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Sergeu [11.5K]
3 years ago
14

Highlight the correct terms:

Physics
1 answer:
larisa [96]3 years ago
8 0
The (more) speed an object has, the (lower) the potential energy and the (higher) the kinetic energy. (I believe that is correct but it’s been a while since I’ve done this)
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A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr
Musya8 [376]

Answer:

162500000.  

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

I=\dfrac{q}{t}

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}

n=162500000 t

\dfrac{n}{t}=16250000

The number of electron passe per second will be 162500000.

4 0
3 years ago
Worth 15 points<br><br> Please answer which one matches for each conversion
FrozenT [24]

Answer:

distance - meters

speed - meters/seconds

time - seconds

velocity - meters/seconds

acceleration - meters/seconds²

5 0
2 years ago
What is the formula for calculating the efficiency of a heat engine
horsena [70]
The answer is Eficiency=T<< Tox100
5 0
3 years ago
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Andrej [43]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

3 0
3 years ago
What is the force of gravity acting on a 1-kg m mass? (g = 9.8 m/s ^ 2)
Ksenya-84 [330]

Answer: Use this F=Ma.

Explanation: So your answer will be

F=1 Kg+9.8 ms-2

So the answer will be

F=9.8N

How'd I do this?

I just used Newton's second law of motion.

I'll also put the derivation just in case.

Applied force α (Not its alpha, proportionality symbol) change in momentum

Δp α p final- p initial

Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)

or then

F α m(v-u)/t

So, as we know v=final velocity & u= initial velocity and v-u/t =a.

So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.

So, F=<em>k</em>ma (where k is the proportionality constant)

<em>k</em> is 1 so you can ignore it.

So, our equation becomes F=ma

7 0
3 years ago
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