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3 years ago

A bullet is shot from a rifle with a velocity of 720 m/s. What is the velocity of the bullet in km/h.

Physics

1 answer:

Elan Coil [88]3 years ago

8 0

Answer:

2592 km/h

Explanation:

Given that a bullet is shot from a rifle with a velocity of 720 m/s. What is the velocity of the bullet in km/h.

The velocity = 720 m/s

Solution

To convert metres per second to kilometer per hour, you will multiply by 3600 and divide by 1000

720 × 3600/1000

720 × 3.6

72 × 36

2592 km/h

Therefore, the velocity of the bullet in kilometer per hour is 2592 km/h

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salantis [7]

Answer:

The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

The density of air is $\rho_a = 1.21 \ kg/m^3$

The diameter of bottom part is $d = 0.15 \ m$

The power trend-line equation is mathematically represented as

$F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

$F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where

$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

$v$ is the flow velocity

A is the area which mathematically evaluated as

$A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

$A = 0.0176 \ m^2$

Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

$D_z = 1.30512$

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2 years ago

The ocean may pull away rom the shore as a tsunami approaches?

ale4655 [162]

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2 years ago

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What angle is formed by the sun, the earth, and the moon during an eclipse?.

Andrew [12]

Answer:

The Sun-Earth-Moon system happens to exhibit a striking geometric coincidence, which we examine in the first problem. PROBLEM 1. To an observer on Earth, the Sun and the Moon subtend almost the same angle in the sky. The average angle is 0.52 degrees for the Moon and 0.53 degrees for the Sun.

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2 years ago

three resistors which values of 15 ohm 40 ohm 60 ohm respectively are connected in parallel. what is their equivalent resistance

VashaNatasha [74]

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7 0

2 years ago

During a workout, the football players at State U ran up the stadium stairs in 61 s. The stairs are 130 m long and inclined at a

NeTakaya

Answer:

P = 1097 Watt

Explanation:

given,

length of stairs, L = 130 m

inclination with horizontal,θ = 30°

mass of the football player = 105 Kg

time = 61 s

we know,

$Power = \dfrac{work}{time}$

Work = change in Potential energy

h = L sin 30°

h = 130 x 0.5

h = 65 m

W = m g h

W = 105 x 9.8 x 65

W = 66885 J

now,

$P = \dfrac{66885}{61}$

P = 1097 Watt

hence, the power output on the way is 1097 W

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2 years ago