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ss7ja [257]
4 years ago
5

You go rock climbing with a pack that weighs 70 N and you reach a height of 30 m. how much work did you do to lift your pack? If

you finished the climb in 10 minutes (600 s), what was your power?
Physics
1 answer:
Vikentia [17]4 years ago
8 0
When you climb, earth exerts gravitational force on pack in downward direction(pointing towards the center of earth).
In order to climb, you need to work against work done by gravity on the pack.
Hence work done by you = work done by gravity on pack 
                                        = Force x displacement = 70 x 30 = 2100 J.
 
So you need to do 2100 joules of work to lift your pack.

Power is the rate of work done.
Therefore power = work done by you/time(in seconds)
                            =     2100/600 =3.5 watts 
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When the distance between two charges is halved, the electrical force between them?
Llana [10]
If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
F= k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
r'= \frac{r}{2}
the magnitude of the force changes as follows:
F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k  \frac{q_1 q_2}{r^2}=4 F
so, the force increases by a factor 4.
3 0
3 years ago
In a pig caller can produce a sound intensity level of 107 dB. How many pig callers would be needed to generate an intensity lev
myrzilka [38]

Answer:

20 pig callers

Explanation:

Given that:

A pig caller produced intensity level of  a sound = 107 dB

To find how many pig callers required to generate an intensity level of 120 dB;

we have:

120 dB - 107 dB = 13 dB

Taking the logarithm function;

10 \ log \bigg(\dfrac{I}{I_o} \bigg) = 13 \ dB

where;

I_o = initial intensity

log \bigg(\dfrac{I}{I_o} \bigg) = 1.3

\dfrac{I}{I_o}=  10^{1.3 }

I = 19.95I_o

I ≅ 20 pig callers

6 0
3 years ago
This problem has been solved!
lana66690 [7]

Answer:

Charge on each metal sphere will be 8\times 10^{8}C

Explanation:

We have given number of electron added to metal sphere A n=10^{12}electron

As both the spheres are connected by rod so half -half electron will be distributed on both the spheres.

So electron on both the spheres =\frac{10^{12}}{2}=5\times 10^{11}electron

We know that charge on each electron e=1.6\times 10^{-19}C

So charge on both the spheres will be equal to q=1.6\times 10^{-19}\times 5\times 10^{11}=8\times 10^{8}C

So charge on each metal sphere will be equal to 8\times 10^{8}C

6 0
3 years ago
) Un círculo de 120 cm de radio gira a 600 rpm. Calcula: a) su velocidad angular
DIA [1.3K]

Responder:

20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz

Explicación:

Dado lo siguiente:

Radio (r) del círculo = 120 cm

600 revoluciones por minuto en radianes por segundo

(600 / min) * (2π rad / 1 rev) * (1min / 60seg)

(1200πrad / 60sec) = 20π rad ^ -1

Velocidad angular (w) = 20πrads ^ -1

Velocidad lineal = radio (r) * velocidad angular (w)

Velocidad lineal = (120/100) * 20πrad

Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1

C.) Período (T):

T = 2π / w = 2π / 20π = 0.1 seg

D.) Frecuencia (f):

f = 1 / T = 1 / 0.1

1 / 0,1 = 10 Hz

5 0
3 years ago
I NEED HELP ASAP!! I NEED IT BEFORE TOMORROW!! D: 10pts+5pts!
Wewaii [24]
9. Compounds can form from two nonmetals by sharing their electrons in a
C) covalent

11. An atom that has an excess positive or negative electrical charge caused by the loss or addition of an electron is called a(n) ______.
B) ion

5 is either A or C
4 0
3 years ago
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