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2 years ago

How do I solve this question

Physics

1 answer:

babunello [35]2 years ago

6 0

Newton would not agree with that explanation, and he would tell my brother to go to his room. Then Newton would turn to me and explain:

Once the soccer ball is in motion, it will continue moving in a straight line with constant speed, until external forces change its motion. (Newton's #1 Law of Motion).

There are external forces on the soccer ball after the kick: gravity, air resistance, and scraping through grass.

Gravity has no effect on its horizontal motion. But air resistance and scraping through grass are both friction ... they both rob kinetic energy from the ball. Once the kinetic energy of the ball is all gone, it doesn't move any more. An observer looking at the ball and measuring its motion would write in his notebook "The ball has stopped." .

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A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th

inn [45]

Answer:

$|\vec r|=339.82\ m$

$\theta=-6.67^o$

Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

$\vec r=\vec r_1+\vec r_2$

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

$x=rcos\theta$

$y=rsin\theta$

And the vector is expressed as

$\vec z==$

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

$\vec r_1==\ km=\ m$

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

$\vec r_2==\ m$

The total displacement is

$\vec r=\ m+\ m$

$\vec r=\ m$

In (magnitude,angle) form:

$|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m$

$\boxed{|\vec r|=339.82\ m}$

$\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169$

$\boxed{\theta=-6.67^o}$

5 0

3 years ago

The PE of the box that is on a 2.0 m high self is 1600 J. What is the power expelled to lift the box to this height in 10.0 seco

Pani-rosa [81]

Answer:

160 W

Explanation:

Power is the ratio of work to time:

(1600 J)/(10 s) = 160 J/s = 160 W

7 0

2 years ago

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes th

Sonbull [250]

Answer:

V = 331.59m/s

Explanation:

First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.

S = ut + 1/2at²

Given height of the cliff S = 80m

initial velocity u = 0m/s²

a = g = 9.81m/s²

Substitute

80 = 0+1/2(9.81)t²

80 = 4.905t²

t² = 80/4.905

t² = 16.31

t = √16.31

t = 4.04s

Next is to get the vertical velocity

Vy = u + gt

Vy = 0+(9.81)(4.04)

Vy = 39.6324

Also calculate the horizontal velocity

Vx = 1330/4.04

Vx = 329.21m/s

Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.

V² = Vx²+Vy²

V² = 329.21²+39.63²

V² = 108,379.2241+1,570.5369

V² = 109,949.761

V = √ 109,949.761

V = 331.59m/s

Hence the speed of the shell as it hits the ground is 331.59m/s

7 0

2 years ago

Archaeologists use the radioactive decay of carbon-14 to estimate the age of fossils and can be used on samples of bone, cloth,

timama [110]

The answer should be 11,460 because the first half-life leaves 50 percent left and the next half-life would leave 25 percent which dates the bones at 11,460 years old.

8 0

2 years ago

Read 2 more answers

Ali is whirling a 2.0 kg bunch of bananas in a circular path having a radius of 0.50 m. The bananas complete 2 revolutions every

skad [1K]

Answer:

1) 2.467 N

2) a) 0.248m

b) 2.3π rad/sec

Explanation:

Given data:

mass of Banana bunch ( m ) = 2.0 kg

radius of circular path ( R ) = 0.5 m

number of revolutions completed = 2

Time to complete 2 revolutions = 6 seconds

1) Determine the force to keep the motion constant for one complete revolution in every 4 seconds

F = mv^2 / r ----- ( 1 )

where V = 2πR/T

where : R = 0.5 , m = 2, T = 4 seconds

Insert values into equation 1

F = 2 * 4π^2 * 0.5/4^2

= 2.467 N

2a) Calculate the maximum distance of coin from center

angular velocity ( w ) = v/r

coefficient of static friction ( μ ) = 0.25

$F_{c} = u mg$ ---- ( 1 )

mv^2/r = μmg --- ( 2 ) cancelling the mass on both sides eqn 2 becomes

v^2 = μ*g*r

dividing both sides of equation by r^2

w^2 = μ*g/r

hence determine distance ( r ) of coin from center

r = 0.25 * 9.81 / π^2 = 0.248 m

2b ) determine the maximum speed of rotation of the turntable for the coin to move relative to the turntable without slipping

distance coin is placed ( r ) = 4.7 cm = 0.047 m

find speed of rotation ( w )

w^2 = μ*g/r

w = √ 0.25 * 9.81/ 0.047

= 7.2236 rad/secs ≈ 2.3π rad/sec

3 0

2 years ago