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Elenna [48]
3 years ago
12

What equation illustrates the law of conservation of matter?

Physics
1 answer:
scoray [572]3 years ago
3 0
 <span>In a correctly balanced chemical equation, there are exactly the same number of atoms for each element found on both side of the equation (that is, in the reactants and the products). 

No new atoms were made. None were destroyed. Matter was conserved.</span>
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Car A has twice the mass of Car B; both travel at the same speed. Compared to Car B. Car A has:
krek1111 [17]

Answer:

Car A has twice the energy of car B.

Explanation:

The kinetic energy of an object is given by :

K=\dfrac{1}{2}mv^2

Where

m is the mass

v is the speed of the object

Car A has twice the mass of Car B. Both travel at the same speed i.e.

m_A=2m_B

So,

\dfrac{K_A}{K_B}=\dfrac{(1/2)m_Av^2}{(1/2)m_Bv^2}\\\\=\dfrac{2m_B}{m_B}\\\\\dfrac{K_A}{K_B}=2\\\\K_A=2\times K_B

So, the kinetic energy of car A is twice of the kinetic energy of car B.

4 0
3 years ago
Read 2 more answers
Express 1010 miners in examiners. Answer in units of Eminer.
Lady_Fox [76]
Since 1 examiner = 10^18miner
then 10^10miners =10^(10-18)=10^-8examiner
7 0
3 years ago
A car travelled 60 km west in 1.33 h along a straight line, then returned back east for 40 km in 0.67 h. Its average velocity is
Sholpan [36]

Answer:

Option A

Explanation:

Velocity is expressed as distance covered per unit time, with respect to direction. Therefore, v=d/t

Given distance west as 60 km and time as 1.33 then velocity will be

V=60/1.33=45.112781954887 km/h

Rounded off as 45.11 km/h West

Velocity in East will also be given by substituting 40 km for d and 0.67 h for h hence

V=40/0.67=59.701492537313 km/h rounded off as 59.70 km/h East

Taking East as positive then West as negative, the sum of two velocities will be (59.70+-45.11)/2=7.295 km/h East

Approximately 10 km/h East since it is positive

7 0
3 years ago
When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and a
ryzh [129]

Complete question:

The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?

Answer:

The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

Explanation:

Given;

combined mass of the shotgun and arm–shoulder, m₁ = 15 kg

mass of the projectile, m₂ = 0.04 kg

speed of the projectile, u₂ = 380 m/s

let the recoil velocity of the shotgun and arm–shoulder combination = u₁

Apply the principle of conservation of linear momentum;

m₁u₁  +  m₂u₂ = 0

m₁u₁ = - m₂u₂

u_1 = -\frac{m_2u_2}{m_1} \\\\u_1 = - \frac{0.04\times 380}{15} \\\\u_1 =-1.013 \ m/s\\\\u_1 = 1.013 \ m/s \ \ \ in \ opposite \ direction

Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s

3 0
3 years ago
A cruise ship made a trip to Guam and back. The trip there took 12 hours and the trip
vekshin1
15 because i did the math, had this assignment, and also watch a video
6 0
3 years ago
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