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Inessa05 [86]
3 years ago
7

Three laws that indicate a chemical change

Physics
1 answer:
alexandr1967 [171]3 years ago
4 0

Temperature, The highness, and the time.

Hope this helps!

C=

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Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.
Korolek [52]

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

\int\limits^{10}_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

4 0
4 years ago
A bullet whose mass is 30.2 g leaves a rifle with a muzzle velocity of 1,000 m/s. It strikes a block of wood (mass 5 kg), initia
AveGali [126]
This problem here is an example of inelastic collision where kinetic energy is not conserved but momentum is. We calculate as follows:

m1v1 + m2v2 = (m1 + m2)v3
v3 = m1v1 + m2v2  / m1 + m2
v3 = (30.2)(1000) + (5000)(0) / (30.2 + 5000)
v3 = 6.00 m/s
5 0
3 years ago
Is what flows and carries energy in a circuit
xenn [34]

The correct answer to this Electric current.

7 0
3 years ago
Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 2.6 m long, its mass is 0.5
Irina-Kira [14]

Answer:

15.13 m/s

Explanation:

The wave speed of the stretched rope can be calculated using the following formula

v = \sqrt{\frac{F_T}{\mu}}

where F_T = 44N is the tension on the rope and \mu = m/L = 0.5 / 2.6 = 0.1923 kg/m is the density of the rope per unit length

v = \sqrt{\frac{44}{0.1923}} = \sqrt{228.8} = 15.13 m/s

6 0
3 years ago
A mixture of helium and argon gas is expanded from a volume of 29.0L to a volume of 82.0L, while the pressure is held constant a
finlep [7]

Answer:

322 kJ

Explanation:

The work is the energy that a force produces when realizes a displacement. So, for a gas, it occurs when it expands or when it compress.

When the gas expands it realizes work, so the work is positive, when it compress, it's suffering work, so the work is negative.

For a constant pressure, the work can be calcutated by:

W = pxΔV, where W is the work, p is the pressure, and ΔV is the volume variation. To find the work in Joules, the pressure must be in Pascal (1 atm = 101325 Pa), and the volume in m³ (1 L = 0.001 m³), so:

p = 60 atm = 6.08x10⁶ Pa

ΔV = 82.0 - 29.0 = 53 L = 0.053 m³

W = 6.08x10⁶x0.053

W = 322x10³ J

W = 322 kJ

8 0
3 years ago
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