Answer:

in t seconds, Car A sweep out t radian { i.e θ = t radian }

Explanation:

Given the data in the question;

4 toy racecars are racing along a circular race track.

They all start at 3 o'clock position and moved CCW

Car A is constantly 2 feet from the center of the race track and moves at a constant speed

so maximum distance from the center = 2 ft

The angle Car A sweeps out increases at a constant rate of 1 radian per second.

Rate of change of angle = dθ/dt = 1

Now,

since dθ/dt = 1

Hence θ = t + C

where C is the constant of integration

so at t = 0, θ = 0, the value of C will be 0.

Hence, θ = t radian

Therefore, in t seconds, Car A sweep out t radian { i.e θ = t radian }

Answer:

a = 6.67 m/s²

Explanation:

F = 10.0 N

m = 1.50 kg

a = ?

F = ma

10.0 = (1.50)a

6.67 = a

Answer:

22.1 years

Explanation:

Since the current in the wire is I = nevA where n = electron density = 8.50 × 10²⁸ electrons/cm³ × 10⁶ cm³/m³= 8.50 × 10³⁴ electrons/m³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.00 cm = 2 × 10⁻² m

Making v subject of the formula, we have

v = I/neA

So, v = I/neπd²/4

v = 4I/neπd²

Since I = 1,015 A, substituting the values of the other variables into the equation, we have

v = 4I/neπd²

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π ×(2 × 10⁻² m)²]

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π × 4 × 10⁻⁴ m²]

v = (1,015 A)/[42.779 × 10¹¹ electronsC/m]

v = 23.73 × 10⁻¹¹ m/s

v = 2.373 × 10⁻¹⁰ m/s

Since distance d = speed, v × time, t

d = vt

So, the time it takes one electron to travel the full length of the cable is t = d/v

Since d = distance moved by free charge = length of transmission line = 165 km = 165 × 10³ m and v = drift velocity of charge = 2.373 × 10⁻¹⁰ m/s

t = 165 × 10³ m/2.373 × 10⁻¹⁰ m/s

t = 69.54 × 10⁷ s

t = 6.954 × 10⁸ s

Since we have 365 × 24 hr/day × 60 min/hr × 60 s/min = 31536000 s in a year = 3.1536 × 10⁷ s

So,  6.954 × 10⁸ s =  6.954 × 10⁸ s × 1yr/3.1536 × 10⁷ s = 2.21 × 10 yrs = 22.1 years

It will take one electron 22.1 years to travel the full length of the cable

Answer:

KE = 2800 J

Explanation:

Usually a velocity is expressed as m/s. Then the energy units are joules.

[tex]\frac{108 km}{hr} * \frac{1000m}{1 km} * \frac{1 hour}{3600 seconds} =\frac{108*1000 m}{3600sec}[/tex]

v = 30 m / sec

KE = 1/2 * 4 * (30)^2

KE =2800 kg m^2/sec^2

KE = 2800 Joules

Answer:

Explanation:

a)

The force on electron acts opposite to the velocity , and direction of force on electron is always opposite to direction of electric field .

Hence direction of electric field must be in the same  in which electrons travels.

Hence option iii is correct.

b )

deceleration a = force / mass

= qE / m

= 1.6 x 10⁻¹⁶ x 5.6 x 10⁵ / 9.1 x 10⁻³¹

= .98 x 10²⁰ m /s²

v² = u² - 2 a s

0 = (8.06 x 10⁶ )² - 2 x .98 x 10²⁰ s

s = 64.96 x 10¹² / 1.96 x 10²⁰

= 33.14 x 10⁻⁸ m

c ) time required

= 8.06 x 10⁶ / .98 x 10²⁰

= 8.22 x 10⁻¹² s .

d ) Its speed will be same as that in the beginning ie 8.06 x 10⁶ m/s .

Answer:

(a) Option (i)

(b) 6.6 x 10^-4 m  

(c) 8.2 x 10^-11 s

Explanation:

initial velocity, u = 8 .06 x 10^6 m/s

Electric field, E = 5.6 x 10^5 N/C

(a) The direction of field is opposite.

Option (i).

(b) Let the distance is s.  

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times \frac{qE}{m}\times s\\\\8.06\times 10^6\times 8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} s\\\\s = 6.6\times 10^{-4} m[/tex]

(c) Let the time is t.

Use first equation of motion.

[tex]v = u + a t \\\\0 = u - \times \frac{qE}{m}\times t\\\\8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} t\\\\t = 8.2\times 10^{-11} s[/tex]

Answer:The answer is C

Explanation:

Answer:

the answer to the question is known as D

10N/m

Explanation:

f=kx

k=f/x

k=20N/0.2m

k=10N/m

Answer:

Explanation:

From the information given:

The motional emf can be computed by using the formula:

[tex]E = L^{\to}*(V^\to*\beta^{\to})[/tex]

[tex]E = L^{\to}*((x+y+z)*\beta^{\to})[/tex]

[tex]E = 0.50*((18\hat i+24 \hat j +72 \hat k )*0.0800)[/tex]

[tex]E = 0.50*((18*0.800)\hat k +0j+(72*0.080) \hat -i ))[/tex]

[tex]E = 0.50*((18*0.800)[/tex]

E = 0.72 volts

According to the question, suppose the wire segment was parallel, there will no be any emf induced since the magnetic field is present along the y-axis.

As such, for any motional emf should be induced, the magnetic field, length, and velocity are required to be perpendicular to one another .

Then the motional emf will be:

[tex]E = 0.50 \hat j *((18*0.800)\hat k -(72*0.080) \hat i ))[/tex]

E = 0 (zero)