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zalisa [80]
2 years ago
9

What was the vertical component of her acceleration during push-off? the positive direction is upward?

Physics
1 answer:
gladu [14]2 years ago
5 0
<span>The answer of these two problems are : A) a = (F-W)/m = (100-500)/500/g = 9.8*500/500 = +9.8 m/sec^2 B) a1 = +9.8-g = 0</span>
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Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b
Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

\Delta ACB

tan 30 = \frac{8.3}{h}

h= \frac{8.3}{tan 30} = 14.376 m

therefore

height of keisha is = h  + 1.5 m

                               = 14.376 + 1.5

= 15.876 \simeq 16 m

therefore option c is correct

5 0
2 years ago
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
kipiarov [429]

Answer:

I=68.31\times 10^{-6}\ A

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

By putting the values

I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A

I=68.31\times 10^{-6}\ A

7 0
2 years ago
Read 2 more answers
Suppose you want to make a scale model of a hydrogen atom. You choose, for the nucleus, a small ball bearing with a radius of 1.
RoseWind [281]

Answer:

A)  x _electron = 0.66 10² m , B)   x _Eart = 1.13 10² m , C)  d_sphere = 1.37 10⁻² mm

Explanation:

A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of

Let's use proportions rule

                x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5

               x _electron = 0.66 10⁵ mm = 0.66 10² m

B) the radii of the Earth and the sun are

               R_{E} = 6.37 10⁶ m

                tex]R_{Sum}[/tex] = 6.96 10⁸ m

                Distance = 1.5 10¹¹ m

                x_Earth = 1.5 10¹¹ / 6.96 10⁸  1.5

                x _Eart = 1.13 10² m

C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions

            d_sphere = 1.5 / 6.96 10⁸  6.37 10⁶

            d_sphere = 1.37 10⁻² mm

5 0
2 years ago
A car came to a stop from a speed of 35 m/s in a time of 8.1 seconds. What was the acceleration of the car?
uranmaximum [27]
Simply subtract the two velocities and divide by 8.1,

\frac{0 - 35}{8.1} = - 4.32

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey
5 0
2 years ago
Read 2 more answers
calculate the workdone to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it
Lera25 [3.4K]
The force of F=10 N produces an extension of
x=4 cm=0.04 m
on the string, so the spring constant is equal to
k= \frac{F}{x}= \frac{10 N}{0.04 m}=250 N/m

Then the string is stretched by \Delta x=40 cm=0.40 m. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
W= \Delta U= \frac{1}{2}k(\Delta x)^2  = \frac{1}{2}(250 N/m)(0.40 m)^2=20 J
5 0
2 years ago
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