All categories

3 years ago

Tyler is in downtown San Antonio entertaining the tourists with amazing feats of physics. He carries a nickel

Physics

1 answer:

professor190 [17]3 years ago

3 0

Answer:

The velocity of the nickel just before it hits the ground is approximately 49.3 m/s

Explanation:

In order to calculate the velocity of the nickel just before it hits the ground, we recall the kinematic equation of motion, v² = u² + 2·g·h, where the variables of the equation are defined as follows;

v = The velocity of the nickel just before it hits the ground after it is pushed off the observation deck

u = The initial velocity of the nickel just before it is pushed off the observation deck = 0 m/s

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which he nickel is pushed off the edge = 124 m

Substituting the given values and the constant, "g", we have;

v² = 0² + 2 × 9.8 × 124 = 2,430.4

v = √2,430.4

Using a graphing calculator, we have;

v = √2,430.4 = 14·√(62/5) ≈ 49.3

The velocity of the nickel just before it hits the ground = v ≈ 49.3 m/s.

You might be interested in

Mendel made conclusions about inheritance without ever seeing chromosomes or knowing about DNA. Which technology has enabled the

kupik [55]

It would most likely be the microscope.

7 0

3 years ago

Read 2 more answers

Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge

kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

$K =1/4 \pi \epsilon_0 = 8.99 \times 10^9$ N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

$E_1 + E_3 = 0$

$E =E_2 + E_4$

$E = 2 E_2$

[ $E_2 =\frac{2\times k \times q}{r^2}$

[ $r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m$ ]

plugging all value

$E = 2 E_2$

$E = 2 E_2 =\frac{2\times k \times q}{r^2}$

$E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}$

E = 440816.32 N/C

3 0

3 years ago

Read 2 more answers

A compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm). The lenses are separated by

Marizza181 [45]

Answer:

-252.52

Explanation:

L = Distance between lenses = 10 cm

D = Near point = 25 cm

$f_o$ = Focal length of objective = 0.9 cm

$f_e$ = Focal length of eyepiece = 1.1 cm

Magnification of a compound microscope is given by

$m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52$

The angular magnification of the compound microscope is -252.52

6 0

3 years ago

which type of lightening device would you use for each of the following needs: an economical light source in a manufacturing pla

Yakvenalex [24]

Btw only someone who is nice will answer tour question. You can't expect for explanition when the question is only worth 5 points. Not trying to be mean sorry if i am being mean

7 0

3 years ago

A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt

Yuri [45]

The total power emitted by an object via radiation is:
$P=A\epsilon \sigma T^4$
where:
A is the surface of the object (in our problem, $A=1.25 m^2$
$\epsilon$ is the emissivity of the object (in our problem, $\epsilon=1$ )
$\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4)$ is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is $T=100^{\circ} C=373 K$

Substituting these values, we find the power emitted by radiation:
$P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW$
So, the correct answer is D.

6 0

3 years ago