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sertanlavr [38]
3 years ago
8

if a person can jump 2m in earth surface how high can he jump in the moon (g of moon = 1.66m/s, g of earth = 9.8 m/s) [hint: use

w=mgh since work done and mass of a person is same everywhere​
Physics
1 answer:
julia-pushkina [17]3 years ago
5 0

Answer:

h_{moon} = 11.8\ m

Explanation:

Since the work done is same everywhere in the universe. Hence, the work done in jumping will be same for the person on moon and earth:

W_{moon} = W_{earth}\\\\P.E_{moon} = P.E_{earth}\\\\mg_{moon}h_{moon} = mg_{earth}h_{earth}\\\\g_{moon}h_{moon} = g_{earth}h_{earth}\\\\(1.66\ m/s^2)h_{moon} = (9.8\ m/s^2)(2\ m)\\\\h_{moon} = \frac{(9.8\ m/s^2)(2\ m)}{1.66\ m/s^2}\\\\h_{moon} =  11.8\ m

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Transverse wave

Explanation:

Its because in transverse wave the particle displacement is perpendicular to the direction of wave propagation..hope it helps you...

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3 years ago
Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m
Alex Ar [27]

Answer:

18 N/C

Explanation:

Given that:

Electric field constant, k = 9*10^9 N/c

Distance, r = 10^-8 m

Dipole moment, p = 10^-33

Using the relation for electric field due to dipole :

E = [2KP / r³]

E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3

E = (18 * 10^9 * 10^-33) ÷ 10^-24

E = [18 * 10^(9-33)] ÷ 10^-24

E = (18 * 10^-24) / 10^-24

E = 18 * 10^-24+24

E = 18 * 10^0

E = 18 N/C

5 0
3 years ago
Can a moving object have gravitational energy?
Evgesh-ka [11]

Answer:

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3 years ago
In tae-kwon-do, a hand is slammed down onto a target at a speed of 14 m/s and comes to a stop during the 3.0 ms collision. Assum
GrogVix [38]

Answer:

A) The impulse is 11,7 kg.m/s .  B) The average force on the hand is 3900N

Explanation:

Givens:

Hand's mass (m)= 0,90kg

Hand's initial velocity (vi)= 14m/s

Time of collision (t)= 3ms

<u>A) Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time. We use the formula:</u>

<em />

<em>/F/ = m. /Δv/</em>

<u>Where:</u>

F= Force of Impulse

m= Mass

Δv= change in speed ( vf-vi)

Using that formula we get:

/F/= 0,90kg . / (0m/s-14m/s) /

/F/=0,90kg . 14m/s

<em>/F/ = 11,7 kg. m/s</em>

*note that (vf) is 0 because the hand stops in the action, so it's final

velocity = 0

<u>B)The average force is equal to the change in the momentum over the change in time. We use the formula:</u>

<em>/F/= m. (Δv) : Δt</em>

<u>Where:</u>

F= the average force from the target

m= the mass of the hand

Δv= hange in speed (vf/vi)

Δt= change in time

*if we look closely, we find the we alredy have half of the equation because we have alredy calculated m.(Δv) in point A.

We boil the equation down to:

/F/= Impulse : Δt

/F/= 11,7kg.m/s : 0,003s

<em>/F/= 3900N</em>

*we use 0,003s as our time because the given time was 3ms.

*the final result is expressed in Newtons because our final result ends up beeing <em>kg.m/s²</em> = N

<u />

8 0
4 years ago
How does electric circuit wor
zaharov [31]
Electricity passes through it
4 0
4 years ago
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