What type of image results when the object is located between 2f (2 focal lengths) and f (the focal point) of a convex lens?

What is the force on a 250 kg elevator that is falling freely at 9.8 m/sec2?

Viefleur [7K]

Answer:

Since it is falling freely, the only force on it is its weight, w.

w = m × g = 250 kg × 9.8 m/s^2 = 2450 Newton/N

6 0

3 years ago

Read 2 more answers

Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

Kay [80]

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

$F=kx$

where, $F$ is the force which is stretching or compressing the spring,

$k$ is the spring constant; and

$x$ is the distance the spring is stretched.

Substituting the given values to find the elastic constant $k$ to get:

$F=kx$

$4=k(2)$

$k=\frac{4}{2}$

$k=2$

Therefore, the elastic constant is 2 Newton/cm.

3 0

2 years ago

A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of –3.00nC/m2

lys-0071 [83]

Answer: sheet of charge

Explanation:

a )

Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.

b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.

C ) Electric field which is almost uniform near the sheet of charge is equal t the following

E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²

E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²

= .3389 x 10³

= 338.9 V / m

spacing between 1 V

= 1 / 338.9 m

= 2.95 X 10⁻3 m

= 2.95 mm.

3 0

3 years ago

Calculate the force of gravity between a comet with a mass of 500kg and a small asteroid with a mass of 20kg that is separated b

givi [52]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The equivalent gravitational force is ~

$F \approx1.48\times 10 {}^{ - 7} \: \: N$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

We know that ~

$\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}$

where,

F = gravitational force

$m_1$ = mass of 1st object = 500 kg

$m_2$ = mass of 2nd object = 20kg

G = gravitational constant = $6.674 × {10}^ {-11}$

r = distance between the objects = 2.12 m

Let's calculate the force ~

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 500 \times 20}{(2.12) {}^{2} }$

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 10 {}^{4} }{4.4944}$

$F = \dfrac{6.674}{4.4944} \times 10 {}^{ - 7}$

$F =1.484 \times 10 {}^{ - 7} \: \: newtons$

7 0

2 years ago

During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal

weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

The Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º

The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

Vectors are decomposed into a coordinate system

The components are added

The resulting vector is constructed

Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

v = $\frac{\Delta d}{t}$

Δd = v t

Where v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

t₁ = 2 min = 120 s

Δd₁ = v₁ t₁

Δd₁ = 600 120

Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s

time t₂ = 1 min = 60 s

Δd₂ = v₂ t₂

Δd₂ = 400 60

Δd₂ = 24 103 m

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

cos (180 -35) = $\frac{x_1}{\Delta d_1}$

sin 145 = $\frac{y_1}{\Delta d1}$

x₁ = Δd₁ cos 125

y₁ = Δd₁ sin 125

x₁ = 72 103 are 145 = -58.98 103 m

y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

cos (90+ 30) = $\frac{x_2}{\Delta d_2}$

sin (120) = $\frac{y_2}{\Delta d_2}$

x₂ = Δd₂ cos 120

y₂ = Δd₂ sin 120

x₂ = 24 103 cos 120 = -12 10³ m

y₂ = 24 103 sin 120 = 20,78 10³ m

The component of the resultant vector are

Rₓ = x₁ + x₂

R_y = y₁ + y₂

Rx = - (58.98 + 12) 10³ = -70.98 10³ m

Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

R = $\sqrt{R_x^2 +R_y^2}$

R = $\sqrt{70.98^2 + 62.08^2}$ 10³

R = 94.30 10³ m

We use trigonometry for the angle

tan θ ’= $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{62.08}{70.98}$

θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

θ = 180 - θ'

θ = 180 -41.2

θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

v = $\frac{\Delta R}{t_3}$