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4 years ago

DUE TONIGHT HURRY PLS I give 30 pts

Physics

1 answer:

Paha777 [63]4 years ago

6 0

The answer is both.

For kinetic energy:

KE = 1/2*m*v^2 = 0.5*20,000 grams*5 = 50,000 J

For gravitational potential energy:

Pe = mgh = 20,000 grams*9.81 m/s^2*2 m = 392.2 J

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A camper walks away from camp. He walks 23.72m north, then 39.25m south and

Arte-miy333 [17]

Answer:

89.53 North

Explanation:

Total N-S displacement

23.72 - 39.25 + 105.06 = + 89.53

5 0

2 years ago

The near point of an eye is 110 cm. A corrective lens is to be used to allow this eye to focus clearly on objects 26.0 cm in fro

irga5000 [103]

Answer:

The focal length of the lens is 34.047 cm

The power of the needed corrective lens is 2.937 diopter.

Explanation:

Distance of the object from the lens,u = 26 cm

Distance of the image from the lens ,v= -110 cm

(Image is forming on the other side of the lens)

Since ,lens of the human eye is converging lens,convex lens.

Using a lens formula:

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

$\frac{1}{f}=\frac{1}{26 cm}+\frac{1}{-110 cm}$

f = 34.047 cm = 0.3404 m

Power of the lens = P

$P=\frac{1}{f}=\frac{1}{0.34047 m}=2.937 diopter$

6 0

3 years ago

A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if

Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if a. The mass is doubled? b. The mass is halved? c. The amplitude is doubled? d. The spring constant is doubled?

We have a block-spring system, whose angular frequency ( $\omega$ ) is defined by the following formula:

$\omega = \sqrt{\frac{k}{m} }$ (1)

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

And the period ( $T$ ), measured in seconds, is determined by the following expression:

$T = \frac{2\pi}{\omega}$ (2)

By applying (1) in (2), we get the following formula:

$T = 2\pi\cdot \sqrt{\frac{m}{k} }$

a) If the mass is doubled, then the period is increased by $\sqrt{2}$ . Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by $\frac{\sqrt{2}}{2}$ . Hence, the period of the system is 1.414 seconds.

8 0

3 years ago

Between which two points of a concave mirror should an object be placed to obtain a magnificati of -3

kobusy [5.1K]

Answer:

When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.

When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.

Explanation:

8 0

3 years ago

An apple with a mass of 0.95 kilograms hangs from 3.0 meters above the ground. What is the potential energy of the apple

Vesnalui [34]

As Potential energy =mgh

m= 0.95kg

h=3 meter

g = 9.8 m/sec^2. ( acceleration due to gravity)

So P.E =(0.95)(9.8)(3)kgm^2/s^2

P.E =27.93 joules

3 0

3 years ago