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2 years ago

7) A force of 500N exists between two identical point charges separated by a dis-

Physics

2 answers:

navik [9.2K]2 years ago

8 0

Answer:

q=9.43×10^-5C

Explanation:

F=kq^2/r^2

500= 9×10^9 × q^2/ (0.4)^2

500N×0.16m=9×10^9Nm^2C^2 × q^2

80/(9×10^9)= q^2

√(8.8889×10^9) = q

q= 9.43×10^-5C

since they are identical both charges, q=9.43×10^-5C

ioda2 years ago

5 0

Explanation:

f=500N

d=40cm=0.4m

$\sqrt{500 \times 500 + .4 \times .4}$

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Recall that the blocks can only move along the x axis. the x components of their velocities at a certain moment are v1x and v2x.

Contact [7]

The center of mass is given with this formula:
$x_c=\frac{\sum_{n=1}^{n=i}m_ix_i}{M}$
Velocity is:
$v=\frac{dv}{dt}$
So, for the velocity of the center of mass we have:
$\frac{dx_c}{dt}=\frac{\sum_{n=1}^{n=i}d(m_ix_i)}{Mdt}\\
v_c=\frac{\sum_{n=1}^{n=i}p_i}{M}\\$
In our case it is:
$v_{xc}=\frac{m_1v_{x1}+m_2v_{x2}}{m_1+m_2}$

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2 years ago

A TV set is pushed a distance of 2 m with a force of 20 N how much work is done on the set

vladimir2022 [97]

Answer:

40 J

Explanation:

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2 years ago

Objects have a tendency to resist changing their motion. This property is called: *

Salsk061 [2.6K]

Answer:

The answer your looking for is option 2 - Inertia

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2 years ago

Many caterpillars construct cocoons from silk, one of the strongest naturally occurring materials known. Each thread is typicall

joja [24]

Answer:

a) N = 145,833,674.52174 = 1.458 × 10⁸ strands

b) diameter of single rope with the same effect = 2.415 cm

Explanation:

Hooke's law explains that stress is directly proportional to strain.

Stress ∝ Strain.

Stress = E × Strain

E = constant of proportionality = Young's Modulus = 4.0 ✕ 10⁹ N/m².

Stress = (Load/Total Cross sectional Area)

Load = a pair of 85 kg mountain climbers = 85 × 2 × 9.8 = 1666 N

Total Cross sectional Area = (Number of strands) × (Area of one strand) = A

Strain = (ΔL/L)

ΔL = 1.00 cm = 0.01 m

L = 11 m

Strain = (0.01/11) = 0.0009091

Stress = (Young's Modulus) × (Strain) = (4.0 ✕ 10⁹) × (0.0009091) = 3,636,363.64 N/m²

(Load/ total Area) = 3,636,363.64

Total area = (Load/3,636,363.64) = (1666/3,636,363.64) = 0.00045815 m²

Recall,

Total Cross sectional Area = (Number of strands) × (Area of one strand)

Area of one strand = (πd²/4)

diameter of one strand = 2 μm = (2×10⁻⁶) m

Area of one strand = (πd²/4)

= π × (2×10⁻⁶)² ÷ 4 = (3.142 × 10⁻¹²) m²

Total Cross sectional Area = (Number of strands) × (Area of one strand)

0.00045815 = N × (3.142 × 10⁻¹²)

N = 145,833,674.52174 = 1.458 × 10⁸ strands

b) If it was a single rope, the cross sectional Area would just be equal to the total cross sectional Area obtained in (a)

A = 0.00045815 m²

A = (πD²/4)

where D = diameter of the single rope

0.00045815 = (πD²/4)

D² = (4×0.00045815) ÷ π = 0.0005833347

D = 0.02415 m = 2.415 cm

Hope this Helps!!!

6 0

2 years ago

Read 2 more answers

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0

2 years ago