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Firdavs [7]
3 years ago
11

A bullet travels at 850 m/s. how long will it take a bullet to go 1 km?

Physics
1 answer:
Dmitrij [34]3 years ago
8 0
The speed of bullet =
850 m/s

Distance given = 1 km = 1000m

S = D/t
t • S = D/t • t
St = D
St/S = D/S
t = D/S
t = 1000m/850m/s
t = 1.176 s

It will take the bullet 1.176 or about 1.18 seconds to go 1 km.
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PHYSICS PLEASE HELP
WINSTONCH [101]

Explanation:

Newton’s First Law of Motion - if an object is at rest, it takes un-

balanced forces to make it move. Conversely, if an object is moving

it takes an unbalanced force to make it change it’s direction or speed.

Newton was the first to see that such apparently diverse phenomena as a satellite moving near the Earth's surface and the planets orbiting the Sun operate by the same principle: Force equals mass multiplied by acceleration, or F=ma.

Mark me as brainlist

7 0
3 years ago
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Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
3 years ago
A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyon
jek_recluse [69]

Answer:

x₂ = 1.33 m

Explanation:

For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall

            Στ = 0

            W₁ x₁ - W₂ x₂ = 0

where W₁ is the weight of the woman, W₂ the weight of the table.

Let's find the distances.

Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.

           x₁ = 2.5 -1.5 = 1 m

The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative

            x₂ = \frac{M_1g  }{m_2 g} \ x_1

           

let's calculate

           x₂ = \frac{100}{75}  \ 1

           x₂ = 1.33 m

7 0
3 years ago
A light bulb does 100 joules of work in 2.5 seconds. how much work power dos it have
Lesechka [4]
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4 years ago
Which of the following are not vector directions?
Svetradugi [14.3K]

Answer:

here north are not vector option b hope ur help

6 0
3 years ago
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