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Pavel [41]
3 years ago
15

A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 \,\dfra

c{\text m}{{\text s}^2}4 s 2 m ? rightward. After 3\,\text s3s, what will be the velocity of the rocket ship? Answer using a coordinate system where rightward is positive.
Physics
1 answer:
Sergio039 [100]3 years ago
6 0

Answer:

+12 m/s

Explanation:

The velocity of an object moving of accelerated motion is given by:

v(t) = v_0 +at

where

v0 is the initial velocity of the object

a is the acceleration

t is the time

In this problem, the rocket starts from rest, so v_0 =0. The acceleration is a=4 m/s^2, so the velocity after t=3 s will be

v(3 s)=0 +(4 m/s^2)(3 s)=+12 m/s

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The achievement of lifting a rocket off the ground and into space can be explained by
lawyer [7]

Answer:

The achievement of lifting a rocket off the ground and into space can be explained by Newton's third law of motion. What is required for a rocket to lift off into space? Thrust is required for a rocket to lift off into space, ... An object that travels around another object in space is called a satellite.

Explanation:

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3 years ago
What would be the daughter product of the beta decay of lead-212 and what would be the daughter product of the alpha decay of po
kati45 [8]

Answer:

lead with Z = 82 is transformed into Bismuth with Z = 73

The Po with Z = 84 becomes Pb with Z = 82

Explanation:

Beta decay occurs when a neutron emits an electron and an anti neutrino from the atomic nucleus, therefore the atomic number of the material increases by one unit.

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In alpha decay, a helium nucleus is emitted from the nucleus of the atom, therefore the atomic number decreases by two units.

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Which compound has ionic bonding? A. Cl2 B. HF C. CaO D. NO2
GrogVix [38]

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7 0
3 years ago
A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with
Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)

Charge density of the shell is,

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}

Here, R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}

B)

The electric field is E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

C)

For R <r <2R According to gauss law

E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)

substitute \rho=\frac{-3Q}{28\piR^3}

E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}

D)

The net charge enclosed for each r in this range is positive and the electric field is outward

E)

For r>2R

Charge enclosed is zero, so electric field is zero

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If you were trying to cross a river with the shortest possible time, would you aim your boat slightly upstream, directly across
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