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maks197457 [2]
4 years ago
13

A person can jump 1.5m on the earth. How high could the person jump on a planet having the twice the mass of the earth and twice

the radius of the earth?
Physics
2 answers:
guapka [62]4 years ago
8 0
I believe you'd have to calculate the different planets info and use the formula you the problem used to find their height of the jump in order to find yours ..
MrMuchimi4 years ago
5 0
F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high

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What is the acceleration of a 0.90g drop of blood in the fingertips at the bottom of the swing?
Scilla [17]
The acceleration of a 0.90 g drop of blood in the fingertips at the bottom of the swing is the sum of the acceleration of the movement of the finger and the acceleration of gravity. In this case, this is different when the finger goes down, since the acceleration now becomes the difference between the two.
5 0
4 years ago
A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d
zvonat [6]

Answer:

<u />D_l=d<u />

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

V_{e1}=V_{e2}

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

D=\frac{at^2}{2}

Where

Acceleration is given as

a=\frac{V_o}{2d}

And

Time

T=\frac{d}{v_0}

Therefore horizontal displacement towards the left is

D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}

<u />D_l=d<u />

5 0
3 years ago
The magnetic field at the center of a 1.40-cm-diameter loop is 2.50 mT . PART A) What is the current in the loop?
NISA [10]

Explanation:

It is given that,

Diameter of loop, d = 1.4 cm

Radius of loop, r = 0.7 cm = 0.007 m

Magnetic field, B=2.5\ mT=2.5\times 10^{-3}\ T

(A) Magnetic field of a current loop is given by :

B=\dfrac{\mu_oI}{2r}

I is the current in the loop

I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.5\times 10^{-3}\times 0.007}{4\pi \times 10^{-7}}

I = 27.85 A

(B) Magnetic field at a distance r from a wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 27.85}{2\pi \times 2.5\times 10^{-3}}

r = 0.00222 m

r=2.2\times 10^{-3}\ m

Hence, this is the required solution.

3 0
3 years ago
Help with 3 please! I will give brainliest!
rusak2 [61]
The answer is point b because vertical velocity is zero at the maximum height
8 0
3 years ago
What is 0.002 represented by
Helga [31]
I think you want 2x10^-2
It’s called Scientific Notation
6 0
3 years ago
Read 2 more answers
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