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Mazyrski [523]
3 years ago
12

g An arrow is shot straight up in the air at an initial speed of 15.5 m/s. After how much time is the arrow moving downward at a

speed of 5.20 m/s
Physics
1 answer:
Sati [7]3 years ago
5 0

Answer:

2.11 seconds

Explanation:

We use the kinematic equation for the velocity in a constantly accelerated motion under the acceleration of gravity (g):

v_f=v_i-g*t\\-5.2 = 15.5 - 9.8\,t\\9.8 \,t= 15.5 + 5.2\\t = 20.7/9.8\\t = 2.11 \,\,sec

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A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of
ehidna [41]

Answer:

Explanation:

Given

Speed of block at bottom is v=3.8\ m/s

Distance traveled s=6\ m

initial velocity is zero

using equation of motion

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

(3.8)^2-0=2\times a\times 6

a=1.203\ m/s^2

when it is 3 m from top then

v^2-u^2=2as

v^2-0=2\times 1.203\times 3

v=2.68\ m/s      

5 0
3 years ago
A segment of DNA that codes for the ability to make one type of protein molecule is known as a(n)____________________. A) axon B
astraxan [27]
The answer is C) Gene


3 0
3 years ago
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The diameter of circular park is 80 m find its area​
Drupady [299]
Explanation:to find the area you must first find the radius given you have the diameter you cut the diameter in half giving you 40 then you follow the formula “A= (pie)r (squared) leaving you with

answer:5024
6 0
3 years ago
The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which
spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒  \frac{1}{v} -\frac{1}{u} =\frac{1}{f}

On putting estimate values, we get

⇒  \frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}

⇒  \frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}

⇒  v=1.94 \ cm

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

m=\frac{v}{u} and m=\frac{h_{1}}{h_{0}}

⇒  \frac{v}{u} =\frac{h_{1}}{h_{0}}

⇒  \frac{1.94}{25}=\frac{{h_{1}}}{15}

⇒  {h_{1}}=1.16 \ mm

8 0
3 years ago
SP 15The magnetic field in a region of space is measured to be:This field is known to be caused by a cluster of long-straight wi
tino4ka555 [31]

Answer:

 i = 0.477 10⁴ B

the current flows in the  counterclockwise

Explanation:

For this exercise let's use the Ampere law

                    ∫ B . ds = μ₀ I

Where the path is closed

Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.

From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.

We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field

Segment on the y axis

        L₀ = (y2-y1)

        L₀ = 3-0 = 3 cm

Segment on the point x = 2 cm

        L₁ = 3-0

        L₁ = 3cm

       B L = μ₀ I

       B 2L = μ₀ I

        i = 2 L B /μ₀

        i= 2 0.03 / 4π 10⁻⁷   B

        i = 4.77 10⁴  B

The current is perpendicular to the magnetic field whereby the current flows in the  counterclockwise

8 0
3 years ago
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