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3 years ago

13

a rocket, initially at rest, steadily gains speed at a rate of 13.0m/s^2 for 6.40 during take-off. How far did the rocket travel

during take off?

1 answer:

Aliun [14]3 years ago

3 0

Explanation:

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A 56.0 kg box hangs from a rope. what is the tension in the rope if: the box has vy = 5.30 m/s and is speeding up at 5.10 m/s2 ?

Minchanka [31]

T = tension force in the rope in upward direction

m = mass of the box attached at end of rope = 56 kg

W = weight of the box in downward direction due to gravity

a = acceleration of the box in upward direction = 5.10 m/s²

weight of the box is given as

W = mg

inserting the values

W = (56) (9.8)

W = 548.8 N

force equation for the motion of the box is given as

T - W = ma

inserting the values

T - 548.8 = (56) (5.10)

T = 834.4 N

5 0

3 years ago

Which among the following statement/s is correct?

seropon [69]

Answer:

(D) None

Explanation:

The force of gravity is the force pulling every element of matter together. The more the matter the higher the force of gravity.

Examples of this force at work are;

The force that causes an apple to fall from the tree
The force that causes a rock to roll downside a hill
The force causing people to walk on the earth surface instead of floating

The force that facilitates a pen on your hand to write on a paper is friction force between the pen and the paper. Gravitational force acts downwards thus force applied on an object beside you is not the force of gravity.

8 0

2 years ago

Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.

AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is $25.86 \times 10^4 ~m/s$ .

Initial velocity of the proton $u = 0$

Given potential difference $\Delta V = 350V$

let's assume that the speed of the proton is $v$ ,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge $q$ when accelerated with a potential difference $\Delta V$ is,

$U = q \Delta V$

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E</em> i.e

$\Delta K = \Delta V$

If the initial and final velocity of the proton is $u$ and $v$ respectively then,

change in Kinetic Energy $\implies \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0$

change in Potential Energy $\implies \Delta U = q\Delta V$

from conservation of energy,

$v= \sqrt{\frac{2q\Delta V}{m}}$

so, $v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}$

$= 25.86 \times 10^4 ~m/s$

To read more about the conservation of energy, please go to brainly.com/question/14668053

7 0

1 year ago

A car is traveling at 70 km/h. It then uniformly decelerates to a complete stop in 12 s. Find its acceleration ( in m/s2 ).

Yakvenalex [24]

Answer:

Acceleration will be $-1.620m/sec^2$

Explanation:

We have given initial speed of the car is 70 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $70km/hr=70\times \frac{1000m}{3600sec}=19.444m/sec$

It is given that car stops in 12 sec

So final speed of the car v = 0 m/sec

Time t = 12 sec

From first equation of motion v = u+at

So $0=19.444+a\times 12$

$a=\frac{-19.444}{12}=-1.620m/sec^2$ ( negative sign indicates that speed of the car will constantly decrease )

5 0

3 years ago

Read 2 more answers

In a photoelectric experiment, you shine light onto an electrode and record a current of 25 μA. When you apply +500 mV to the el

kkurt [141]

Answer:

2.083 V.

Explanation:

Stopping potential is the potential that is required to stop the current to zero . This potential is applied externally to oppose the potential created by the photoelectric effect . It gives the measure the photoelectric potential being generated .

Here current drops to 25 μA to 19 μA by a potential of 500mV

Change in current

= 25 - 19 = 6 μA

Voltage requirement for unit reduction in current

= 500 / 6 μA

To reduce current 0f 25 μA

requirement of V = (500 / 6 ) x 25 = 2083.33 mV = 2.083 V.

7 0

3 years ago