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2 years ago

15

If it takes 3.5 hours for the hogwarts express, moving at a speed of 120 mi/hr, to make it from platform 9 and 3/4 to hogwarts,

how far apart are they?

1 answer:

jeka942 years ago

7 0

Answer:

Given:

Time taken: 3.5 hrs

Speed: 120mi/HR

Now we know that

distance = speed x time

Substituting the given values in the above formula we get

Distance= 120mi/hr x 3 hrs

Distance = 360 mi= 579.36 Km

Explanation:

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What is the equator ?????e define

pav-90 [236]

$\bold{ANSWER}$

EQUATOR IS THE LINE THAT DIVIDES EARTH INTO TWO HEMISPHERES .

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3 years ago

Two 0.50 g spheres are charged equally and placed 2.5 cm apart. When released, they begin to accelerate at 170 m/s^2 .What is th

vitfil [10]

Answer:

$q=7.65*10^{-8}C$

Explanation:

Using Newton's second law, we calculate the magnitude of the electric force between the spheres:

$F=ma\\F=0.5*10^{-3}kg(170\frac{m}{s^2})\\F=0.085N$

The magnitude of the charge in both spheres is the same. So, we calculate the charge, using Coulomb's law:

$F=\frac{kq^2}{d^2}\\q=\sqrt\frac{Fd^2}{k}\\q=\sqrt\frac{(0.085N)(2.5*10^{-2}m)^2}{8.99*10^9\frac{N\cdot m^2}{C^2}}\\q=7.65*10^{-8}C$

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3 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

coldgirl [10]

Answer:

Explanation:

charge, q = 1.6 x 10^-19 C

distance, r = 911 nm = 911 x 10^-9 m

The Coulomb's force is given by

$F=\frac{Kq_{1}q_{2}}{r^{2}}$

$F=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{\left (911\times 10^{-9} \right )^2}$

F = 2.78 x 10^-16 N

The force between the electron and the proton is 2.78 x 10^-16 N.

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3 years ago

Explain why you can only change one variable in an experiment

ohaa [14]

The purpose of an experiment is to LEARN the EFFECT of something.

The way you do that is to CHANGE the thing and see what happens.

You can change as many things as you want to. But If you change
TWO things and observe the result, then you don't know which one
of them caused the effect you see.

Or maybe BOTH of them working together caused it. You don't know.

So your experiment is not really much good. You need to do it again.

5 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

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3 years ago