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White raven [17]

4 years ago

7

8. How does inertia act on an object that is moving in a circle? A. It causes the object to speed up as it moves. B. It causes t

he object to move along a straight path. C. It directs the object toward the center of the curve.

2 answers:

valentinak56 [21]4 years ago

7 0

Answer: its D

Explanation:

took on study island

elena-s [515]4 years ago

5 0

I think my answer is C

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A balloon filled with 2.00 L of helium initially at 1.25 atm of pressure rises into the atmosphere. When the surrounding pressur

Ierofanga [76]

Answer:

final volume the balloon occupy in instant before it burst is 4.634 L

Explanation:

given data

helium filled V1 = 2 L

initially pressure P1 = 1.25 atm

finally pressure P2 = 410 mmHg

1 atm = 760 mmHg

to find out

at what volume the balloon occupy in instant before it burst

solution

we have given

760 mmHg = 1atm

and 1 mmHg = 1/760

so 410mmHg = 410/760 atm

so that final pressure is = 410/760 atm

we use here Boyle's Law that is

Boyle's law states that "the volume of an ideal gas is inversely proportional to the pressure of the gas at constant temperatures"

so P ∝ 1/V

P1V1 = P2V2

V2 = P1V1 / P2

put all value

V2 = 1.25 (2) / (410/760)

V2 = 2.5 / 0.53947 = 4.634

so final volume the balloon occupy in instant before it burst is 4.634 L

4 0

3 years ago

1. Consider a 1000 kg car rounding a curve on a flat road of radius 50 m at a speed of

raketka [301]

Answer:

The car will make the turn perfectly

Explanation:

Given that the centripetal force= mv^2/r

M= mass of the car

v = speed of the car

r= radius

Hence;

F = 1000 × (14)^2/50

F= 3920 N

The frictional force = μmg

μ = coefficient of static friction

m= mass

g = acceleration due to gravity

Frictional force= 0.6 × 1000× 10

Frictional force = 6000 N

The car will not skid off the curve because the frictional force is greater than the centripetal force.

6 0

3 years ago

A magnesium surface has a work function of 2.65 eV. Electromagnetic waves with a wavelength of 280 nm strike the surface and eje

Neko [114]

Answer:

Explanation:

Energy of radiation in eV = 1237.5/280 = 4.42 eV.

Work function = 2.65 eV.

Maximum kinetic energy = 4.42 - 2.65 = 1.77 eV.

8 0

4 years ago

In the photoelectric effect, it is found that incident photons with energy 5.00ev will produce electrons with a maximum kinetic

Yakvenalex [24]

From Literature:

The amount of energy in the photons is given by this equation:

E = hf

where E = energy
h = Planck's constant = 6.63 * 10^-34 Joule seconds
f = frequency of the light, Hz

Given:

E= 3.00 eV and Planck's constant

To solve for the frequency, E = 3.00 eV

1 electronvolt = 1.60218 x 10^-19 Joules

3 * 1.60218 x 10^-19 Joules = 6.63 * 10^-34 Joule seconds * f
f = 7.25 x 10^14 /second or hertz

Therefore, the threshold frequency of the material is 7.25 x 10^14 Hertz.

7 0

3 years ago

Read 2 more answers

A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the partic

olga_2 [115]

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma

$a = \frac{F}{m}$ ...... (1)

acceleration (a) $= \frac{dv}{dt}$ ......(2)

substituting (2) into (1)

Hence, F $= \frac{mdv}{dt}$

$\frac{dv}{dt} = \frac{F}{m}$

$dv = \frac{F}{m} dt$

$dv = \frac{1}{m}Fdt$

Integrating both sides

$\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt$

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

$v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt$ ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

$v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt$

$v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}$

$v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|$

$v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |$

$v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |$

$v = \frac{1}{5} | 250 - 50 + 15 |$

$v = \frac{215}{5}$

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s

6 0

3 years ago