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3 years ago

If you had a planet to choose from to live in what would it be?

Physics

2 answers:

Alja [10]3 years ago

8 0

I would personally like to live on mars because it is like our temperature, but a little bit colder.

postnew [5]3 years ago

4 0

I personally would live on Mars cuz that is red cuz

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ANSWER QUCIK<br> PLS THANK YOU

Reil [10]

Electrical energy in the charger and cable

Chemical energy in the battery of the mobile phone

8 0

3 years ago

A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular

ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

$\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0$

Where,

$\theta =$ Angular Displacement

$\alpha =$ Angular Acceleration

$\omega_0 =$ Angular velocity

$\theta_0 =$ Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

$\theta = \frac{1}{2}\alpha t^2$

Re-arrange for $\alpha,$

$\alpha = \frac{2\theta}{t^2}$

Replacing our values,

$\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}$

$\alpha = 0.139rad/s$

Therefore the ANgular acceleration of the mass is $0.139rad/s^2$

4 0

3 years ago

Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5

schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0

3 years ago

What is the acceleration of a dog that runs from 3 m/s to 6 m/s over a distance of 90m?

KonstantinChe [14]

Answer:

solution given:

acceleration (a)=?

initial velocity (u)=3m/s

final velocity (v)=6m/s

distance (s)=90m

we have

v²=u²+2as

substituting value

6²=3²+2*a*90

36=9+180a

36-9=180a

a=25/180

<u>a=0.1388m/s²</u>

6 0

3 years ago

Machmer Hall is 400 m North and 180 m West of Witless.

yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

$D = \sqrt{(d_{m})^2+(d_{w})^2}$

Where, $d_{m}$ =distance of Machmer Hall

$d_{w}$ =distance of Witless

Put the value into the formula

$D = \sqrt{(400)^2+(180)^2}$

$D=438.63\ m$

Hence, The distance from Witless to Machmer is 438.63 m.

5 0

3 years ago