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ANTONII [103]
3 years ago
11

Can someone define Newton's 1st 2nd and 3rd law​

Physics
1 answer:
Mrrafil [7]3 years ago
7 0

Answer:

<u>HEY</u><u> </u><u>THERE</u><u>!</u><u> </u><u>HERE</u><u>'S</u><u> YOUR</u><u> ANSWER</u><u>!</u><u> </u>

Explanation:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

<u>HOP</u><u>E</u><u> THIS</u><u> HELPS</u><u> YOU</u><u> BUDDY</u><u>!</u>

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A 250. mL sample of gas at 1.00 atm and 20.0°C has the temperature increased to 40.0°C and the volume increased to 500. mL. What
ladessa [460]

Answer:

New pressure is 0.534 atm

Explanation:

Given:

Initial volume of the gas, V₁ = 250 mL

Initial pressure of the gas, P₁ = 1.00 atm

Initial temperature of the gas, T₁ = 20° C = 293 K

Final volume of the gas, V₂ = 500 mL

Final pressure of the gas = P₂

Final temperature of the gas, T₁ = 40° C = 313 K

now,

we know for a gas

PV = nRT

where,

n is the moles

R is the ideal gas constant

also, for a constant gas

we have

(P₁V₁/T₁) = (P₂V₂/T₂)

on substituting the values in the above equation, we get

(1.00 × 250)/293 = (P₂ × 500)/313

or

P₂ = 0.534 atm

Hence, the <u>new pressure is 0.534 atm</u>

5 0
4 years ago
. A toy rocket has a mass of 350 g at launch. The force it produces
nydimaria [60]

Answer:

the acceleration is reduced by gravity

a = (15 / .35) - [9.8 * sin(65º)]

Explanation:

break the launch vector into two components, vertical and horizontal

Force Net Vertical=-9.8*.350+15cos65 N

force net horizonal=15sin65

initial acceleration= force/mass= (-9.8+15/.350*cos65)j+(15/.350*sin65)i

using i,j vectors..

5 0
3 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
4 years ago
The electromagnetic waves that have the lowest frequency is called
Luda [366]
Radar waves are the waves with the lowest energy.
5 0
3 years ago
A firecracker breaks up into three pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0
Mamont248 [21]

Answer:

|P_3|=21.242m/s

Explanation:

From the question we are told that:

Mass m=200g=0.2

Speed v=82.0m/s

Mass M_2=300g=0.3kg

Speed V_2=45.0m/s

Generally the equation for Magnitude of the Third piece is mathematically given by

P_1+P_2+P_3=0

|P_3|=|P_1+P_2|

Where

P_1=m_1v_1

P_1=0.2*82

P_1=16.4kgm/s

And

P_2=0.3*45

P_2=13.5kgm/s

Therefore

|P_3|=\sqrt{16.4^2+13.5^2}

|P_3|=21.242m/s

8 0
3 years ago
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