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devlian [24]
2 years ago
13

What is a

Physics
1 answer:
LiRa [457]2 years ago
4 0

Answer:

I think there six points

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Please help me fill out the chart.??
LekaFEV [45]

Answer:unbalanced: have direction,Change an objects motion, causes object to accelerate

Balanced:Do not change an objects motion, Net forces equal sum of all forces on object, and Does not equal 0 N

Explanation: Thats all I know Hoped I helped sum

5 0
3 years ago
Un globo de aire caliente tiene un volumen de 560 mL a la presión atmosférica normal y una temperatura del aire de 400 ºC. Cuand
dalvyx [7]

Explanation:

charles law V1/T1 =V2/T2

560 x 673 =V2/973

376880 = V2/973

V2 = 376880 x 973 = 366704240mL

4 0
2 years ago
Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a
allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2

Acceleration of the frog is 2874.33 m/s²

6 0
3 years ago
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several
mamaluj [8]

Answer:

The  charge on the dust particle is  q_d  = 6.94 *10^{-13} \  C

Explanation:

From the question we are told that

    The length is  l = 2.0 \ m

    The width is  w = 4.0 \ m

   The charge is  q =  -10\mu C= -10*10^{-6} \ C

    The mass suspended in mid-air is m_a =  5.0 \mu g =  5.0 *10^{-6} \ g =  5.0 *10^{-9} \  kg

   

Generally the electric field on the carpet is mathematically represented as

           E =  \frac{q}{ 2 *  A  *  \epsilon _o}

Where \epsilon _o is the permittivity of free space with value \epsilon_o = 8.85*10^{-12}  \ \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

           E =  \frac{-10*10^{-6}}{ 2 *  (2 * 4 )  *  8.85*10^{-12}}

           E = -70621.5 \  N/C

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        F__{E}} =  F__{G}}

=>     q_d *  E  =  m * g

=>      q_d  =  \frac{m * g}{E}

=>      q_d  =  \frac{5.0 *10^{-9} * 9.8}{70621.5}

=>     q_d  = 6.94 *10^{-13} \  C

4 0
3 years ago
10. A boat traveling at 9.5 m/s reduces its acceleration at a rate of 2.5 m/s2. What is the final speed of the boat
Fiesta28 [93]

Answer:

6.75m/s

Explanation:

using the second equation of motion, the time is calculated.

and with the formula a= (v - u)/t

where a is acceleration but in this case it's deceleration (and should be negated as you solve it ) .

v is final velocity

u is initial velocity

t is time taken

7 0
2 years ago
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