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3 years ago

14

Please help me link - https://syx48hz22ou.typeform.com/to/TIIxnYFr tell me when your finished

1 answer:

Blababa [14]3 years ago

6 0

Answer:

the link doesn't work

Explanation:

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The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n

monitta

Answer:

i know the questin but i got to try and find it

Explanation:

5 0

3 years ago

What gravitational force does the moon produce on the earth is their centers are 3. 88x10^8 m apart and the moon has a mass of 7

vitfil [10]

The Moon is 3.8 108 m from Earth and has a mass of 7.34 1022 kg. 5.97 1024 kg is the mass of the Earth.

<h3>What kind of gravitational pull does the moon have on the planet?</h3>

On the surface of the Moon, the acceleration caused by gravity around 1.625 m/s2 which is 16.6% greater than on the surface of the Earth 0.166.

<h3>What does the Earth's center's gravitational pull feel like?</h3>

Gravity is zero if you are in the centre of the earth since everything around you is pulling "up" (up is the only direction).

<h3>Where is the Earth's and the moon's gravitational centre?</h3>

It is around 1700 kilometres below Earth's surface.

To know more about gravitational force visit:-

brainly.com/question/12528243

#SPJ4

6 0

1 year ago

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

Molodets [167]

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 jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e 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 fina 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 ony 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)

5 0

3 years ago

Please help have due very soon?thank you

Andru [333]

The answer is :78 I think

8 0

3 years ago

You have two resistors with the same cross sectional area and resistivity. Resistor A has length L1 and resistor B has length L2

oee [108]

Answer:

Explanation:

Given

Resistor A has length $L_1$

and Resistor B has Length $L_2$

and Resistance is given by

$R=\frac{\rho L}{A}$

Considering $\rho$ and A to be constant thus

$R_2>R_1$ because $L_2>L_1$

(a)When they are connected in series

As the current in series is same and power is $i^2R$

therefore $P_2>P_1$ as R is greater for second resistor

(b)if they are connected in Parallel

In Parallel connection Voltage is same

$P=\frac{V^2}{R}$

resistance of 2 is greater than 1 thus Power delivered by 1 is greater than 2

8 0

3 years ago