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-Dominant- [34]

2 years ago

13

Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a

nd Jackson’s is 54.0 kg. Jackson is running at a velocity of 3.0 m/s. What must Tammy’s minimum velocity be in order to stop forward momentum?

1 answer:

Naily [24]2 years ago

8 0

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

$m_1v_1 + m_2v_2 = 0$

final they both comes to rest so here we can say that final momentum must be zero

now we have

$34 v + 54 (3 m/s) = 0$

$v = -4.76 m/s$

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your drop a coin from the top of a hundred-story building(1000m). If you ignore air resistance, how fast will it be falling righ

Ede4ka [16]

consider the motion of con from top to bottom

Y = vertical displacement = 1000 m

a = acceleration due to gravity = 9.8 m/s²

v₀ = initial velocity at the top = 0 m/s

v = final velocity at the bottom = ?

using the kinematics equation

v² = v²₀ + 2 aY

v² = 0² + 2 (9.8) (1000)

v = 140 m/s

t = time taken to hit the ground

Using the equation

v = v₀ + at

140 = 0 + 9.8 t

t = 14.3 sec

7 0

2 years ago

Read 2 more answers

A grasshopper jumps at a 65.0 degree angle at 5.42m/s. At what time does it reach its maximum height?

crimeas [40]

When the grasshoppers vertical velocity is exactly zero.
v = -g•t + v0.
v: vertical part of velocity. Is zero at maximum height.
g: 9.81
t: time you are looking for
v0: initial vertical velocity
Find the vertical part of the initial velocity, by using the angle at which the grasshopper jumps.

6 0

2 years ago

Read 2 more answers

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed

$\alpha = 0.745 rad/s2 \rightarrow$ The angular acceleration

$r = 4.65 m \rightarrow$ The distance

The relation between the linear velocity and angular velocity is

$v = r\omega$

Where,

r = Radius

$\omega =$ Angular velocity

At the same time we have that the centripetal acceleration is

$a_c = \frac{v^2}{r}$

$a_c = \frac{(r\omega)^2}{r}$

$a_c = \frac{r^2\omega^2}{r}$

$a_c = r \omega^2$

$a_c = (4.65 )(1.25 rad/s)^2$

$a_c = 7.265625 m/s^2$

Now the tangential acceleration is given as,

$a_t = \alpha r$

Here,

$\alpha =$ Angular acceleration

r = Radius

$\alpha = (0.745)(4.65)$

$\alpha = 3.46425 m/s^2$

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

$|a| = \sqrt{a_c^2+a_t^2}$

$|a| = \sqrt{(7.265625)^2+(3.46425)^2}$

$|a| = 8.049 m/s^2 \approx 8.05 m/s2$

Therefore the correct answer is C.

7 0

2 years ago

A spring balance consists of a pan that hangs from a spring. A damping force Fd = −bv is applied to the balance so that when an

Citrus2011 [14]

Answer:

b ≈ 64 Kg/s

Explanation:

Given

Fd = −bv

m = 2.5 kg

y = 6.0 cm = 0.06 m

g = 9.81 m/s²

The object in the pan comes to rest in the minimum time without overshoot. this means that damping is critical (b² = 4*k*m).

m is given and we find k from the equilibrium extension of 6.0 cm (0.06 m):

∑Fy = 0 (↑)

k*y - W = 0 ⇒ k*y - m*g = 0 ⇒ k = m*g / y

⇒ k = (2.5 kg)*(9.81 m/s²) / (0.06 m)

⇒ k = 408.75 N/m

Hence, if

b² = 4*k*m ⇒ b = √(4*k*m) = 2*√(k*m)

⇒ b = 2*√(k*m) = 2*√(408.75 N/m*2.5 kg)

⇒ b = 63.9335 Kg/s ≈ 64 Kg/s

5 0

2 years ago

A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylin

Leya [2.2K]

Answer:

v = 13.19 m / s

Explanation:

This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical

X axis

N = m a

Centripetal acceleration is

a = v² / r

Y Axis

fr -W = 0

fr = W

The force of friction is

fr = μ N

Let's calculate

μ (m v² / r) = mg

μ v² / r = g

v² = g r / μ

v = √ (g r /μ)

v = √ (9.8 11 / 0.62)

v = 13.19 m / s

7 0

2 years ago