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sammy [17]
3 years ago
9

Which accounts for an increase in the temperature of a gas that is kept a constant volume

Physics
1 answer:
oksian1 [2.3K]3 years ago
7 0

Answer:

An increase in pressure

Explanation:

The ideal gas law states that:

pV=nRT

where

p is the gas pressure

V is the volume

n is the number of moles

R is the gas constant

T is the temperature of the gas

in the equation, n and R are constant. For a gas kept at constant volume, V is constant as well. Therefore, from the formula we see that if the temperature (T) is increase, the pressure (p) must increase as well.

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A +3.4 x 10-6 C test charge experiences forces from two other nearby charges: a 3 N force due east and a 15 N force due west. Wh
Alecsey [184]

Answer:

3.53×10⁶ N/c due west

Explanation:

From the question

E = F'/q........................ Equation 1

Where E =  Electric Field, F = Net Force, q = Charge.

But,

F' = F₂-F₁...................... Equation 2

Substitute equation 2 into equation 1

E = (F₂-F₁)/q................ Equation 3

Given: F₁ = 3 N due east, F₂ = 15 N due west,  q = 3.4×10⁻⁶ C

Substitute these values into equation 1

E = (15-3)/(3.4×10⁻⁶)

E = 12/(3.4×10⁻⁶)

E = 3.53×10⁶ N/c due west

4 0
3 years ago
How can you measure mold on food?
Vilka [71]
Trimming the mold of and putting it on a weight scale
6 0
3 years ago
20 kg who is running at a speed of 4.0 m/s jumps onto a stationary sled of mass 5.0 kg on a frozen lake. the speed at which the
Step2247 [10]

Here we can use momentum conservation as there is no external force on sled and child while he jump on sled

by momentum conservation equation

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v{2f}

20*4 + 5*0 = 20*v + 5 *v

since sled and child both moves with same speed so here they both will have same final speed "v"

by solving above equation we will have

25 v = 80

v = 3.2 m/s

So they will move together with speed 3.2 m/s

6 0
3 years ago
A 37 kg object has an applied force of 85N [R] acting on it. The coefficient of
Sliva [168]

Answer:

Explanation:

This is quite tricky! You need to do 2 different equations to solve all the parts of this problem. First is finding the acceleration in one dimension, which has an equation of

F - f = ma

where F is the applied Friction,

f is the frictional force acting against F,

m is the mass of the object, and

a is the acceleration of the object (NOT the velocity!)

This is Newton's Second Law expanded on a bit. The sum of the forces working on an object is equal to the object's mass times its acceleration. We have F, but we need f which is found in the equation

f = μF_n which is the coefficient of kinetic friction times the weight of the object. Weight is found in the equation

w = mg where m is mass and g is the pull of gravity. Let's start there and work backwards:

w = 37(9.8) to 2 sig figs so

w = 360N. Now fill that in to find f:

f = (.17)(360) to 2 sig figs so

f = 61. Now for the final answer in the original equation way back up at the top:

85 - 61 = 37a and do the subtraction on the left side first:

24 = 37a and then we divide to 2 sig figs to get

a = .65 m/s/s

Since we are moving in a straight line (as opposed to on an angle) the displacement is found in

d = rt which simply says that the distance an object moves is equal to its rate times the time. Therefore,

d = 2.2(3.4) to 2 sig figs so

d = 7.5 m

6 0
3 years ago
Let's apply Snell's law to the refraction of light across a water–air interface. Suppose you kneel beside the fishpond in your b
Nataliya [291]

Answer:

The angle of refraction is 41.68°.

Explanation:

The refractive index for water is n_2 = 1.333, and for air n_1 = 1.00: the angle of light with the normal is 90^o-60^o = 30^o; therefore Snell's law gives

n_1sin(\theta_1)= n_2sin(\theta_2)

1.00*sin(\theta_1) = 1.33 sin(30^o)

sin (\theta_1) = \dfrac{1.33sin(30^o)}{1.00}

sin (\theta_1) = 0.665

\theta _1 = sin^{-1}(0.665)

\boxed{\theta_1 = 41.68^o}

4 0
3 years ago
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