Name three mass units.! THANKS

A 2800 kg truck moving at 12 m/s to the right hits a stopped 1100 kg car. What is the combined velocity the moment they stick to

leva [86]

Answer:

The combined velocity is 8.61 m/s.

Explanation:

Given that,

The mass of a truck, m = 2800 kg

Initial speed of truck, u = 12 m/s

The mass of a car, m' = 1100 kg

Initial speed of the car, u' = 0

We need to find the combined velocity the moment they stick together. Let it is V. Using the conservation of momentum.

$m_1v_1+m_2v_2=(m_1+m_2)V\\\\V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\V=\dfrac{2800\times 12+0}{2800+1100}\\\\V=8.61\ m/s$

So, the combined velocity is 8.61 m/s.

5 0

3 years ago

What is the time period and frequency of a sound wave if it completes 500 vibrations in 50 seconds

Degger [83]

Answer:

10 vibrations per second

Explanation:

6 0

3 years ago

Capacitors C1 = 5.85 µF and C2 = 2.80 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon

posledela

Answer:

Q1_new = 515.68 µC

Q2_new = 246.82 µC

Explanation:

Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.

Then it is only necessary to calculate the charge on each capacitor:

Q1 = 5.85 µF * 250 V = 1462.5 µC

Q2 = 2.8 µF * 250 V = 700 µC

Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.

When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:

1462.5 µC - 700 µC = 762.5 µC

Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.

This 762.5 µC will be divided proportionately between the two capacitors.

Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC

Q2_new = 762.5 µC * (2.8/(5.85 + 2.8) = 246.82 µC

4 0

4 years ago

An elevator accelerates upward at 1.2 m/s 2 . the acceleration of gravity is 9.8 m/s 2 . what is the upward force exerted by the

Marysya12 [62]

We can find the force by using the following formula;
N = ma + mg
Fa = ma = 76 x 1.2 = 91.2
Fg = mg = 76 x 9.8 = 744.8
N = 91.2 + 744.8 = 836
So, the force is 836 N.

8 0

3 years ago

A force of 6.0 N is applied horizontally to a 3.0 kg crate initially at rest on a horizontal frictionless surface. After the cra

Savatey [412]

Answer:

a. Yes, because the acceleration of the crate is 2.0 m/s².

Explanation:

Given

$Force = 6N$ --- f