If a rock has a mass of 0.15 kg on the moon what will its mass be on earth

Sunny_sXe [5.5K]

The wavelength is the spatial period of a periodic wave—the distance over which the wave's shape repeats.

6 0

3 years ago

A man hits a 50 grams golf ball such that it leaves the tee at an angle of 40o with the horizontal and strikes the ground at the

lukranit [14]

Answer:

Explanation:

Range of projectile R = 20 m

formula of range

R = u² sin2θ / g

u is initial velocity , θ is angle of projectile

putting the values

20 = u² sin2x 40 / 9.8

u² = 199

u = 14.10 m /s

At the initial point

vertical component of u

= u sin40 = 14.1 x sin 40

= 9.06 m/s

Horizontal component

= u cos 30

At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .

Horizontal component of velocity

u cos 30

Vertical component

= - u sin 30

= - 9.06 m /s

So its horizontal component remains unchanged .

change in vertical component = 9.06 - ( - 9.06 )

= 18.12 m /s

change in momentum

mass x change in velocity

= .050 x 18.12

= .906 N.s

Impulse = change in momentum

= .906 N.s .

3 0

3 years ago

A cyclist reduces his speed from 6.5 m/s to 0.0 m/s with an acceleration of

kolbaska11 [484]

Answer:

a= -1.2 m/s^2

Vi= 6.5 m/s

Vf= 0 m/s

t= 0-6.5/-1.2= <u>5.45 Sec</u>

Explanation:

4 0

2 years ago

An inductor is connected in series to a fully charged capacitor. Which of the following statements are true? Check all that appl

Luba_88 [7]

Answer:

As the capacitor is discharging, the current is increasing

Explanation:

Lets take

C= Capacitance

L=Inductance

V=Voltage

I= Current

The total energy E given as

$E=\dfrac{IL^2}{2}+\dfrac{CV^2}{2}$

We know that total energy E is conserved so when electric energy 1/2 CV² decreases then magnetic energy 1/2 IL² will increases.

It means that when charge on the capacitor decreases then the current will increase.

As the capacitor is discharging, the current is increasing

3 0

3 years ago

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago