An average force of 328 N acts for a time interval of 0.05s on a golf ball

Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while block B , with a mass of 8.0 kg, is moving in the opposi

Anon25 [30]

Consider velocity to the right as positive.

First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right

Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left

Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s

Let v (m/s) be the velocity of the center of mass of the 2-block system.

Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s

Answer:
The center of mass is moving at 1.33 m/s to the left.

5 0

3 years ago

How do defy gravity?

kaheart [24]

Exert force upward.
Like when you pick something up from the floor, or walk up the stairs.

6 0

3 years ago

Why do elements need to be heated to emit color?

kirill115 [55]

Light carries away the energy. At a room temperature they are at their lowest energy

7 0

3 years ago

Elect the correct answer. Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of laun

allsm [11]

Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction $= u\cos\theta$

and the speed in the vertical direction is $= u\sin\theta$ upward.

For A:

The speed in the horizontal direction $= u\cos75^{\circ}$

and the speed in the vertical direction is $= u\sin75^{\circ}$ upward.

For B:

The speed in the horizontal direction $= u\cos15^{\circ}$

and the speed in the vertical direction is $= u\sin15^{\circ}$ upward.

Let $t_A$ and $t_B$ are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A $= u\cos75^{\circ}\times t_A\cdots(i)$

The range for the projectile B = $u\cos15^{\circ}\times t_B\cdots(ii)$

At the highest point, the vertical velocity is 0.

Bu using the equation of motion $v^2=u^2 +2a s.$

Here, the final velocity v=0, the initial velocity $u = u \sin \theta$ , h= vertical distance up to the highest point, and $a= -g$ (as per sign convention).

So, $s= \frac{u^2\sin^2 \theta}{2g}$

For projectile A: The maximum height attained.

$s_A= \frac{u^2\sin^2 75^{\circ}}{2g}$

For projectile B: The maximum height attained.

$s_B= \frac{u^2\sin^2 15^{\circ}}{2g}$

As $\sin^2 75^{\circ} > \sin^2 15^{\circ}$ , the height of A is attained by A is more than the heigHt attained by B.

Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.

So, the total time of flight = 2 x (Time to reach the highest point)

In a similar way, by using the equation of motion v=u+at,

The time to reach the highest point $=\frac {u\sin\theta}{g}$

where g is the acceleration due to gravity.

So, the total time of flight

$= 2 \times \frac {u\sin\theta}{g}$

The total time of flight for A

$=2 \times \frac {u\sin75^{\circ}}{g}$

The total time of flight for A

$=2 \times \frac {u\sin15^{\circ}}{g}$

Now, from equations (i) and (ii),

The range for the projectile A =

$u\cos75^{\circ}\times \frac {2u\sin75^{\circ}}{g}=\frac {u^2 \sin 150^{\circ}}{g}= \frac {u^2 \sin 30^{\circ}}{g}$

The range for the projectile B =

$u\cos15^{\circ}\times \frac {2u\sin15^{\circ}}{g}=\frac {u^2 \sin 30^{\circ}}{g}.$

Both the projectile have the same range.

Hence, option (A) is correct.

7 0

3 years ago

Shortly after receiving a traffic ticket for speeding, Fred made numerous comments about the road signs being inadequate and is

k0ka [10]

Answer:

External locus of control

Explanation:

External locus of control is an attitude people possess that makes them attribute their failures or successes to factors other than themselves. The opposite of this type of attitude is the Internal locus of control where the individuals take responsibility for the outcomes of their actions whether good or bad. One good thing about the external locus of control is that when the individuals with this characteristic record successes, they attribute it to others and this presents them as people with team spirit. However, when they record failures, they do not want to take the blame, but rather attribute it to others.

Fred exhibits an external locus of control because he attributed his speeding to other factors like the road signs and GPS instead of fully admitting that it was his fault.

7 0

3 years ago