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Marta_Voda [28]

3 years ago

6

What happens to the force of gravity between two masses if one mass is decreased?

2 answers:

Andrews [41]3 years ago

7 0

<span>B. It stays the same</span>

Cloud [144]3 years ago

4 0

B. It stays the same

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PLSSSSSSSS SOMEONE HELP ME WITH THIS ONE!!

zloy xaker [14]

Answer:

D

Explanation:

At the other end of the spectrum toward red, the wavelengths are longer and have lower energy.

8 0

2 years ago

First law of equilibrium

valentina_108 [34]

Answer:

For an object to be an equilibrium it must be experiencing no acceleration.

Explanation:

Hope it helps.

5 0

2 years ago

A woman on a bicycle traveling at 10 m/s on a horizontal road stops pedaling as she starts up a hill inclined at 4.0º to the hor

IrinaK [193]

The kinetic energy K = 0.5 * m * v² must be equal to the potential energy U = m * g * h.

m mass
v velocity
h height
g = 9.81m/s²

The mass m cancels out:
0.5 * v² = g * h
Solve for height h and transform to distance traveled.
(sin (4°) = height / distance)

6 0

3 years ago

After a check up, a person now has a far point of 100 cm, but with good near point vision. He needs to wear a new pair of correc

lakkis [162]

Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

$P = -0.5 dioptre$

now we have

$f = \frac{1}{P}$

$f = -2 m$

$f = -200 cm$

now we know that for normal vision the maximum distance of vision is for infinite distance

so by lens formula we have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{d_i} + 0 = \frac{1}{200}$

$d_i = 200 cm$

so his far point according to this pair of glass is 200 cm

7 0

2 years ago

The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol

Lesechka [4]

Answer:

$\Delta V = 0.053 A$

Explanation:

Electric field in a given region is given by equation

$E = \frac{A}{r^4}$

as we know the relation between electric field and potential difference is given as

$\Delta V = -\int E. dr$

so here we have

$\Delta V = - \int (\frac{A}{r^4}) .dr$

$\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}$

here we know that

$r_1 = 1.71 m$ and $r_2 = 2.89 m$

so we will have

$\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})$

so we will have

$\Delta V = 0.053 A$

8 0

2 years ago