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Sav [38]
2 years ago
15

A space rocket is launched and accelerates uniformly to 160 m/s in 4.5 s. Calculate the acceleration of the rocket.​

Physics
1 answer:
-BARSIC- [3]2 years ago
6 0

Answer:

<h2>35.56 m/s²</h2>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

a =  \frac{v}{t}  \\

v is the velocity

t is the time

From the question we have

a =  \frac{160}{4.5}  = 35.55555...

We have the final answer as

<h3>35.56 m/s²</h3>

Hope this helps you

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Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe
Colt1911 [192]

Answer:

Explanation:

First of all we shall calculate electric field near charge 2Q .

electric field due to charge Q = K x Q /  (5 x 10⁻² )²

E₁  = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis

E₁  = KQ x 10⁴  i / 25  

Similarly electric field due to charge 3Q near 2Q

=  3KQ x 10⁴  i / 25 . It is acting along y-axis

E₂ = 3KQ x 10⁴  j / 25

Similarly electric field due to charge 4Q near 2Q

=  4KQ x 10⁴  j / (25 x 2 )

= 2 KQ x 10⁴  / 25 . It is acting acting along north east direction

unit vector in north east direction = ( i + j )/ √2

So E₃ can be represented by

E₃ = 2 KQ x 10⁴  ( i + j )  / 25 x √2

Total field =  KQ x 10⁴  i / 25 + 3KQ x 10⁴  j / 25 + 2 KQ x 10⁴  ( i + j )  / ( 25 x √2 )

= KQ x 10⁴  [ i + 3 j + √2 i + √2 j ) / 25

= 400 KQ ( 2.414 i + 4.414 j )  N / C

Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j )  x 2Q  N

= 800 KQ² ( 2.414 i + 4.414 j ) N

= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N

= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

4 0
3 years ago
Can you help me with these physics questions
Andre45 [30]

Answer:

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Explanation:

5 0
3 years ago
What is the frequency of a tuning fork that vibrates 518 times in 3.50 s?
Serga [27]

Given:

The number of vibrations of the tuning fork is,

51\text{8 times}

in,

t=3.50\text{ s}

To find:

The frequency of the fork

Explanation:

The frequency is the number of vibrations in 1 second

The frequency for the given fork is,

\begin{gathered} \frac{The\text{ total vibrations}}{The\text{ time}} \\ =\frac{518}{3.50} \\ =148\text{ Hz} \end{gathered}

Hence, the frequency is 148 Hz.

7 0
1 year ago
What is an object's change in position, or displacement, and the change in time over which that displacement occurred used to ca
nikklg [1K]
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7 0
3 years ago
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ivanzaharov [21]

Answer:

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