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sattari [20]
3 years ago
7

A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v0 = 22 m/

s at an angle θ = 23 ° below horizontal. The stone moves without air resistance; use a Cartesian coordinate system with the origin at the stone's initial position.

Physics
1 answer:
Alina [70]3 years ago
5 0

Answer:

V = 25.3 , θf = 36.7° below the horizontal

Explanation:

Given Vo = 22m/s , θ = 23°

Vx = VoCosθ

Vy = VoSinθ – gt

t = total time of flight

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A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i
Veseljchak [2.6K]

a. The magnitude of the change of the momentum of the 5-kg ball is equal to the magnitude of the change of momentum of the 10-kg ball.

c. The magnitude of the change of velocity the 5-kg ball experiences is greater than that of the 10-kg ball.

Explanation:

For an inelastic collision:

  • The total momentum of the system is conserved
  • The total kinetic energy of the system is not conserved

Using these facts, let's now analyze each statement given.

a. The magnitude of the change of the momentum of the 5-kg ball is equal to the magnitude of the change of momentum of the 10-kg ball.  --> TRUE. Since the total momentum is conserved, we can write:

p_1 = p_1'+p_2'

where

p_1 is the initial momentum of the 5-kg ball

p_1' is the final momentum of the 5-kg ball

p_2' is the final momentum of the 10-kg ball

The equation can be rewritten as

p_1-p_1'=p_2'

which is equivalent to

-\Delta p_1 = \Delta p_2

which means that the magnitude of the change of momentum of the two balls is the same.

b. Both balls lose all their momentum since the collision is inelastic.  --> FALSE, the 10-kg ball gains momentum, so it does not lose it.

c. The magnitude of the change of velocity the 5-kg ball experiences is greater than that of the 10-kg ball.  --> TRUE. We already said that the magnitude of the change in momentum of the two balls is the same. However, it can be written as

\Delta p = m\Delta v

where m is the mass of the ball and \Delta v its change in velocity. Therefore, the 5-kg ball (which has smaller mass) will have a larger \Delta v, so a larger change in velocity.

d. The magnitude of the change of velocity the 5-kg ball experiences is equal to that of the 10-kg ball.  --> FALSE, as we discussed in c).

e. The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball. --> FALSE, as we discussed in c).

Learn more about change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

6 0
4 years ago
Any help please, physics is not my strong suit?
frez [133]

Answer:

2000mg = 2g

5L = 5000mL

16cm = 160mm

5 0
3 years ago
Read 2 more answers
find the average velocity of a bicycle that starts 100km south and is 120km south of town after 0.4 hours​
dusya [7]

Answer:

The average velocity is 50 km/h south

Explanation:

The average velocity of an object is its total displacement divided by

the total time taken.

That means it is the rate at which an object changes its position from

one place to another.

Average velocity is a vector quantity.

The SI unit is meters per second.

A bicycle that starts 100 km south and is 120 km south of town after

0.4 hour​.

The displacement = 120 - 100 = 20 km south

The time = 0.4 hour

The average velocity = \frac{D}{T}, where D is the displacement

and t is the time

The average velocity of the bicycle = \frac{20}{0.4}=50 km/h

<em>The average velocity is 50 km/h south</em>

If you want it in meter per second, change the kilometer to meter

and change the hour to seconds

1 km = 1000 m

1 hour = 60 × 60 = 3600 seconds

The average velocity of the bicycle = \frac{50(1000)}{3600}=13.89 m/s south

5 0
3 years ago
a 70 kg man standing on ice throws a 3 kg body horizontally at 8 m/s. the friction coefficient between the ice and his feet is 0
GalinKa [24]

The distance at which the man slips is 0.3 m

Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.

Given-

mass of man= 70 kg

frictional coefficient μ=0.02

mass of body thrown= m2 = 3kg

let s be the stopping distance

we know that frictional force = F= μN

=μMg= 0.02 x 70 x 10

=14 N

∴acceleration, a= 14/70 = 0.2 m/s²

now on applying conservation of linear momentum

pi=pf            pi=0 (initially at rest)

0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s

v1= m2v2 /m1= 0.3 m/s

we know,

v²- u² = -2as

0- (0.3) ²= -2 x 0.2 x 5

s= 0.09/0.4 ≈ 0.3 m

Learn more about distance here-

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6 0
2 years ago
What is the minimum force require to move a 5kg wooden crate on a wooden floor?
kolbaska11 [484]

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with <em>µ</em>. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ <em>F</em> = <em>p</em> - <em>f</em> = 0

• net vertical:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

where

<em>p</em> = magnitude of the <u>p</u>ushing force

<em>f</em> = mag. of <u>f</u>riction

<em>n</em> = mag. of the <u>n</u>ormal force

<em>w</em> = <u>w</u>eight of the crate

The second equation gives

<em>n</em> = <em>w</em> = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of <em>µ</em>, so

<em>f</em> = <em>µ</em> (49 N) = 49<em>µ</em> N

To overcome static friction, the push has to exceed this in magnitude, so that

<em>p</em> > 49<em>µ</em> N

For instance, if <em>p</em> = 0.25, then <em>p</em> would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

7 0
3 years ago
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