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worty [1.4K]
2 years ago
12

I''l give brainliest. Please help. I hold 2 objects about 0.1 meters apart. What is the electrostatic force between the two obje

cts if each has a charge of 4.6 x 10^-16 C?
Physics
1 answer:
matrenka [14]2 years ago
7 0

Answer:

F = K Q1 Q2 / R^2       where K = 9 * 10E9  (1 / 4 pi ∈0)

F = 9.00E9 * (4.6E-16)^2 / .01 = 1.90E-19 N

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A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this dista
Nastasia [14]

Answer:A)u =4.295m/s  , B)a = 29.746m/s²   C) F=3,153N

Explanation:

Using the kinematic expression  

v² = u² - 2as

where

u = initial velocity

v = final velocity

s = distance

g = acceleration due to gravity .

Given that he reaches a height of 0.940 m above the floor,

the final velocity  = 0

Here, acceleration due to gravity is acting in  opposite the initial direction of motion. So, a=-9.81 m/s.

v² = u² + 2as

0² - u² = 2 (- 9.81) × 0.940

- u² = 2 × - 9.81 × 0.920

- u² = -18.4428

cancelling the minus in both sides , we have that  

u² = 18.4428

u = √18.4428

u =4.295m/s

(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2

Using v² = u² + 2as

where u = initial speed of basketball player before lengthening = 0 m/s,

v = final speed of basketball player after lengthening =  4.295m/s,

a = acceleration while  straightening his legs

s = distance moved during lengthening = 0.310m

v² = u² + 2as  

 a = (v² - u²)/2s

a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)

a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)

a = (18.4428 m²/s²/(0.62 m)

a = 29.746m/s²

c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N

Force= mass x acceleration.

F = 106 kg X 29.746m/s²

 F = 3,153.076 rounded to  3,153N

8 0
3 years ago
The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag
allochka39001 [22]
In order to get the propoerty of work you need to use the following formula
 <span>work = force times distance
</span>replacing data you will get:
W = (1.500) (3)
W =  4.500 NM
The answer should be in NM. So it will be 4500 NM againts the force of gravity
4 0
3 years ago
A car (mass 1200 kg, speed 100 km/h) and a truck (mass 2800 kg, speed 50 km/h) are moving in the same direction along a highway.
Sloan [31]

Answer:

Speed of the wreck after the collision is 65 km/h

Explanation:

When a car hits truck and sticks together, the  collision would be totally inelastic.  Since the both the vehicles  locked  together, they have the same final velocity.

Mass of car  m_{1}=1200 kg

Mass of truck m_{2}=2800 kg

Initial speed of the car u_{1}=100 km /h

Initial speed  of the truck u_{2}=50 km /h

The final velocity of the wreck will be

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

since final speed are same, v_{1}=v_{2}=v

m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v

1200\times 100 +2800 \times 50 =(1200+2800)v\\v=\frac{260000}{4000} \\v=65 km/h

5 0
3 years ago
Convertir 500 f a grado celsios
Allushta [10]

Answer:

260° C

Explanation:

si

5 0
2 years ago
Read 2 more answers
A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug
natali 33 [55]

Answer:

Approximately 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}, assuming friction between the vehicle and the ground is negligible.

Explanation:

Let m denote the mass of the vehicle. Let v denote the initial velocity of the vehicle. Let k denote the spring constant (needs to be found.) Let x denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}.

Initial kinetic energy ({\rm KE}) of the vehicle:

\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}.

When the vehicle is brought to a rest, the elastic potential energy (\text{EPE}) stored in the spring would be:

\displaystyle \frac{1}{2}\, k\, x^{2}.

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial \text{KE} of the vehicle should be equal to the {\rm EPE} of the vehicle. In other words:

\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}.

Rearrange this equation to find an expression for k, the spring constant:

\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}.

Substitute in the given values m = 1508\; {\rm kg}, v \approx 1.908\; {\rm m\cdot s^{-1}}, and x = 6.87\; {\rm m}:

\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}

8 0
2 years ago
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