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Brut [27]
3 years ago
13

I NEED AN ANSWER ASAP! PLEASE

Physics
1 answer:
Y_Kistochka [10]3 years ago
4 0

i dont know

Answer:

Explanation:

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Scientists are expected to share their results by
shepuryov [24]

Answer:

Publishing results of research projects in peer-reviewed journals enables the scientific and medical community to evaluate the findings themselves. It also provides instructions so that other researchers can repeat the experiment or build on it to verify and confirm the results.

7 0
4 years ago
An 80kg parachutist jumps out of an airplane. Neglecting air friction, how fast will he be going after a 10 second free fall?
olganol [36]
If he is jumping you are adding force which means that you will be falling twice as fast
6 0
3 years ago
A mass of 2.82 kg is hung from a spring, causing the spring to stretch 0.331 m. If a second mass of 3.09 kg is now added to the
irina1246 [14]

Answer:

0.694 m

Explanation:

Case 1 : When only mass of 2.82 kg is hanged from spring

m = mass hanged from the spring = 2.82 kg

x = stretch caused in the spring = 0.331 m

k = spring constant

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

k (0.331) = (2.82) (9.8)

k = 83.5 N/m

Case 2 : When both masses are hanged from spring

m = mass hanged from the spring = 3.09 + 2.82 = 5.91 kg

x = stretch caused in the spring = ?

k = spring constant = 83.5 N/m

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

(83.5) x = (5.91) (9.8)

x = 0.694 m

4 0
3 years ago
Key Stage 3 Science - Physics
stira [4]

Answer:

F = 3750 N

Explanation:

Given that,

Pressure, P = 150 Pa

Area, a = 25m²

We need to find the force applied. We know that, pressure is equal to the force acting per unit area. It can be given by :

P=\dfrac{F}{A}\\\\F=P\times A\\\\F=150\ Pa\times 25\ m^2\\\\F=3750\ N

So, the required force is 3750 N.

5 0
3 years ago
1.00kg of ice at -24 degrees Celsius is placed in contact with a 1.00kg block of a metal at 5.00 degrees Celsius. They come to e
Alona [7]

Answer:

C₂ = 2.22 KJ/kg °C

Explanation:

Since, both objects are in thermal contact. Therefore, the law of conservation of energy tells us that:

Heat\ Lost\ by\ Metal\ Block = Heat\ Gained\ by\ Ice\\m_{1}C_{1} \Delta T_{1} = m_{2}C_{2} \Delta T_{2}

where,

m₁ = mass of ice = 1 kg

C₁ = specific heat of ice = 2.04 KJ/kg.°C

ΔT₁ = Change in Temperature of Ice = -8.88°C - (-24°C) = 15.12°C

m₂ = mass of metal block = 1 kg

C₂ = specific heat of metal = ?

ΔT₂ = Change in Temperature of Metal Block = 5°C - (-8.88°C) = 13.88°C

Therefore, using these values in the equation, we get:

(1\ kg)(2.04\ KJ/kg.^0C)(15.12\ ^0C) = (1\ kg)C_{2}(13.88\ ^0C) \\C_{2} = \frac{30.84\ KJ}{13.88\ kg/^0C}

<u>C₂ = 2.22 KJ/kg °C</u>

4 0
3 years ago
Read 2 more answers
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