Identify the machines in which both a lever and a wedge are used

A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg

alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = - $\frac{v}{u}$

where

u = object distance

v = image distance

Now,

1.90 = $\frac{v}{u}$

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

$\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}$

$\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}$

$\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}$

$\frac{1}{16.5} = \frac{ 0.90}{1.90u}$

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0

3 years ago

4. What is the magnitude and direction of the gravitational force that acts on a man

Cloud [144]

The gravitational force acting on the man is 800 N towards the Earth's centre

Explanation:

The weight of an object on the Earth is exactly the gravitational force exerted by the Earth on the object.

The gravitational force exerted by the Earth on an object located at the Earth's surface is given by:

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

R is the radius of the Earth

And the direction of the force is towards the Earth's centre.

Since G, M and R are constant, they are grouped into a single constant called g, acceleration of gravity:

$g=\frac{GM}{R^2}$

therefore the gravitational force can be rewritten as

$F=mg$

And this is the usual equation that we use to calculate the weight of an object.

Therefore, weight and gravitational force acting on an object on Earth are the same thing: so, the gravitational force acting on the man is equal to his weight, 800 N, and it acts towards the Earth's center.

Learn more about weight and gravitational force:

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#LearnwithBrainly

4 0

3 years ago

You punt a ball straight up at 20 m/s. What is the balls hangtime

Natalija [7]

The hang time of the ball is 4.08 s

Explanation:

The ball is in free fall motion: this means that it is acted upon gravity only, so its acceleration is the acceleration of gravity,

$a=g=-9.8 m/s^2$

downward (the negative sign refers to the downward direction).

Since this is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

First we calculate the time it takes for the ball to reach the maximum height, where the velocity is zero:

v = 0

Substituting:

u = +20 m/s

$a=-9.8 m/s^2$

we find t

$t=\frac{v-u}{a}=\frac{0-20}{-9.8}=2.04 s$

The motion of the ball is symmetrical, so the total time of flight is just twice the time needed to reach the maximum height, therefore:

$T=2t=2(2.04)=4.08 s$

Learn more about free fall:

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4 0

3 years ago

A battery-operated car utilizes a 120.0 V battery with negligible internal resistance. Find the charge, in coulombs, the batteri

fgiga [73]

Answer:

4.29×10⁵ C

Explanation:

From the question,

The energy stored in the battery = Kinetic energy of the car+ Energy needed to make the car climbed the hill+Energy required to exert a force.

E = 1/2mv²+mgh+Fd.................... Equation 1

Where E = Energy stored in the battery, m = mass of the car, v = velocity of the car, h = height of the hill, F = force exerted on the car, d = distance traveled by the car.

But,

d = vt.................... Equation 2

Where v = velocity, t = time.

Substitute equation 2 into equation 1

E = 1/2mv²+mgh+F(vt)................... Equation 3

Given: m = 770 kg, v = 26 m/s, h = 2.15×10² m = 215 m, F = 5.3×10² N = 530 N, t = 1 hour = 3600 s, g = 9.8 m/s²

Substitute into equation 1

E = 1/2(770)(26²)+(770)(9.8)(215)+(530)(26)(3600)

E = 260260+1622390+49608000

E = 51490650 J

Using,

E = qV................. Equation 4

Where q = charge of the battery, V = Voltage.

make q the subject of the equation

q = E/V............... Equation 5

Given: E = 51490650 J, V = 120 V

Substitute into equation 5

q = 51490650/120

q = 429088.75 C

q = 4.29×10⁵ C

7 0

3 years ago

Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil

MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

$emf_{max}=NBA\omega$

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

$A=\pi r^2 =\pi(0.23m)^2=0.16m^2$

$emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V$

Finally, the peak current is given by:

$I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA$

b)

we have that

$I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A$

$P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W$

hope this helps!!

6 0

3 years ago