Which liquid has a higher density; shampoo or rubing (isopropyl) alcohol?

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim

Nonamiya [84]

Answer:

The distance covered by puck A before collision is $z = 8.56 \ m$

Explanation:

From the question we are told that

The label on the two hockey pucks is A and B

The distance between the two hockey pucks is D 18.0 m

The speed of puck A is $v_A = 3.90 \ m/s$

The speed of puck B is $v_B = 4.30 \ m/s$

The distance covered by puck A is mathematically represented as

$z = v_A * t$

=> $t = \frac{z}{v_A}$

The distance covered by puck B is mathematically represented as

$18 - z = v_B * t$

=> $t = \frac{18 - z}{v_B}$

Since the time take before collision is the same

$\frac{18 - z}{V_B} = \frac{z}{v_A}$

substituting values

$\frac{18 -z }{4.3} = \frac{z}{3.90}$

=> $70.2 - 3.90 z = 4.3 z$

=> $z = 8.56 \ m$

8 0

3 years ago

A concrete column has a diameter of 380 mm and a length of 2.6 m . if the density (mass/volume) of concrete is 2.45 mg/m3, deter

grigory [225]

The volume of the column is

(π) · (r²) · (length) =

(π) · (0.19 meter)² · (2.6 meters) =

(π) · (0.036 m²) · (2.6 m) =

0.294 m³ .

The density is 2,450 kg/m³ (VERY very dense, heavy concrete)

so the weight of the column is (mass)·(gravity) or

(density) · (volume) · (gravity) =

(2,450 kg/m³) · (0.294 m³) · (9.81 m/s²) =

(2,450 · 0.294 · 9.81) (kg · m³· m) / (m³ · s²) =

7,066 kg-m/s² = 7,066 Newtons .

But 9.81 Newtons = 2.20462 pounds on Earth (the weight of 1 kilogram of mass), so we have

(7,066 N) · (2.205 pound/9.81 N) =

(7,066 · 2.205 / 9.81) pounds =

1,588 pounds .

5 0

3 years ago

The strong interaction is necessary for the formation of which of the following?

sleet_krkn [62]

Answer:

Electrons and protons

5 0

3 years ago

A bullet with mass m = 5.21 g is moving horizontally with a speed v = 443 m/s when it strikes a block of hardened steel with mas

AlladinOne [14]

Answer:

0.312 m/s

Explanation:

Elastic collisions conserve momentum and kinetic energy

The velocity of the center of mass will not change. It continues at

0.00521(443) / 14.80521 = 0.155893... ≈ 0.156 m/s

To conserve kinetic energy we can think of the center of mass (CoM) as an ideal spring returning to each mass that strikes it an identical speed of collision in the opposite direction.

The CoM sees the target approach at - 0.156 and will see it depart at 0.156 m/s

A ground based observer sees the target depart at the velocity of the CoM plus the relative velocity .

v = 0.156 + 0.156 = 0.312 m/s

6 0

3 years ago

A stone is dropped from the upper observation deck of a tower, 250 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dista

Vitek1552 [10]

(a) $y(t)=250 - 4.9 t^2$

For an object in free-fall, the vertical position at time t is given by:

$y(t) = h + ut - \frac{1}{2}gt^2$

where

h is the initial vertical position

u is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

In this problem,

h = 250 m

u = 0 (the stone starts from rest)

So, the vertical position of the stone is given by

$y(t) = 250 - \frac{1}{2}(9.8) t^2 = 250 - 4.9 t^2$

(b) 7.14 s

The time it takes for the stone to reach the ground is the time t at which the vertical position of the stone becomes zero:

y(t) = 0

Which means

$y(t) = h - \frac{1}{2}gt^2=0$

So for the stone in the problem, we have

$250 - 4.9 t^2 = 0$

Solving for t, we find:

$t=\sqrt{\frac{250}{4.9}}=7.14 s$

(c) -70.0 m/s (downward)

The velocity of an object in free fall is given by the equation

$v(t) = u - gt$

where

u is the initial velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have

u = 0

So if we substitute t = 7.14 s, we find the velocity of the stone at the time it reaches the ground:

$v=0-(9.8 m/s^2)(7.14 s)=-70.0 m/s$

The negative sign means the direction of the velocity is downward.

(d) 6.94 s

In this situation, the stone is thrown downward with an initial speed of 2 m/s, so its initial velocity is

u = -2 m/s

So the equation of the vertical position of the stone in this case is

$y(t) = h + ut - \frac{1}{2}gt^2=250 - 2t - 4.9 t^2$

By solving the equation, we find the time t at which the stone reaches the ground.

We find two solutions:

t = -7.35 s

t = 6.94 s

The first solution is negative, so it has no physical meaning, therefore we discard it. So, the time it takes for the stone to reach the ground is:

t = 6.94 s

6 0

3 years ago