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2 years ago

Friction is a force in which two objects __________.

Physics

2 answers:

balu736 [363]2 years ago

7 0

The answer should be C) Slide against each other.

Kazeer [188]2 years ago

4 0

Answer:

Explanation:

You might be interested in

A certain frictionless simple pendulum having a length l and mass m swings with period t. If both l and m are doubled, what is t

iVinArrow [24]

If l and m both are doubled then the period becomes √2*T

what is a simple pendulum?

It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.

A pendulum is a weight suspended from a pivot so that it can swing freely.

Here,

A certain frictionless simple pendulum having a length l and mass m

mass of pendulum = m

length of the pendulum = l

The period of simple pendulum is:

$T = 2\pi \sqrt{\frac{l}{g} }$

Where k is the constant.

Now the length and mass are doubled,

m' = 2m

l' = 2l

$T' = 2\pi \sqrt{\frac{2l}{g} }$

$T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }$

$T' = \sqrt{2} * T$

Hence,

If l and m both are doubled then the period becomes √2*T

Learn more about Simple Harmonic Motion here:

<u>brainly.com/question/17315536</u>

#SPJ4

8 0

1 year ago

At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl

bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C) the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

4 0

3 years ago

A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c

satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

Given the data in the question;

First course : $a_{x}$ = 0.75 m/s², $d_{x}$ = 20 m, $u_{x}$ = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

$v_{x}$ ² = (10)² + (2 × 0.75 × 20)

$v_{x}$ ² = 100 + 30

$v_{x}$ ² = 130

$v_{x}$ = √130

$v_{x}$ = 11.4 m/s

for the Second Course:

$u_{y}$ = 11.4 m/s, $a_{y}$ = -1.15 m/s², $v_{y}$ = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) × $d_{y}$ )

0 = 129.96 - 2.3 $d_{y}$

2.3 $d_{y}$ = 129.96

$d_{y}$ = 129.96 / 2.3

$d_{y}$ = 56.5 m

so;

|d| = √( $d_{x}$ ² + $d_{y}$ ² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

7 0

2 years ago

JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back

zzz [600]

(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>

3 0

3 years ago

Please, Help!!

Elenna [48]

Let, 1st force = a

2nd force = b

A.T.Q,

a+b = 10

a-b = 6

Calculate for a & b, you'll get a=8 & b= 2

After increasing by 3, it'll be a = 8+3 = 11 & b=2+3 = 5

Resultant force at 90 degree angle = 11+5 = 16 Newtons

7 0

2 years ago

Read 2 more answers