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3 years ago

a car accelerates from 2 m/s to 28m/s at a constant rate of 3 m/s^2. How far does it travel while accelerating?

Physics

1 answer:

xxTIMURxx [149]3 years ago

3 0

Answer:

Distance, S = 130m

Explanation:

Given the following data;

Initial velocity = 2m/s

Final velocity = 28m/s

Acceleration = 3m/s²

To find the distance, we would use the third equation of motion.

V² = U² + 2aS

Substituting into the equation, we have;

28² = 2² + 2*3*S

784 = 4 + 6S

6S = 784 - 4

6S = 780

S = 780/6

Distance, S = 130m

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A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

Olegator [25]

Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

$m_1$ = Mass of block = 0.072 kg

$m_2$ = Mass of bullet = 4.67 g

$u_1$ = Initial Velocity of block = 0

$u_2$ = Initial Velocity of bullet = 629 m/s

$v_1$ = Final Velocity of block = 17 m/s

$v_2$ = Final Velocity of bullet

In this system the linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s$

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

$K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J$

The final kinetic energy

$K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J$

Initial Kinetic energy > Final kinetic energy

3 0

3 years ago

A 5 kg ball is sitting on the floor without moving. What is the momentum of the ball?

natita [175]

Answer:

The ball has no momentum

Explanation:

The given parameters are;

The mass of the ball = 5 kg

The velocity of the ball = 0 (The ball is sitting on the floor without moving)

The momentum of the ball = The mass of the ball × Velocity of the ball

Therefore, the momentum of the ball = 5 kg × 0 m/s = 0

The momentum of the ball is zero, the ball has no momentum.

5 0

2 years ago

If the final speed of an object as is strikes the ground is 77 m/s and it was in the air for 6.5 seconds. What was the initial d

azamat

Answer:

The answer to your question is: 13.2 m/s

Explanation:

final speed (fs) = 77 m/s

t = 6.5 s

gravity (g) = 9.81 m/s2

initial speed (is) = ?

Formula

fs = is + gt from this equation we clear "is" = fs - gt

Substitution is = 77 - (9,81)(6.5)

Process is = 77 - 63.8

is = 13.2 m/s

8 0

3 years ago

What was your train of thought as you navigated the picture of the candle?

defon

Answer:

Where is the picture

Explanation:

WHERE IS THE PICTURE

4 0

3 years ago

The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 310 grams, and it is released at an angle of 12 degree

lina2011 [118]

A) the periodic time is given by the equation;
T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
= 6.28 × √0.0673
= 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
thus; frequency = 0.614 Hz

b) The vertical distance, the height is given by
h= 0.66 cos 12
h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
mgh = 1/2mv²
v² = 2gΔh = 2×9.81 × 0.01
= 0.1962
v = 0.443 m/s

c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
= 1/2 × 0.310 × 0.443²
= 0.0304 J

4 0

3 years ago