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2 years ago

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A soccer ball is kicked horizontally off a 25 m high cliff and lands at a distance of 40.0 m from the edge of the cliff. Determi

ne the initial horizontal velocity of the soccer ball.

1 answer:

tankabanditka [31]2 years ago

8 0

Answer:

1

Explanation:

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3 years ago

A hockey puck with a mass of 0.16 kg travels at a velocity of 40 m/s toward a goalkeeper. The goalkeeper has a mass of 120 kg an

Ainat [17]

Answer:

Explanation:

Momentum is the product of mass of a body and its velocity.

Given the mass of the puck m1 = 0.16kg

velocity of the puck v1 = 40m/s

Given the mass of the goalkeeper m2 = 120kg

velocity of the goalkeeper v2= 0m/s (goal keeper at rest)

The total momentum of the goalkeeper and puck after the puck is caught by the goalkeeper is expressed as:

m1v1 + m2v2 (their momentum will be added since they collide)

= 0.16(40) + 120(0)

= 0.16(40) + 0

= 6.4kgm/s

Let us calculate their common velocity using the conservation of momentum formula;

m1u1 + m2u2 = (m1+m2)v

6.4 = (0.16+120)v

6.4 = 120.16v

v = 6.4/120.16

v = 0.053m/s

Hence after collision, both objects move at a velocity of 0.053m/s

Momentum of the puck after collision = m1v

Momentum of the puck after collision = 0.16*0.053m/s

Momentum of the puck after collision = 0.0085kgm/s

Momentum of the keeper after collision = m2v

Momentum of the keeper after collision = 120*0.053m/s

Momentum of the keeper after collision = 6.36kgm/s

From the calculation above, it can be seen that the keeper has the greater momentum after the puck was caught since the momentum of the keeper after collision is greater than that of the puck

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2 years ago

A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00

ANTONII [103]

Answer:

a) 200m, 100m/s

b) 710.20m

c) -117.98 m/s

d) 26.24 s

Explanation:

To solve this we have to use the formulas corresponding to a uniformly accelerated motion problem:

$V=Vo+a*t$ (1)

$X=Xo+Vo*t+\frac{1}{2}*a*t^2\\$ (2)

$V^2=Vo^2+2*a*X$ (3)

where:

Vo is initial velocity

Xo=intial position

V=final velocity

X=displacement

a)

$X=0+0*4+\frac{1}{2}*25*4^2$

the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.

$X=200m$

$V=0+25*4$

$V=100m/s$

b)

The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8 $\frac{m}{s^2}$ because it is going downward.

$0=100^2+2*(-9.8)*X\\X=510.20\\totalheight=200+510.20=710.20m$

c)

In order to find the velocity when it crashes, we can use the formula (3).

the initial velocity is 0 because in that moment is starting to fall.

$V^2=0^2+2*(-9.8)*(710.20)\\V=-117.98 m/s$

the minus sign means that the object is going down.

d)

We can find the total amount of time adding the first 4 second and the time it takes to going down.

to calculate the time we can use the formula (2) setting the reference at 200m:

$-200=0+100*t+\frac{1}{2}*(-9.8)*t^2$

solving this we have: time taken= 22.24 seconds

total time is:

total=22.24+4=26.24 seconds.

3 0

2 years ago