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Elanso [62]
3 years ago
11

I dont understand physics.

Physics
1 answer:
Elanso [62]3 years ago
8 0

Answer:

same but from what i know is the newtons thingy

Explanation:

hope that helps!

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An object moves in uniform circular motion at 25 m/s and takes 1.0 second to go a quarter circle. What is the radius of the circ
FromTheMoon [43]
Object Motion: 25 m/s

Circumference of Circle: 
1/4 Circumference of Circle in 1 second = 25 meters
25 meters times 4 = Circumference of Circle 
Circumference = 100 meters

Formula to Find Circumference of Circle: (work opposite)
C = 2<span>πr

100 = </span>2πr          divided
100/2π = r         simplify
50/π = r         (exact radius)

Answer:
50/π meters = r         (exact radius)

4 0
3 years ago
Read 2 more answers
At what time after being ejected is the boulder moving at a speed 20.7 m/s upward?
Svetlanka [38]

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a = g = -9.8 $$m / s^2

downward (acceleration due to gravity).

By using Suvat equation:

v = u + at

where: v is the velocity at time t

u = 40.0 m/s is the initial velocity

a = g = -9.8 $$m/s^2 is the acceleration

To find the time t at which the velocity is v = 20.7 m/s

Therefore,

$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

To learn more about uniformly accelerated motion refer to:

brainly.com/question/14669575

#SPJ4

4 0
2 years ago
An object is placed 50.0 cm in front of a convex mirror. where can be the image located if the focal length is 40 cm from the mi
sergiy2304 [10]

Answer:

Image will form at distance 22.22 cm behind the mirror

Explanation:

As we know that the mirror formula is given as

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

now we know that

object distance from mirror is

d_o = -50 cm

Focal length of the mirror is given as

f = 40 cm

now we have

\frac{1}{-50} + \frac{1}{d_i} = \frac{1}{40}

\frac{1}{d_i} = \frac{1}{50} + \frac{1}{40}

d_i = 22.22 cm

6 0
3 years ago
Time measurements from a stopwatch are not precise. Why not ?
fgiga [73]
Human error (average human reaction time is .2 seconds)
5 0
3 years ago
A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri
s344n2d4d5 [400]

Answer:

The\quad magnitude\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad \quad =2.52m/{ s }^{ 2 }\\ The\quad direction\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad =15.{ 72 }^{\o}\\The\quad car\quad begins\quad to\quad slide\quad out\quad \quad of\quad the\quad circle\quad after\quad 26.09s.\quad \quad \quad \quad

Explanation:

8 0
3 years ago
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