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3 years ago

Help me please, MathPhys be my hero :)

Physics

1 answer:

Drupady [299]3 years ago

8 0

Answer:

T = 261 K

T = -12.2°C

Explanation:

PV = nRT

If P and n are constant:

V / T = V / T

2.75 L / (273.15 + 21) K = 2.44 L / T

T = 261 K

T = -12.2°C

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An object, initially at rest, is subject to an acceleration of 34 m/s2. How long will it take for that

SIZIF [17.4K]

Answer:

If it is moving 34 m/s it will take 100 seconds, or 1:40 to reach 3400 meters.

Explanation:

I found this answer by dividing 3400 by 34 and converting seconds to minutes

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3 years ago

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What does the equation n =Pout/Pin mean

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When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards

aksik [14]

Answer:

51.85m/s

Explanation:

Given parameters:

Mass of ball = 0.0459kg

Force = 2380N

Time taken = 0.001s

Unknown:

Speed of the ball afterwards = ?

Solution:

To solve this problem, we use Newton's second law of motion:

F = m x $\frac{v - u}{t}$

F is the force

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

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A damped harmonic oscillator consists of a mass on a spring, with a small damping force that is proportional to the speed of the

exis [7]

Answer:

2.19 N/m

Explanation:

A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:

T = 2π√(m/K)

Where T is the period, m is the mass (in kg), and K is the damping constant. So:

2.4 = 2π√(0.320/K)

√(0.320/K) = 2.4/2π

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(√(0.320/K))² = (0.38197)²

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3 years ago

Um corpo de massa 50g recebe 300 calorias e sua temperatura sobe de -10°C até 20°C. Determine a capacidade térmica do corpo e o

kykrilka [37]

The specific heat capacity of the substance is $0.963 J/g^{\circ}C$

Explanation:

When an object of mass m is supplied with a certain amount of energy Q, its temperature increases according to the equation:

$Q=mC_s \Delta T$

where

m is the mass of the object

$C_s$ is its specific heat capacity

$\Delta T$ is the increase in temperature of the object

In this problem, we have

$Q=300 cal \cdot 4.814 = 1444.2 J$

m = 50 g

$\Delta T = 20C-(-10C)=30^{\circ}C$

Therefore, we can solve for $C_s$ to find its specific heat capacity:

$C_s = \frac{Q}{m\Delta T}=\frac{1444.2}{(50)(30)}=0.963 J/gC$

Learn more about specific heat capacity:

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3 years ago