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3 years ago

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The protons in a nucleus are approximately 2 ✕ 10^−15 m apart. Consider the case where the protons are a distance d = 1.93 ✕ 10^

−15 m apart. Calculate the magnitude of the electric force (in N) between two protons at this distance.

2 answers:

Ad libitum [116K]3 years ago

8 0

Answer:

61.8 N

Explanation:

Given data

Charge of the protons (q): 1.60 × 10⁻¹⁹ Coulomb

Distance between the protons (d): 1.93 × 10⁻¹⁵ meters

Coulomb's constant (k): 8.99 × 10⁹ N.m².C⁻²

We can find the magnitude of the electric force (F) between the two protons using Coulomb's law.

$F=k.\frac{q_{1}q_{2}}{d^{2} } \\F=8.99 \times 10^{9} N.m^{2} .C^{-2} .\frac{(1.60 \times 10^{-19}C)^{2} }{(1.93 \times 10^{-15}m)^{2} } \\F=61.8N$

The magnitude of the electric force is 61.8 N.

Bond [772]3 years ago

6 0

Answer: 61.2N

F=kq1q2/r^2

F=8.98E9(1.6E-19)(1.6E-19)/(1.93E-15)^2

F=61.2N

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The left face of a biconvex lens has a radius of curvature of magnitude 11.7 cm, and the right face has a radius of curvature of

SIZIF [17.4K]

Answer:

Explanation:

R1 = + 11.7 cm

R2 = - 18.7 cm

n = 1.52

(a) Use lens maker formula

$\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$

$\frac{1}{f}=\left ( 1.52-1 \right )\left ( \frac{1}{11.7}}+\frac{1}{18.7} \right )$

$\frac{1}{f}=\frac{0.52\times 30.4}{218.79}$

f = 13.84 cm

(b)

R1 = + 18.7 cm

R2 = - 11.7 cm

n = 1.52

(a) Use lens maker formula

$\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$

$\frac{1}{f}=\left ( 1.52-1 \right )\left ( \frac{1}{18.7}}+\frac{1}{11.7} \right )$

$\frac{1}{f}=\frac{0.52\times 30.4}{218.79}$

f = 13.84 cm

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3 years ago

What is the gravitational force that two objects would feel if they are 3.5 meters apart? Object 1 has a mass of 10x10^5 kg and

polet [3.4K]

Answer:

1.63 N

Explanation:

F = GMm/r^2

= (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2

= 1.63 N ( 3 sig. fig.)

6 0

2 years ago

PLEASE HELP ASAP!!!<br> GIVING BRAINLIEST!!!<br> 40 points!!

Nana76 [90]

W = Fd = 4(2100) = 8400 J

So the answer is A) 8400 J

I was just rewriting my notes on the work lesson I did in class today, so I saw this question at the perfect time!! :)

Hope it helps!! :)

4 0

2 years ago

Read 2 more answers

The position-time graph for a bug crawling along a line is shown in item 4 below. Determine whether the velocity is positive, ne

Naddika [18.5K]

Answer: The velocity at different marked time points are given as

t1 = -

t2 = +

t3 = +

t4 = -

t5 = 0

Explanation:

The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal

axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.

When the tangent of the line is parallel to the horizontal axis, the velocity is 0.

From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below

t1 = -

t2 = +

t3 = +

t4 = -

t5 = 0

QED!

5 0

2 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

3 years ago