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3 years ago

PLEASE HELP WILL GIVE BRAINLIEST AND EXTRA POINTS!! Super simple I’ve just been sick so I have no clue what’s going on.

Physics

1 answer:

Tju [1.3M]3 years ago

4 0

Answer: 12) 1.07 m/s (right) 13) 4.05 m/s 14) 73 m/s 15) 10.9 m/s

Explanation:

12) Conservation of momentum. Momentum is the produce of mass and velocity.

13(2) + 15(-5) = 13(-5) + 15v

v = 1.06666... ≈ 1.07 m/s (right)

13) 18(9) + 22(0) = 18v + 22v

v = 18(9)/40 = 4.05 m/s

14) 0.65(35) + 0.08(0) = 0.65(26) + 0.08v

v = 73.125

15) This is a bit trickier. Let's ASSUME you jump off at 7 m/s relative to the truck. Doing this, we can assume that the reference frame is moving along with the truck at 10 m/s

the conservation of momentum equation becomes

600(0) + 80(0) = 600v + 80(-7)

v = 0.9333333... m/s

adding back the velocity of the reference frame means the truck is now traveling.

10.9333333... ≈ 10.9 m/s

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In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago

A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?

stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, $m_1 = 300\ g = 0.3\ kg$

Initial speed of the cart, $u_1=1.2\ m/s$

Mass of the larger cart, $m_2 = 2\ kg$

Initial speed of the larger cart, $u_2=0$

After the collision,

Final speed of the smaller cart, $v_1=-0.81\ m/s$ (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let $v_2$ is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+m_2u_2-m_1v_1=m_2v_2$

$v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}$

$v_2=0.301\ m/s$

So, the speed of the large cart after collision is 0.301 m/s.

4 0

3 years ago

Which characteristics of an area could best identify the type of biome it is

jeka94

Answer: average annual rainfall, average annual temperatures, types of plants and animals native to the area

Explanation: the best way you can identify a biome is by telling which animal or species are native to the certain area

6 0

2 years ago

A person has a mass of 60 kg. What is the person’s weight in Newtons and in pounds?

liubo4ka [24]

Answer:

137.2 in pounds and in Newton's it's 588.399

3 0

3 years ago

Accomplished silver workers in india can pound silver into incredibly thin sheets, as thin as 3.00 10-7 m (about one-hundredth o

postnew [5]

The density of silver is ρ = 10500 kg/m³ approximately.

Given:
m = 1.70 kg, the mass of silver
t = 3.0 x 10⁻⁷ m, the thickness of the sheet

Let A be the area.
Then, by definition,
m = (t*A)*ρ

Therefore
A = m/(t*ρ)
= (1.7 kg)/ [(3.0 x 10⁻⁷ m)*(10500 kg/m³)]
= 539.7 m²

Answer: 539.7 m²

8 0

3 years ago