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g100num [7]
3 years ago
9

BRAINLIESTTTT!!!!! THIS IS DUE IN 5 MINUTES!​

Physics
1 answer:
Gnesinka [82]3 years ago
8 0

Equation: -10 40/4^2

Explanation: I multiplied the -10, but idk if that' what you do cause I've never done this. So if you do do that it's -1000, but that is probably wrong. Sorry, I tried.

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B) Calculate the net work required to bring a 1300 kg car moving at 30 m/s to rest.
prohojiy [21]

I Lolo my name is keshav and

5 0
3 years ago
billiardballsnew2 A white billiard ball with mass mw = 1.49 kg is moving directly to the right with a speed of v = 3.09 m/s and
Kobotan [32]

Answer:

Final velocity of white ball is 0m/s

Final velocity of black ball is 3.09m/s

Explanation:

An elastic collision is one that conserves internal kinetic energy

An internal kinetic energy is the sum of kinetic energies of objects in the system

Initial kinetic energy of white ball is Vi1 = 3.09m/s

Final kinetic energy of white ball is Vf1 = ?

Initial kinetic energy of black ball is Vi2 = 0m/s

Final kinetic energy of black ball is Vf2 = ?

m1 = 1.49kg mass of white ball

m2 = 1.49kg mass of black ball

The formula to calculate internal kinetic energy is

1/2m1Vf1^2 + 1/2m2Vf2^2 = 1/2m1Vi1^2

Solving the equation

1.Vf1 = (m1 - m2)Vi1/m1+m2

Vf1 = (1.49-1.49)*3.09/1.49+1/49

Vf1 = 0m/s

2. Vf2 = 2m1Vi1/m1+m2

Vf2 = 2*1.49*3.09/1.49+1.49

Vf2 = 3.09m/s

N:B following the general principle of collision when 2 bodies of same masses collide in elastic collision they exchange velocities.

7 0
4 years ago
It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet
Amiraneli [1.4K]

Answer:

354.72 m/s

Explanation:

m = mass of lead bullet

c = specific heat of lead = 128 J/(kg °C)

L = Latent heat of fusion of lead = 24500 J/kg

T_{i} = initial temperature = 27.4 °C

T_{f} = final temperature = melting point of lead =  327.5 °C

v = Speed of lead bullet

Using conservation of energy

Kinetic energy of bullet = Heat required for change of temperature + Heat of melting

(0.5) m v^{2} = m c (T_{f} - T_{i}) + m L\\(0.5) v^{2} = c (T_{f} - T_{i}) + L\\(0.5) v^{2} = (128) (327.5 - 27.4) + 24500\\(0.5) v^{2} = 62912.8\\v = 354.72 ms^{-1}

3 0
3 years ago
Calculate the heat energy conducted per hour through the side walls of a cylindrical steel
Maru [420]

Explanation:

heat caoacity and heat is difference

7 0
3 years ago
Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1
Nonamiya [84]

Answer:

a_x=6\ \text{m/s}^2 and a_y=0\ \text{m/s}^2

Magnitude of accleration is 6\ \text{m/s}^2 and the direction is 0^{\circ}

Explanation:

t_1=2\ \text{s}

v_x=1\ \text{m/s}

v_y=3\ \text{m/s}

t_2=2.5\ \text{s}

v_x=4\ \text{m/s}

v_y=3\ \text{m/s}

Average acceleration in the different axes

a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2

a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2

The components of the acceleration is a_x=6\ \text{m/s}^2 and a_y=0\ \text{m/s}^2

The magnitude of acceleration

a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2

Direction

\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}

The magnitude of accleration is 6\ \text{m/s}^2 and the direction is 0^{\circ}.

7 0
3 years ago
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