Energy input remains constant and voltage remains the same in a circuit, but current decreases. Which must be happening?

An object is placed at a far distance in front of a thin converging lens. Then the object (1 point) moves gradually toward the l

xeze [42]

Answer:

real, and then virtual

Explanation:

A converging lens is known as convex lens. This lens is called converging lens because it converges all light rays incident on the lens and parallel to the principal axis at the focus.

The nature of image formed by objects placed in front of this lens as mostly REAL IMAGES. The image formed becomes virtual only when the object is almost in close contact with the lens.

Based on the explanation, it can be deduced that an object placed far from a convex lens forms real images but as we move closer to the lens (almost touching the lens), the image formed overtime tends to be virtual.

8 0

3 years ago

Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it

stich3 [128]

Answer:

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

Data given, acceleration,a=35m/s^2

distance covered,d=64cm=0.64m,

a. to determine the speed, we use the equation of motion

initial velocity,u=0m/s

if we substitute values we arrive at

$v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\$

b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2

and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.

Hence we can write the equation above again

$v^{2}=u^{2}-2a(y-2.2)\\$

if we substitute values we have

$v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m$

c. the time it takes to arrive at 1.83m is obtain by using the equation below

$1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37$

if we insert the values, we solve for t , hence t=1.43secs

6 0

3 years ago

The block shown above has a mass of 105 g.<br> What is the density of the block? SC.8.P.8.3

zaharov [31]

The density of the block is 1.25 cm³

The correct answer to the question is Option B. 1.25 cm³

To solve this question, we'll begin by calculating the volume of the block. This can be obtained as follow:

Length = 7 cm

Height = 4 cm

Width = 3 cm

<h3>Volume =? </h3>

Volume = Length × Width × Height

Volume = 7 × 3 × 4

<h3>Volume = 84 cm³</h3>

Thus, the volume of the block is 84 cm³

Finally, we shall determine the density of the block. This can be obtained as follow:

Density is defined as mass per unit volume i.e

$Density = \frac{mass}{volume} \\\\$

Mass of block = 105 g

Volume of block = 84 cm³

<h3>Density of block =? </h3>

$Density = \frac{mass}{volume}\\\\Density = \frac{105}{84}\\\\$

<h3>Density of block = 1.25 cm³</h3>

Therefore, the density of the block is 1.25 cm³.

Hence, Option B. 1.25 cm³ gives the correct answer to the question.

Learn more: brainly.com/question/2040396?referrer=searchResults

4 0

1 year ago

What two states of matter are pictured in the image below?

Anuta_ua [19.1K]

Answer:

volume liquid

Explanation:

6 0

2 years ago

Jeff of the Jungle swings on a 7.6-m vine that initially makes an angle of 32 ∘ with the vertical.

shtirl [24]

To solve this problem we will use the trigonometric concepts to find the distance h, which will allow us to find the speed of Jeff and that will finally be the variable that will indicate the total tension, since it is the variable of the centrifugal Force given in the vine at the lowest poing of the swing.

From the image:

$cos (32) = \frac{(7.6-h)}{7.6}$