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Liono4ka [1.6K]

2 years ago

11

If the temperature is lowered from 60 °c to 30 °c, the volume of a fixed amount of gas will be one half the original volume. if

the temperature is lowered from 60 °c to 30 °c, the volume of a fixed amount of gas will be one half the original volume.
a. True
b. False

1 answer:

bazaltina [42]2 years ago

8 0

I believe it should be FALSE

reply if i´m wrong.

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Ram has power of 550 watt. What does it mean?

WARRIOR [948]

It means you can do 550 Newton Meters of work every second. Power is the rate of doing work, I hope this helps

4 0

2 years ago

Determine whether each substance will sink or float in corn syrup, which has a density of 1.36 g/cm3. Write “sink” or “float” in

Tatiana [17]

Explanation:

A substance will floats if it is having lower density than the density of the liquid in which it is placed.
A substance will sink if it is having density greater than the density of the liquid in which it is kept.

Density of corn syrup = $1.36 g/cm^3$

1) Density of gasoline = $0.748 g/cm^3$

Density of the gasoline is less than the the density of corn syrup which means it will float in corn syrup.

2) Density of water = $1 g/cm^3$

Density of the water is less than the the density of corn syrup which means it will float in corn syrup.

3) Density of honey = $1.45 g/cm^3$

Density of the gasoline is more than the the density of corn syrup which means it will sink in corn syrup.

4) Density of titanium = $4.506 g/cm^3$

Density of the titanium is more than the the density of corn syrup which means it will sink in corn syrup.

3 0

2 years ago

Read 2 more answers

What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?

Stolb23 [73]

Answer:

$1.1\cdot 10^{-10}C$

Explanation:

The electric field produced by a single point charge is given by:

$E=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

$q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C$

8 0

2 years ago

A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the

likoan [24]

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

$a\sin \theta_n=n\lambda$

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

$a\sin \theta_1=(1)\lambda$

solving for θ1:

$\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_1=0.63 degrees$

for the second minimum:

$a\sin \theta_2=(2)\lambda$

$\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_2=1.26 degrees$

So, the angular separation between them is the rest:

$\Delta \theta =1.26-0.63$

$\Delta \theta=0.63$

4 0

2 years ago

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The

Reika [66]

Complete question:

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The

particle is momentarily at rest at t is:

Select one:

a. 9.3s

b. 1.3s

C. 0.75s

d.5.3s

e. 7.3s

Answer:

b. 1.3 s

Explanation:

Given;

position of the particle, x(t)=1 6t- 3.0t³

when the particle is at rest, the velocity is zero.

velocity = dx/dt

dx /dt = 16 - 9t²

16 - 9t² = 0

9t² = 16

t² = 16 /9

t = √(16 / 9)

t = 4/3

t = 1.3 s

Therefore, the particle is momentarily at rest at t = 1.3 s

6 0

2 years ago