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3 years ago

How much energy is required to raise the temperature of 5g of air by 10°C?

Physics

1 answer:

Alex777 [14]3 years ago

4 0

You need to know the specific heat capacity of air.
Then energy needed = 0.005 x sp.heat.cap x 10

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What is weight vs mass

Ray Of Light [21]

Weight is the measurement of the pull of gravity on an object, while mass is the measurement of the amount of matter that an object contains.

8 0

3 years ago

What is the net force on this object?

Sergio039 [100]

200N

Explanation:
600N-400N = 200N

6 0

3 years ago

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge

xxMikexx [17]

Answer:

new force is 6 times of the initial force.

Explanation:

Let the charges on two objects is q₁ and q₂. The electric force between charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Objects 1 and 2 attract each other with a electrostatic force of 18.0 units

If the charge of Object 1 is doubled and the charge of object 2 is tripled, it means, $q_1'=2q_1$ and $q_2'=3q_2$ . New force is given by :

$F'=\dfrac{kq_1'q_2'}{r^2}\\\\F'=\dfrac{k(2q_1)(3q_2)}{r^2}\\\\F'=6\dfrac{kq_1q_2}{r^2}\\\\F'=6F$

So, the new electrostatic force between objects will become 6 times of the initial force.

5 0

2 years ago

A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

6 0

2 years ago

Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the

IRISSAK [1]

Answer:

a₁ = 0.63 m/s² (East)

a₂ = -1.18 m/s² (West)

Explanation:

m₁ = 95 Kg

m₂ = 51 Kg

F = 60 N

a₁ = ?

a₂ = ?

To get the acceleration (magnitude and direction) of the man we apply

∑Fx = m*a (⇒)

F = m₁*a₁ ⇒ 60 N = 95 Kg*a₁

⇒ a₁ = (60N / 95Kg) = 0.63 m/s² (⇒) East

To get the acceleration (magnitude and direction) of the woman we apply

∑Fx = m*a (⇒)

F = -m₂*a₂ ⇒ 60 N = -51 Kg*a₂

⇒ a₂ = (60N / 51Kg) = -1.18 m/s² (West)

For every case we apply Newton’s 3 d Law

8 0

3 years ago