Gold's natural state has a definite shape and a definite volume. What is gold's natural state(s)?

Two rocks are at the top of a building. Rock 1 is dropped from rest while Rock 2 is thrown horizontaly at a velocity of 5 ms.

LuckyWell [14K]

Answer:

I'm pretty sure the answer is 0 m/s²

Explanation:

The horizontal velocity of the second rock is 5 m/s, so if we pretend air resistance doesn't exist, it will maintain that horizontal velocity, meaning that there is no horizontal acceleration.

7 0

3 years ago

Find the voltage change when: a. An electric field does 12 J of work on a 0.0001-C charge. b. The same electric field does 24 J

kondaur [170]

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

$V=\dfrac{W}{q}$

$V=\dfrac{12}{0.0001}$

$V=12\times 10^4\ Volt$

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

$V=\dfrac{W}{q}$

$V=\dfrac{24}{0.0002}$

$V=12\times 10^4\ Volt$

Therefore, this is the required solution.

7 0

3 years ago

A box is sliding up an incline that makes an angle of 14.0° with respect to the horizontal. the coefficient of kinetic friction

mezya [45]

A box is sliding up an incline that makes an angle of 14.0° with respect to the horizontal. the coefficient of kinetic friction between the box and the surface of the incline is 0.180. the initial speed of the box at the bottom of the incline is 2.20 m/s. how far does the box travel along the incline before coming to rest?

6 0

3 years ago

The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the

FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

$B=\frac{2\pi*a }{u*I}$

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

$I=\frac{2\pi*a*B }{u}$ Formula(2)

$B=0.2*10^{-5} T = 0.2*10^{-5} \frac{weber}{m^{2} }$

a =8cm=0.08m

$u=4*\pi *10^{-7} \frac{Weber}{A*m}$

We replace the known information in the formula (2)

$I=\frac{2\pi*0.08*0.2*10^{-5} }{4\pi *10x^{-7} }$

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0

3 years ago

you ride your bike for a distance of 30km. you travel ata a speed of 0.75km/minute.how many minutes does it take

mihalych1998 [28]

Distance for which the bike is ridden = 30 km
Speed at which the bike is driven = 0.75 km/minute
Let us assume the number of minutes taken to travel the distance of 30 km = x
Now we already know the formula of speed can be written as
Speed = Distance traveled/ Time taken
0.75 = 30/x
0.75x = 30
x = 30/0.75
= 40 minutes
So the time taken for riding a distance of 30 km will be 40 minutes. I hope this procedure is simple enough for you to understand.

7 0

3 years ago

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