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2 years ago

HELP ME

Physics

1 answer:

Mazyrski [523]2 years ago

3 0

Answer:

I THINK IT'S <em>D.</em><em>.</em><em>.</em><em>.</em>

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A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b

daser333 [38]

Answer:

<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b) The bullet travels horizontally 110.6 m</h2>

Explanation:

a) Consider the vertical motion of bullet

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 1.5 m

Substituting

s = ut + 0.5 at²

1.5 = 0 x t + 0.5 x 9.81 xt²

t = 0.553 s

Time elapsed before the bullet hits the ground is 0.553 seconds.

b) Consider the horizontal motion of bullet

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 200 m/s

Acceleration, a = 0 m/s²

Time, t = 0.553 s

Substituting

s = ut + 0.5 at²

s = 200 x 0.553 + 0.5 x 0 x 0.553²

s = 110.6 m

The bullet travels horizontally 110.6 m

6 0

2 years ago

The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

3 0

2 years ago

_________ changes from day to day, while ____________ changes from region to region. A) climate, weather B) weather, climate C)

vladimir2022 [97]

B) Weather changes day to day, while climate changes region to region.

Climate is the weather in a certain area. It's usually the average weather over a long period of time

Weather is in shorter terms then climate

Hope this helped!

~Just a girl in love with Shawn Mendes

7 0

2 years ago

Read 2 more answers

Why a plastic pen which is rubbed with hair,is able to attract small pieces of papers

AURORKA [14]

Answer:

Explanation:

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5 0

2 years ago

An object is located 50.0 cm from a concave mirror. The magnitude of the mirror focal length is 25.0 cm. What is the image dista

ioda

Answer:

Correct answer: C. 50 cm

Explanation:

Given data:

The distance of the object from the top of the concave mirror o = 50.0 cm

The magnitude of the concave mirror focal length 25.0 cm.

Required : Image distance d = ?

If we know the focal length we can calculate the center of the curve of the mirror

r = 2 · f = 2 · 25 = 50 cm

If we know the theory of spherical mirrors and the construction of figures then we know that when an object is placed in the center of the curve, there is also a image in the center of the curve that is inverted, real and the same size as the object.

We conclude that the image distance is 50 cm.

We will now prove this using the formula:

1/f = 1/o + 1/d => 1/d = 1/f - 1/o = 1/25 - 1/50 = 2/50 - 1/50 = 1/50

1/d = 1/50 => d = 50 cm

God is with you!!!

6 0

3 years ago

Read 2 more answers