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2 years ago

As a bullet shot vertically upward rises, the kinetic energy of the bullet

Physics

1 answer:

algol [13]2 years ago

3 0

The kenitc energy of the bullet lowers as it keeps going up.

Because gravity is pushing the bullet down as the bullet goes up.

I'm pretty sure that the way to put this answer.

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Luda [366]

Explanation:

heat engine is correct answer

3 0

3 years ago

Read 2 more answers

What pressure expressed in pascals equals 70 N/m2?

12345 [234]

Pascals = N/m2
so it is the same unit
the answer is 70 Pa

4 0

3 years ago

Read 2 more answers

A cannon fires a 40.5kg shell toward a target and the shell moves with a velocity of 120 m/s. Calculate the shells momentum

kvv77 [185]

Answer:

4860 kg m/s

Explanation:

P = mv

P = 40.5 x 120

P = 4860 kg m/s

8 0

2 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J = 120 J decrease

The missing energy has turned into interned energy in the completely inelastic collision

3 0

3 years ago

During the first 14 minutes of a 1.0 hour trip, a car has an average speed 36 km/h. What must the average speed in km/h of the c

oksian1 [2.3K]

Answer:

85.5 km/h

Explanation:

$t_{1}$ = time interval for first phase = 14 min = $\frac{14}{60}$ h = 0.233 h

$t_{2}$ = time interval for second phase = 46 min = $\frac{46}{60}$ h = 0.767 h

$v$ = average speed for the entire trip = 74 km/h

$v_{1}$ = average speed in first phase = 36 km/h

$v_{2}$ = average speed in second phase

$d_{1}$ = distance traveled in first phase

$d_{2}$ = distance traveled in first phase

average speed is given as

$v = \frac{d_{1} + d_{2}}{t_{1} + t_{2}}$

$v = \frac{v_{1} t_{1} + v_{2} t_{2}}{t_{1} + t_{2}}$

$74 = \frac{(36) (0.233) + v_{2} (0.767)}{0.233 + 0.767}$

$74 = (36) (0.233) + v_{2} (0.767)$

$v_{2} = 85.5$ km/h

4 0

3 years ago