The tendency for objects to resist acceleration is called :

The number of employees for a certain company has been decreasing each year by 4%. If the company currently has 650 employees an

nika2105 [10]

The answer would probably be 649.96

8 0

3 years ago

MOST elements on the periodic table are

lidiya [134]

Answer:

B

Explanation:

6 0

3 years ago

Read 2 more answers

Which piece of evidence did Alfred Wegener use to develop the theory of continental drift?

Eduardwww [97]

The right answer for the question that is being asked and shown above is that: "A.tectonic activity concentrated in certain areas." A piece of evidence did Alfred Wegener use to develop the theory of continental drift is that A.tectonic activity concentrated in certain areas

3 0

3 years ago

If the acceleration of a motorboat is 4.0 m/s2, and the motorboat starts from rest, what is its velocity after 6.0 s?

umka2103 [35]

Answer:

The velocity of the motorboat after 6s is 24 m/s.

Explanation:

Given;

acceleration of the motorboat, a = 4.0 m/s²

initial velocity of the motorboat, u = 0

time of motion of the motorboat = 6s

Apply the following kinematic equation to determine the velocity of the motorboat after 6 ;

v = u + at

v = 0 + (4 x 6)

v = 24 m/s

Therefore, the velocity of the motorboat after 6s is 24 m/s.

6 0

3 years ago

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

bulgar [2K]

Answer:

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

Explanation:

Law Of Conservation Of Linear Momentum

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