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2 years ago

A 250 mL sample of nitrogen(N2) has a pressure of 745 mm Hg at 30 C.What is the mass of nitrogen?

1 answer:

3 0

We assume that nitrogen gas sample here is an ideal gas. So that, we can use the ideal gas equation which is expressed as follows:

PV = nRT

First, we calculate the number of moles from the equation above. Then, use molar mass to calculate mass.

n = PV/RT
n = (745/760) (0.250) / 0.08206 (30+273.15)
n = 0.01 mol

m = 0.01 ( 28) = 0.28 g

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A car accelerates uniformly from rest at a speed of 1.67 ft s^2 over a distance of 5 yards.What is the acceleration of a car?

Nina [5.8K]

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

acceleration= 23/1.30

accelaration= 17.69 m/s^2

5 0

2 years ago

Read 2 more answers

NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

2 years ago

A standing wave has a frequency of 471 Hz and a wavelength of 1.9. What is the speed of the

Brums [2.3K]

Answer:

c = 894.90 m/s

Explanation:

Given data:

Frequency of wave = 471 Hz

Wavelength of wave = 1.9 m

Speed of wave = ?

Solution:

Formula:

Speed of wave = frequency × wavelength

c = f×λ

c = 471 Hz × 1.9 m

Hz = s⁻¹

c = 471s⁻¹ × 1.9 m

c = 894.90 m/s

The speed of wave is 894.90 m/s.

6 0

2 years ago

if you jump from a window that is 2 meters up you will most likely break the bones in your legs when you land however on a tramp

Gemiola [76]

The de-acceleration or negative acceleration of stopping is what damages bones. The ground is rigid and therefore the change in momentum when striking the ground will be large. On the trampoline, the elasticity of the material means that the momentum changes more slowly, resulting in smaller accelerations.

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3 years ago

On a horizontal surface is located

Ierofanga [76]

By Newton's second law, the net vertical force acting on the object is 0, so that

n - w = 0

where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².

The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of static friction µ is such that

80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is

40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

(Note: µ and ν are the Greek letters mu and nu)

3 0

2 years ago