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aleksklad [387]
2 years ago
5

Can yall please help!!

Physics
1 answer:
andreyandreev [35.5K]2 years ago
7 0

The compasses would start point towards the wire. When you have a current going through wire it messes with the electrical field and messes up the compasses so they point towards the wire.

Hope this helps!

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Which physical activity would a global positioning system resource help you with?
PolarNik [594]

The physical activity that a global positioning system resource can help me with is : ( A )  Hiking

<h3>Function of Global positioning system ( GPS ) </h3>

GPS helps with the accurate measurement of physical activities  and factors such as location, time, elevation and so on. When hiking, the time, distance covered and location of the hiker  can be accurately measured with the use of the GPS.

Hence we can conclude that The physical activity that a global positioning system resource can help me with is  Hiking.

Learn more about GPS : brainly.com/question/9795929

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4 0
1 year ago
what occurs in a chemical reaction? A. Reactants are formed without chemical bonds being broken B. Reactants are formed from pro
Ann [662]
The answer is D. Products are formed from reactants by the breaking and forming of new bonds. 
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2 years ago
Read 2 more answers
A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
2 years ago
Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.
Alecsey [184]

Time taken by the water balloon to reach the bottom will be given as

h = \frac{1}{2} gt^2

here we know that

h = 18 m

g = 9.8 m/s^2

now by the above formula

18 = \frac{1}{2}*9.8* t^2

18 = 4.9 t^2

t = 1.92 s

now in the same time interval we can say the distance moved by it will be

d = v_x * t

d = 8.2 * 1.92 = 15.7 m

so it will fall at a distance 15.7 m from its initial position

5 0
3 years ago
To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i
ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

m_1v_1+m_2v_2=(m_1+m_2)v

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

m_1(-v_1)+m_2v_2=(m1+m2)(-v2)

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s

hence, the time is t=0.42 s

4 0
3 years ago
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