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3 years ago

Which statements describe the image produced by a concave lens?check all that apply.

Physics

2 answers:

ikadub [295]3 years ago

8 0

Answer:

A,E

Explanation:

A-the image is right-side up

B-the image is virtual

kotegsom [21]3 years ago

7 0

Answer:

The answers are:

A-which is the image is always right side up.

E-the image is virtual

Explanation: MY EXPLANATION IS YOU ARE WELCOME BIG DOG 100..

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During take-off a 5kg model rocket, initially at rest, burns fuel for 4.6s causing its speed to increase from rest to 39m/s duri

MrRa [10]

Answer:

The height is "89.61 m". A further explanation is given below.

Explanation:

According to the question,

Mass of rocket,

m = 5 kg

Time,

t = 4.6 s

Initial speed of rod

u = o m/s

Final speed,

v = 39 m/s

drag,

= 60 N

So,

⇒ acceleration, $a=\frac{v-u}{t}$

$=\frac{39-0}{4.6}$

$= 8.478 \ m/s^2$

⇒ $F_{net}= Net \ force$

$=ma$

$=5\times 8.478$

$=42.89 \ N$

Now,

The thrust will be:

⇒ $Thrust-weight-drag=Net \ force$

⇒ $Thrust-(5\times 9.8)-60=42.89$

⇒ $Thrust-49-60=42.89$

⇒ $Thrust-109=42.89$

⇒ $Thrust=42.89+109$

⇒ $Thrust=151.89 \ N$

The height will be:

⇒ $h=\frac{1}{2} at^2$

$=\frac{1}{2}\times 8.47\times (4.6)^2$

$=\frac{1}{2}\times 8.47\times 21.16$

$=89.61 \ m$

5 0

3 years ago

A 0.18 m radius pulley is free to rotate about a horizontal axis. A mass and a mass are attached by a massless string, which is

Musya8 [376]

Answer:

T = 1.766(M-m) Nm where M and m are the 2 masses of the objects

Explanation:

Let m and M be the masses of the 2 objects and M > m so the system would produce torque and rotational motion on the pulley. Force of gravity that exert on each of the mass are mg and Mg. Since Mg > mg, the net force on the system is Mg - mg or g(M - m) toward the heavier mass.

Ignore friction and string mass, and let g = 9.81 m/s2, the net torque on the pulley is the product of net force and arm distance to the pivot point, which is pulley radius r = 0.18 m

T = Fr = g(M - m)0.18 = 0.18*9.81(M - m) = 1.766(M-m) Nm

8 0

2 years ago

You place the spring vertically with one end on the floor. You then lay a 1.60 kg book on top of the spring and release the book

malfutka [58]

Answer:

Compression in the spring, x = 3.7 cm

Explanation:

It is given that,

Mass of the book, m = 1.6 kg

It can be assumed the spring constant of the spring is, k = 840 N/m

As the book moves down, the change in potential energy of the book is converted to spring potential energy of compression. The mathematical expression is as follows :

$\dfrac{1}{2}kx^2=mgx$

$x=\dfrac{2mg}{k}$

$x=\dfrac{2\times 1.6\ kg\times 9.8\ m/s^2}{840\ N/m}$

x = 0.037 meters

x = 3.7 cm

So, the spring is compressed to a distance of 3.7 cm. Hence, this is the required solution.

4 0

2 years ago

Two forces act at a point in the plane. The angle between the two forces is given. Find the magnitude of the resultant force. fo

Zielflug [23.3K]

Answer:

408N at 89.89°

Explanation:

This problem requires that we resolve the force vectors into

x- and y

-componentsOnce this is done, we can add the components easily, as the one 2-dimensional problem will be two 1-dimensional problems.

Finally, we will convert the resultant force into standard form and find the equilibrant.

Resolve into components:

F1x =F1cos 180°= 232(−1)=−232N

F1y=F1sin180°=0N

F2x=F2cos(−140°)=194(−0.766)=−148.6N

F1y=F1sin(−140°)=232(−0.643)=−149.17N

Note the change of the angle used to give the direction of

F2. Standard angles (rotation from thex

-axis; counterclockwise is +) should be used to avoid sign errors in the results.

Now, we add the components:

Fx=F1x+F2x=−380.6N

Fy=F1y+F1y=−148.17N

Technically, this is the resultant force. However, it should be changed back into standard form. Here's how:

F=√(Fx)2(Fy)2=√(−380.6)^2(−148.17)^2=408N

θ=tan−1(−148.17−380.6)

=89.89°

4 0

3 years ago

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×10^{4} 4 m/s^{2} 2 , and 1.85 ms (1

Paha777 [63]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 = u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

6 0

2 years ago