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zhannawk [14.2K]
2 years ago
15

In vacuum, red light travels at a greater speed than blue light does. a. true b. false?

Physics
1 answer:
3241004551 [841]2 years ago
8 0
B)False.......they travels at the same speed
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A 1090 kg car moving +11.0 m/s
stiks02 [169]

The mass of the second car is 1434.21 kg

<u>Explanation:</u>

Using law of conservation of momentum,

          m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v

Given:

m_{1} = 1090 kg

u_{1} = 11 m/s

u_{2} = 0

v = 4.75 m/s

We need to find m_{2}

When substituting the given values in the above equation, we get

(1090 \times 11)+\left(m_{2} \times 0\right)=\left(1090+m_{2}\right) 4.75

11990=5177.5+4.75 m_{2}

4.75 m_{2}=11990-5177.5

4.75 m_{2}=6812.5

m_{2}=\frac{6812.5}{4.75}=1434.21 \mathrm{kg}

6 0
2 years ago
Which action will increase the gravitational force between two objects?
schepotkina [342]
The answer is D have a nice day!
8 0
3 years ago
Read 2 more answers
What is the average acceleration of a car that starts from
MAXImum [283]

Answer:

1.5m/s^2

Explanation:

In the picture above.

4 0
3 years ago
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A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
2 years ago
The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 310 grams, and it is released at an angle of 12 degree
lina2011 [118]
A) the periodic time is given by the equation;
 T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
   = 6.28 × √0.0673
    = 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
 thus; frequency = 0.614 Hz

b)  The vertical distance, the height is given by
 h= 0.66 cos 12
 h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
 mgh = 1/2mv²
  v² = 2gΔh = 2×9.81 × 0.01 
                   = 0.1962
v = 0.443 m/s

c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
   = 1/2 × 0.310 × 0.443²
   = 0.0304 J


4 0
3 years ago
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