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3 years ago

How much to the left of the lens is the object?

Physics

1 answer:

stich3 [128]3 years ago

4 0

An object is 16.0 cm to the left of a thins lens. The lens forms an image 36.0 cm to the right of the lens.

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In the problem below, what is the student showing?

Phantasy [73]

Answer:

S<em>tudent showing the volume of cube</em>

V = a³

if a = 3 cm

Then volume is a³ = 3 ×3 × 3

= 27 cm³

6 0

3 years ago

The acceleration due to gravity on the Moon's surface is

Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

$g'=\frac{g}{6}$

We know that weight of an object on Earth is,

$w = m\times g$

$m = \frac{w}{g}$

Similarly, weight on Moon will be

$w' = m\times g'$

$w' = \frac{w}{g}\times\frac{g}{6}$

$w' = \frac{300}{6}$

$w' = 50$

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0

3 years ago

A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.

alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

$F_g = F_n + F_a$

given that

$F_g = mg$

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

$F_n = 150 N$

now from above equation

$30*10 = 150 + F_a$

$F_a = 300 - 150 = 150 N$

So force applied by the person must be 150 N

7 0

3 years ago

A motorist driving at 25 meters/second decelerates to

Ilia_Sergeevich [38]

Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.

Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,

Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules

Just ignore the negative value for the final result because work is a scalar quantity.

8 0

3 years ago

Read 2 more answers

A roller coaster car of mass m= 300 kg is released from rest at the top of a 60 m high hill (position A), and rolls with a negli

Andrews [41]

Answer: The principle of conservation of energy, angular speed and centripetal force

Explanation:

At point A, the car experienced maximum of potential energy

As it moves down the hill, the potential energy decreases while the kinetic energy increases.

The maximum kinetic energy of the car is needed for the attainment of enough centripetal force to help the car move through the loop without falling .

4 0

3 years ago