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Mama L [17]
3 years ago
5

The velocity of a mass moving in a circular path is..

Physics
1 answer:
Alenkasestr [34]3 years ago
5 0
Constantly changing.
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A net force of 25N causes an object to accelerate at 4m/s^2. what is the mass of the object?
Doss [256]

Answer:

6.25 kg

Explanation:

Fnet=ma

m=Fnet/a

m=25/4

m=6.25kg

3 0
3 years ago
Help with 1 2 and 3 please
geniusboy [140]

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. I = \frac{q}{t} Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3.  So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

\frac{p1L1}{A1}  = \frac{p2L2}{A2}\\

We are looking for L2 so we can isolate using algebra to get:

\frac{A2(\frac{P1L1}{A1}) }{P2} = L2

If you fill in those values you get 0.0205

or 2.05 cm



6 0
3 years ago
Earth’s atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by Earth’s atmosphere?
GarryVolchara [31]

Answer:

Earth's atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by earth's atmosphere? Earth has moderate temperatures.

Explanation:

7 0
3 years ago
it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J

Therefore, the additional work needed is

\Delta E=E'-E=810-90=720 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

7 0
3 years ago
What current is needed in the solenoid's wires?
marta [7]
Using Ampere's Law, the magnetic field produced inside this solenoid is given by
B = uo N I / h
where uo is the vacuum permeability, N is the number of turns in the solenoid and h is the length of the solenoid. Earth's magnetic field is around 50 microteslas in North America thus the current needed in the solenoid is
I = B h / (uo N) = (50 E-6 ) (4) / ((4 pi E-7)(6000) ) = 0.026 A
I = 26 mA
So you need a current of around 26 mA.
5 0
3 years ago
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