PLEASEEE HELPPPPP does anyone know these answers?

The superheroine Xanaxa, who has a mass of 68.1 kg , is pursuing the 75.3 kg archvillain Lexlax. She leaps from the ground to th

Alexandra [31]

The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

$U = mg\Delta h$

Here,

$\Delta h$ = Change in height

m = mass of super heroine

g = Acceleration due to gravity

The change in height will be,

$\Delta h = h_f - h_i$

The final position of the heroin is below the ground level,

$h_f = -16.1m$

The initial height will be the zero point of our system of reference,

$\Delta h = -16.1m-0m$

$\Delta h = -16.1m$

Replacing all this values we have,

$U = mg\Delta h$

$U = (68.1kg)(9.8m/s^2)(-16.1m)$

$U = -10744.81J$

Since the final position of the heroine is located below the ground, there will net loss of gravitational potential energy of 10744.81J

4 0

3 years ago

In the united states a standard letter sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for

Andre45 [30]

This question is incomplete

Complete Question

In the United States, a standard letter-sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for a letter-sized piece of paper is different. The international standard is based on SI units: 21.0 cm wide by 29.7 cm long.

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

Answer:

a) 8.267721 inches ≈ 8.3 inches

b) 11.6929197 inches ≈ 11.7 inches

c) It's dimensions in inches for the international standard for letter - sized for paper = 8.3 wide inches by 11.7 inches long

d) The International standard letter - sized paper is longer.

Explanation:

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

21 cm =

Cross Multiply

21 cm × 0.393701inch/ 1 cm

= 8.267721 inches

Approximately 8.3 inches

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

29.7 cm =

Cross Multiply

29.7 × 0.393701 inch/ 1 cm

= 11.6929197 inches

Approximately 11.7 inches

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

The international standard for a letter-sized has dimensions 21.0 cm wide by 29.7 cm long.

Where

21.0cm = 8.267721 inches

≈ 8.3 inches

29.7cm = 11.6929197 inches

≈ 11.7 inches

Hence, it's dimensions in inches = 8.3 inches by 11.7 inches.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

U.S letter - sized paper = 8.5 inches wide by 11 inches long

International standard letter- sized paper = 8.3 wide inches by 11.7 inches long.

Hence, the International standard letter - sized paper is longer.

5 0

3 years ago

A 5kg object is moving at a height of 2 m. The potential energy of the object is closest to ___ j

elena55 [62]

Answer:

In this case, a body of mass 5 kg kept at a height of 10 m. So the potential energy is given as 5 * 10 *10 = 500 J.

6 0

3 years ago

Does the moon fall partly into earth's shadow when it is n "full"?

murzikaleks [220]

Yes, <span> the moon fall partly into earth's shadow when it is in its full size</span>

5 0

3 years ago

A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate

kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid $R=1.06 m$

Emf induced $E=8.50\times 10^{-6} V/m$

no of turns per meter n=450

we know Induced EMF is given by

$\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A$

Magnetic Field is given by

$B=\mu _0ni$

thus $\frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}$

Area of cross-section

$A=\pi R^2$ where

solving integration we get

$E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2$

where r=distance from axis

R=radius of Solenoid

$\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}$

$\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}$

$\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s$

4 0

3 years ago