All categories

1 year ago

The Blue Origin rocket ship starts vertically upward from the launch pad a constant

Physics

1 answer:

Flura [38]1 year ago

6 0

The rocket ship rises to a height of 27.83 km above the launch pad after 94 seconds of flight.

We know that,

s = u t + 1 / 2 a t²

where,

s = Displacement

u = Initial velocity

a = Acceleration

t = Time

Given that,

u = 0

a = 6.3 m / s²

t = 94 s

s = 0 + 1 / 2 ( 6.3 * 94² )

s = 27833.4 m

s = 27.83 km

Displacement the distance between the initial point and final point. It is denoted using s.

Therefore, the rocket ship rises to a height of 27.83 km above the launch pad after 94 seconds of flight.

To know more about Displacement

brainly.com/question/11934397

#SPJ4

You might be interested in

What is another name for a "causal loop"?

OleMash [197]

B I believe but hey we also have Google to double check

8 0

4 years ago

Read 2 more answers

The unit of work is a fundamental derived unit.<br>

PIT_PIT [208]

Answer:

Explanation:

Joules

7 0

3 years ago

A 3 Q and a 1.5 0 resistor are wired in parallel and the combination is wired in

alekssr [168]

If a 3 Ω and a 1.5 Ω resistor are wired in parallel and the combination is wired in series to a 4 Ω and a 10 V emf device then the current in the 3 Q resistor is 0.667 Ampere

<h3>What is resistance?</h3>

Resistance is the obstruction of electrons in an electrically conducting material.

The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

As given in the problem 3 Ω and a 1.5 Ω resistor are wired in parallel

Their equivalent resistance would be

1/Re= 1/R1 + 1/R2

1/Re = 1/3 +1/1.5

Re= 1 Ω

Now this parallel equivalent is connected with a series combination with a 4Ω resistor

For calculating equivalent resistance in series combination.

R = R1 + R2

R = 1 + 4

R = 5 Ω

As the 10 V emf device is connected then the current flowing can be calculated by using Ohms law

V=IR

10 = 5×I

I = 2 ampere

as the voltage is divided between parallel resistor combination and series resistor

The voltage drop across parallel combination = 10×(1/1+4) volt

= 2 volt

Now the current flows through a 3 Ω resistor by using Ohms law

I = V/R

I= 2/3

I = 0.667 Ampere

Thus, the current flowing through the 3 Ω resistor comes out to be 0.667 Ampere

Learn more about resistance from here

brainly.com/question/14547003

#SPJ1

3 0

2 years ago

A charge of 0.20uC is 30cm from a point charge of 3.0uC in vacuum. what work is required to bring the 0.2uC charge 18cm closer t

vlada-n [284]

Answer:

The correct answer is " $4.49\times 10^{10} \ joules$ ".

Explanation:

According to the question,

The work will be:

⇒ $Work=-\frac{kQq}{R}$

$=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}$

$=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}$

$=\frac{0.3978}{\varepsilon }$

$=4.49\times 10^{10} \ joules$

Thus the above is the correct answer.

3 0

3 years ago

Consider light that has its third minimum at an angle of 23.3° when it falls on a single slit of width 4.05 μm. Find the wavelen

fredd [130]

Answer:

The wavelength of light is 533 nm.

Explanation:

It is given that,

Width of a single slit, $d=4.05\times 10^{-6}\ m$

Light has its third minimum at an angle of 23.3° when it falls on a single slit. For destructive interference, the equation for minima is given by:

$d\ sin\theta=n\lambda$

Here, n = 3

$\lambda=\dfrac{d\ sin\theta}{n}$

$\lambda=\dfrac{4.05\times 10^{-6}\times sin(23.3)}{3}$

$\lambda=5.33\times 10^{-7}\ m$

$\lambda=533\ nm$

So, the wavelength of the light is 533 nm. Hence, this is the required solution.

6 0

3 years ago