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3 years ago

Object a travels in the +x-direction before hitting a stationary object

2 answers:

8 0

The object’s resultant angle of motion with the +x-axis after the collision is 47°

From object A:

1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.

Now, we know, tanФ = $\frac{y}{x}$

⇒tanФ = $\frac{6.2 × 10^4 }{5.7 × 10^4}$

⇒tanФ = 1.088

⇒ Ф = $tan^{-1}$ 1.088
= 47.4 ≈ 47

barxatty [35]3 years ago

8 0

Answer: C

Explanation:

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A swimming pool measures 5.0 m long x 4.0 m wide x 3.0 mdeep.

QveST [7]

Answer:

a) 588,000 N

b) 294000 N

Explanation:

Given that

Density of water = 1000kg/m3

(g) = 9.8m/s2

volume is given as (V)= 5m*4m*3m

a) force will be equal to weight of water

$W = mg = \rho g V = 1000\times 9.8 \times (5*4*3) = 588,000 kg m/s^2$

b) at either end

$P = \frac{F}{A} = P_o + \rho gh$

$dF = PdA$

$dF = \rho g w \int h dh$

$F = \rho g w \frac{h^2}{2}$

$F = \rho g A \frac{h}{2}$ [A = wh]

$F = 1000\times 9.8 \times 5\times 4 \times\frac{3}{2}$

F = 294000 N

3 0

3 years ago

Throw a paper ball towards a hard wall.

Len [333]

The answer is d hope this helps

7 0

3 years ago

Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o

Ierofanga [76]

C. Sample C would be best, because the percentage of the energy
in an incident wave that remains in a reflected wave from this material
is the smallest.

The coefficient of absorption is the percentage of incident sound
that's absorbed. So the highest coefficient of absorption results in
the smallest percentage of the energy in an incident wave that remains.
That's what you want.

6 0

3 years ago

Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially

mixer [17]

Answer:

D. The momentum of Car B is three times as great in magnitude as that of car A.

Explanation:

I majored in Physics

3 0

3 years ago

What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t

Gekata [30.6K]

Start by facing East. Your first displacement is the vector

d₁ = (225 m) i

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

d₂ = (350 m) j

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

d₃ = (125 m) (cos(60º) i + sin(60º) j )

d₃ ≈ (62.5 m) i + (108.25 m) j

The net displacement is

d = d₁ + d₂ + d₃

d ≈ (287.5 m) i + (458.25 m) j

and its magnitude is

|| d || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

7 0

3 years ago