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Goshia [24]
3 years ago
14

What is the acceleration of a rocket if 200 Newtons are applied to its 5.5 kg​

Physics
1 answer:
snow_tiger [21]3 years ago
7 0
Given
Force=200N
Mass=5.5kg

F=ma
a=F/m=200/5.5=38.18metre per second
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A charge -353e is uniformly distributed along a circular arc of radius 5.30 cm, which subtends an angle of 48°. What is the line
vladimir2022 [97]

Answer:

- 1.3 x 10⁻¹⁵ C/m

Explanation:

Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C

r = Radius of the arc = 5.30 cm = 0.053 m

θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad        (Since 1 deg = 0.0175 rad)

L = length of the arc

length of the arc is given as

L = r θ

L = (0.053) (0.84)

L = 0.045 m

λ = Linear charge density

Linear charge density is given as

\lambda =\frac{Q}{L}

Inserting the values

\lambda =\frac{-564.8\times 10^{-19}}{0.045}

λ = - 1.3 x 10⁻¹⁵ C/m

4 0
3 years ago
In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n
Mazyrski [523]

Answer:

54 g of oxygen ...

4 0
3 years ago
Experiment 1: A study is done to determine which of two fuel mixtures allows a rocket to travel farther over a period of time. R
leva [86]

Answer: D

Experiment 1 has a confounding variable related to the mass of the rockets. Any variation in mass may cause a discrepancy in the distance traveled.

This is the answer to the question because:

  • Both experiments do have a confounding variable.
  • Experiment 1 doesn't have to stay constant.
  • A double-blind experiment will not do anything to the placebo.
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The answer has to be the option D. Hope this helps you!

3 0
4 years ago
What is the formula for conservation of momentum
olga55 [171]

Answer:

The general equation for conservation of momentum during a collision between n number of objects is given as: [m i ×v i a ] = [m i ×v i b ] Where m i is the mass of object i , v i a is the velocity of object i before the collision, and v i b is the velocity of object i after the collision.

Explanation:

6 0
3 years ago
What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne
irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance h= 1790 km = 1.790\times10^{6}\ m

Magnetic field B=4\times10^{-8}\ T

Mass of proton m=1.673\times10^{-21}\ Kg

Radius of earth R =6.38\times10^{6}\ m

Radius of orbit r=R+h

r=6.38\times10^{6}+1.790\times10^{6}

r=8170000\ m

We need to calculate the speed

Using formula of magnetic field

Bvq=\dfrac{mv^2}{r}

v=\dfrac{Bqr}{m}

Put the value into the formula

v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}

v=31.25\ m/s

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0
3 years ago
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