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3 years ago

10m= (5.0) + (.5)(9.8)(5.0)

Physics

1 answer:

Norma-Jean [14]3 years ago

8 0

M=2.45 because you multiply out the equation on the right and divide by 10

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what occurs when force is applied over a certain distance and motion is in the same direction as the force?

faltersainse [42]

An object in motion will continue to move in the same direction and with the same speed unless acted upon by an unbalanced force. states that forces occur as equal and opposite pairs. The strength of the force is related to the mass of the objects and the distance between them.

8 0

3 years ago

Read 2 more answers

-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a

bulgar [2K]

Answer:

<em>The velocity of the carts after the event is 1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

$\displaystyle v'=\frac{2*5+8*0}{2+8}=\frac{10}{10}=1$

The velocity of the carts after the event is 1 m/s

3 0

2 years ago

A 1.80-m string of weight 0.0126 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

Veseljchak [2.6K]

Answer:

W = 0.135 N

Explanation:

Given:

- y (x, t) = 8.50*cos(172*x -2730*t)

- Weight of string m*g = 0.0126 N

- Attached weight = W

Find:

The attached weight W given that Tension and W are equal.

Solution:

The general form of standing mechanical waves is given by:

y (x, t) = A*cos(k*x -w*t)

Where k = stiffness and w = angular frequency

Hence,

k = 172 and w = 2730

- Calculate wave speed V:

V = w / k = 2730 / 172 = 13.78 m/s

- Tension in the string T:

T = Y*V^2

where Y: is the mass per unit length of the string.

- The tension T and weight attached W are equal:

T = W = Y*V^2 = (w/L*g)*V^2

W = (0.0126 / 1.8*9.81)*(13.78)^2

W = 0.135 N

4 0

3 years ago

a shell fired from a cannon at 60 ° from horizontal strikes a target 20m high at a distance 80m. Calculate the initial velocity

stich3 [128]

consider the motion along the X-direction

X = horizontal displacement = 80 m

$V_{ox}$ = initial velocity along the x-direction = v Cos60

t = time of travel

using the equation

X = $V_{ox}$ t

80 = (v Cos60) (t)

t = 160/v eq-1

consider the motion in vertical direction :

Y = vertical displacement = 20 m

$V_{oy}$ = initial velocity in Y-direction = v Sin60

a = acceleration = - 9.8 m/s²

t = time of travel = 160/v

using the equation

Y = $V_{oy}$ t + (0.5) a t²

20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²

v = 32.5 m/s

4 0

3 years ago

It's not D

DanielleElmas [232]

A. it provides support

5 0

2 years ago