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3 years ago

Two moles of an ideal gas at 3.0 atm and 10 °C are heated up to

Physics

1 answer:

ololo11 [35]3 years ago

5 0

Answer:

Explanation:

Two moles of an ideal gas at 3.0 atm and 10°C are heated up to 150 °C. If the volume is held constant during this heating, what is the final pressure? a. 4.5 atm.

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A 10-kg wagon is pulled by a force of 70 Newtons while friction provides 20 Newtons of force. How much acceleration will the wag

kkurt [141]

Answer:

5m/s²

Explanation:

Given parameters:

Mass of wagon = 10kg

Force of pull = 70N

Frictional force = 20N

Unknown:

Acceleration of the wagon = ?

Solution:

Frictional force is a force that opposes motion.

The net force is given as:

Net force = mass x acceleration

Force of pull - Frictional force = mass x acceleration

Insert the parameters and solve;

70 - 20 = 10 x acceleration

50 = 10 x acceleration

Acceleration = 5m/s²

5 0

3 years ago

A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of

Blizzard [7]

Answer:

the strength of the magnetic field is 3 x 10⁻⁵ T

Explanation:

Given;

velocity of the cosmic ray, v = 5 x 10⁷ m/s

force experienced by the ray, f = 1.7 x 10⁻¹⁶ N

angle between the ray's velocity and the magnetic field, θ = 45⁰

The strength of the magnetic field is calculated as;

$F = qvB \ sin(\theta)\\\\B = \frac{F}{qv\times sin(\theta)} \\\\where;\\\\B \ is \ the \ strength \ of \ the \ magnetic \ field\\\\q \ is \ the \ charge \ of \ the \ cosmic \ ray \ proton = 1.602 \times 10^{-19} \ C\\\\B = \frac{1.7\times 10^{-16}}{(1.602 \times 10^{-19})\times (5\times 10^7) \times sin \ (45)} \\\\B = 3 \times 10^{-5} \ T$

Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T

7 0

3 years ago

A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area

Airida [17]

For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
$p_1 = p_2$
which becomes
$\frac{F_1}{A_1}= \frac{F_2}{A_2}$
where
$F_1=3600 N$ is the force on the first piston
$A_1=0.030 m^2$ is the area of the first piston
$F_2$ is the force on the second piston
$A_2=0.015 m^2$ is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
$F_2=F_1 \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N$

7 0

3 years ago

Read 2 more answers

A meteorologist tracks the movement of a thunderstorm with Doppler radar. At 8:00pm the storm was 55 mi northeast of her station

Rus_ich [418]

At 8:00 pm, the velocity of the storm is 55 mi northeast. Assuming that the direction is exactly northeast, the angle is 45°
At 11:00 pm, the velocity is 75 mi north. The angle is 90°
In vector form
55 ∠ 45°
and
75 ∠ 90°
The magnitude and direction of the average velocity is
(55 ∠ 45° + 75 ∠ 90° ) / 3

4 0

3 years ago

A top of rotational inertia 4.0 kg m2 receives a torque of 2.4 nm from a physics professor. the angular acceleration of the body

Tanzania [10]

Angular acceleration is simply the ratio of the Torque over the rotation inertia, that is:

Angular acceleration = Torque / Rotational inertia

So substituting the values:

Angular acceleration = 2.4 N m / 4.0 kg m2

<span>Angular acceleration = 0.7 rad/s^2</span>

5 0

3 years ago