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3 years ago

A speeder passes a stationary police officer at a constant speed of 45 m/s. The police officer

Physics

1 answer:

weqwewe [10]3 years ago

6 0

Answer:

a) 9

b) 45

I am not sure of the b part though :)

Explanation:

a) a= v /t

t= 45/5

= 9

b) intial v= 0 t=9 s a= 5

5x 9 = 45

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An acorn with a mass of .0300 kg is hanging from a branch in a tree it is 2.50 M off the ground what is the potential energy of

Alekssandra [29.7K]

According to another source this is what I got
<span>0.735 J ( Ep-potential energy, m-mass,g-gravitational acceleration = 9.81m/s², h-height; Ep = m * g * h; Ep = 0.0300 kg * 9.81 m/s² * 2.5 m )
</span>Hope it helps

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3 years ago

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Where would the barycenter of these two bodies be located given their masses?

n200080 [17]

It's kinda tough, since we don't know the actual numerical locations of the points, only an approximate picture.

The big ball has 11 times the mass of the small ball, so the small ball is 11 times as far from the barycenter as the big ball is.

If any of the points is marked at the actual barycenter, it can only be point-A .

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3 years ago

What does the "c^2" mean in E=mc^2? *

Korolek [52]

Answer:

C.speed of light times 2

4 0

3 years ago

A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in

Kruka [31]

Answer:

1020g

Explanation:

Volume of can= $1100cm^3=1100\times 10^{-6}m^3$

$1cm^3=10^{-6}m^3$

Mass of can=80g= $\frac{80}{1000}=0.08kg$

1Kg=1000g

Density of lead= $11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3$

By using $1g/cm^3=10^3kg/m^3$

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can and lead inside it they are floating in the water.

Buoyancy force = $F_b=Weight of can+weight of lead$

$\rho_wV_cg=m_cg+m_lg$

Where $\rho_w=10^3kg/m^3=$ Density of water

$m_c=$ Mass of can

$m_l=$ Mass of lead

$V_c=$ Volume of can

Substitute the values then we get

$1000\times 1100\times 10^{-6}=0.08+m_l$

$1.1-0.08=m_l$

$m_l=1.02 kg=1.02\times 1000=1020g$

$1 kg=1000g$

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0

4 years ago

A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at

VARVARA [1.3K]

Answer:

The percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

Explanation:

Given;

mass of bullet, m₁ = 4g = 0.004kg

initial velocity of bullet, u₁ = 589 m/s

mass of block of wood, m₂ = 2.3 kg

initial velocity of the block of wood, u₂ = 0

let the final velocity of the system after collision = v

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

0.004(589) + 2.3(0) = v(0.004 + 2.3)

2.356 = 2.304v

v = 2.356 / 2.304

v = 1.0226 m/s

Initial kinetic energy of the system

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(0.004)(589)² = 693.842 J

Final kinetic energy of the system

K.E₂ = ¹/₂v²(m₁ + m₂)

K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)

K.E₂ = 1.209 J

The kinetic energy left in the system = final kinetic energy of the system

The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%

= (1.209 / 693.842) x 100%

= 0.174 %

Therefore, the percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %

6 0

4 years ago