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3 years ago

8

plastic bags have replaced for the packing of the most materials. give these advantages and two disadvantages of using plastic o

ver paper

1 answer:

Nezavi [6.7K]3 years ago

5 0

It is waterproof and more durable

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The velocity versus time graph of particle A is tangent to the velocity versus time graph for particle B at point O. What is the

Fudgin [204]

Answer: C. -1.16 meters/second2

Explanation:

A= v/t (velocity/time)

in this case: v=7 and t=6

So, A= 7/6

A=1.16

The graph is decreasing so accelleration would be negative

A= <u>-1.16 meters/second2</u>

<u>Option C!</u> ; )

<u></u>

3 0

2 years ago

You hold glider AA of mass 0.125 kgkg and glider BB of mass 0.375 kgkg at rest on an air track with a compressed spring of negli

andrew11 [14]

Answer:

The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction

Explanation:

Inelastic Collision

Given data

mass of glider A m1= 0.125kg

initial velocity u1=0

final velocity v1= 0.600 m/s

mass of glider B m2= 0.375kg

initial velocity u2=0

final velocity v2=?

We know that the expression for the conservation of momentum is given as

m1u1+m2u2=m1v1+m2v2

since u1=u2=u=0m/s

u(m1+m2)=m1v1+m2v2

substituting we have

0(0.125+0.0375)=0.125*0.6+0.375*v2

0=0.075+0.375v2

0.375v2=-0.075

v2=-0.075/0.375

v2=-0.2m/s

The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction

3 0

3 years ago

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

larisa [96]

Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

6 0

3 years ago

The magnetic field at the equator points north. If you throw a positively charged object (for example, a baseball with some elec

PilotLPTM [1.2K]

Answer:

The magnetic force points in the positive z-direction, which corresponds to the upward direction.

Option 2 is correct, the force points in the upwards direction.

Explanation:

The magnetic force on any charge is given as the cross product of qv and B

F = qv × B

where q = charge on the ball thrown = +q (Since it is positively charged)

v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)

B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)

F = qv × B = (+qvî) × (Bj)

F =

| î j k |

| qv 0 0|

| 0 B 0

F = i(0 - 0) - j(0 - 0) + k(qvB - 0)

F = (qvB)k N

The force is in the z-direction.

We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.

Hope this Helps!!!

5 0

3 years ago

9- Under what circumstances would a vector have components that are equal in

valkas [14]

Explanation:

c. if the vector is oriented at 0° from the X -axis.

6 0

2 years ago