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nexus9112 [7]
2 years ago
10

An unspecified force causes a 0.20-kg object to accelerate at 0.40 m/s2. If 0.30 kg is added to the 0.20-kg object and the force

remains the same, what is the acceleration (in m/s2) of the heavier object? (NEVER include units with the answer to a numerical question.)
Physics
1 answer:
Naddik [55]2 years ago
4 0

Answer:

a = 0.16

Explanation:

given,

mass of the object 1 = 0.2 kg

mass of the object 2 = 0.3 kg

acceleration when force is on 0.2 kg = 0.4 m/s²

acceleration when both mass are combine = ?

F = m a

F = 0.2 × 0.4

F = 0.08 N

force acting is same and total mass  = 0.2 + 0.3 = 0.5 Kg

F = m a

a = \dfrac{F}{m}

a = \dfrac{0.08}{0.5}

a = 0.16 m/s²

the acceleration  acting when both the body is attached is a = 0.16

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The speed of light is constant in ALL frames of reference. That is 2.9 x10^8 m/s (290,000,000 m/s). It is independent of the motion of the light source.
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2 years ago
A ball rolls 6.0 meters as its speed changes from 15 meters per second to 10 meters per second. What is the average speed of the
antoniya [11.8K]
Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m

The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
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a = -10.4167 m/s²

The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s

The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s

Answer: 12.5 m/s

6 0
2 years ago
A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
GREYUIT [131]

Answer:

The flux through the surface of the cube is 2.314\ Nm^{2}/C

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = 8.0\times 10^{- 2}\ m

Volume Charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}

Now,

To calculate the electric flux:

\phi = \frac{q}{\epsilon_{o}}                                                      (1)

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of free space  

Volume Charge density for the given case is given by the formula:

\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}                  (2)

Volume of cube, V = a^{3}

Thus

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}

Thus from eqn (2), the total charge is given by:

q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}

q = 2.048\times 10^{-11}\ F = 20.48\ pF

Now, substitute the value of 'q' in eqn (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C

5 0
2 years ago
What is the total energy q released in a single fusion reaction event for the equation given in the problem introduction? use c2
kkurt [141]

Answer:

4.3\cdot 10^{-12} J

Explanation:

The fusion reaction in this problem is

4^1_1H \rightarrow ^4_2He +2e^+

The total energy released in the fusion reaction is given by

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where

c=3.0\cdot 10^8 m/s is the speed of light

\Delta m is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

m(^1_1H)=1.007825u is the mass of one nucleus of hydrogen

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So the mass defect is:

\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u

The conversion factor between atomic mass units and kilograms is

1u=1.66054\cdot 10^{-27}kg

So the mass defect is

\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg

And so, the energy released is:

\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J

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Answer:

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