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Sveta_85 [38]
2 years ago
9

Help me please thank you !

Physics
1 answer:
son4ous [18]2 years ago
3 0

Answer:

one is photosynthesis ik

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Choose all the answers that apply. Oxygen is _____. transported by blood picked up in the alveoli released into the air by the l
aleksklad [387]
ANSWER:

- Transported by blood picked up in the alveoli
- Pumped to cells by ventricles

Hope this helps! :)
4 0
2 years ago
Read 2 more answers
Imagine that the apparent weight of the crown in water is Wapparent=4.50N, and the actual weight is Wactual=5.00N. Is the crown
9966 [12]

Answer:

Explanation:

Actual weight, Wo = 5 N

Apparent weight, W = 4.5 N

density of water = 1 g/cm^3 = 1000 kg/m^3

density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3

Buoyant force = Actual weight - Apparent weight

Volume x density of water x g = 5 - 4.5

V x 1000 x 9.8 = 0.5

V = 5.1 x 10^-6 m^3

Weight of gold = Volume of gold x density of gold x gravity

W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N

As W' is less than W so, it is not pure gold.

4 0
3 years ago
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl
Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0
2 years ago
How much force would be needed to cause a 4.6kg object to accelerate at 9.2m/s/s? *
poizon [28]

Answer:

<h2>42.32 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 4.6 × 9.2

We have the final answer as

<h3>42.32 N</h3>

Hope this helps you

5 0
2 years ago
To warm 2.0 l of tea (d = 1.01 g/ml; sp. heat = a cook places a 500 g block of stone at a temperature of 200f into the teapot. a
IRISSAK [1]
Volume of tea V = 2.0L = 2000 mL density d = 1.01 g/ mL mass of tea m = V * d = 2000mL * 1.01g/mL = 2020 gWhen we assume that the tea was initially at 72, the final temperature of the tea in F is 91. 
The answer in this question is B. 91
4 0
3 years ago
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