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drek231 [11]
3 years ago
8

An isotope is a version of the same element that differs in the composition of the

Physics
1 answer:
Anestetic [448]3 years ago
8 0
Answer: B) Nucleus
Located in the nucleus, neutrons are the particles in an atom that have a neutral charge. Isotopes have different numbers of neutrons.
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Sean pushes a box with a force of 100 N. Christian pushes an identical box with a force of
kap26 [50]

Answer:

Christian

Explanation:

Given that

Force applied by Sean ,F₁=100 N

Force applied by Christian ,F₂=200 N

Lets take mass of the box =  m kg

As we know that from second law of Newton's

F= m a

F=Force,m=mass ,a=acceleration

F_1=ma_1

100=m\times a_1

a_1=\dfrac{100}{m}\ m/s^2

F_2=ma_2

200=m\times a_2

a_2=\dfrac{200}{m}\ m/s^2

From the above we can say that acceleration a₂ is greater than a₁ is for the same mass of the box.

Therefore Christian will acceleration more for the box.

5 0
3 years ago
A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is
lbvjy [14]

Answer:

a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\

b)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}

8 0
3 years ago
Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electr
svlad2 [7]

Answer:

Electric field, E = 45.19 N/C

Explanation:

It is given that,

Surface charge density of first surface, \sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2

Surface charge density of second surface, \sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2

The electric field at a point between the two surfaces is given by :

E=\dfrac{\sigma}{2\epsilon_o}

E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}

E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}

E = 45.19 N/C

So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.

6 0
3 years ago
I really need help with this problem ;)
Nesterboy [21]

They're never the same ray.

4 0
3 years ago
Read 2 more answers
A standing wave of 603 Hz is produced on a string that is 1.33 m long and fixed on both ends. If the speed of the waves on this
kondaur [170]

Answer:

4.

Explanation:

Given,

frequency of standing wave = 603 Hz

length of string,L = 1.33 m

speed of the wave, v = 402 m/s

number of antinodes = ?

Wavelength of the standing wave

\lambda = \dfrac{v}{f}

\lambda = \dfrac{402}{603}

\lambda = 0.67\ m

Number of anti nodes in the standing wave

n=\dfrac{l}{\frac{\lambda}{2}}

n=\dfrac{2l}{\lambda}

n=\dfrac{2\times 1.33}{0.67}

n =3.97= 4.

Number of antinodes is equal to 4.

7 0
3 years ago
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