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Rus_ich [418]
3 years ago
11

What would be Kelley's weight be in newtons if her mass was 70 kilograms

Physics
2 answers:
ad-work [718]3 years ago
8 0

Answer:

Kelly's weight would be 688.47 Newtons.

Explanation:

1 Kilogram would be 9.81 Newtons.

xxMikexx [17]3 years ago
5 0

It all depends on where Kelley is at the moment.

-- If she's on Mars, she weighs 229 N.

-- If she's on the Moon, she weighs 113N.

-- If she's on Earth, she weighs 686 N.

-- If she's in a space capsule coasting from one body to another, then there's no instrument that can measure any weight of her.

So it completely depends on where she happens to be at the moment.

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A mosquito flies toward you with a velocity of 2.4 km/h [E]. If a distance of 35.0 m separates you and the mosquito initially, a
tangare [24]

First convert the speed of mosquito to m/s:

So the mosquito is flying at (2,400/3,600) m/s, or ⅔ m/s. 

<span>

Since you are moving at 2m/s, so this makes the closing velocity between you and the mosquito to be 2⅔ m/s. </span>

Therefore the mosquito will hit your sunglasses at:<span>

35 m / (2⅔ m/s) = 13⅛ seconds. 

2.0 m/s * 13⅛ s = 26¼ m from your initial position. 

<span>⅔ m/s * 13⅛ s = 8¾ m from the mosquito's initial position. </span></span>
7 0
3 years ago
Draw the Lewis dot Structure for the molecule C2H6​
zzz [600]

Answer:

This means that the Lewis dot structure for C2H6 must account for 14 valence electrons, either through bonding between atoms, or through lone pairs. So, the two C atoms are placed in the center of the molecule.

6 0
3 years ago
What quantities belong in cells X and Y?
satela [25.4K]

Answer:

X: period

Y: tangential speed

Explanation:

100% on quiz your welcome(:

6 0
2 years ago
The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an
yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}

\frac{1}{f}=0.028

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

p=\frac{1}{f}

Here, p is the power of the lens.

Put f= 35.71 cm.

p=\frac{1}{35.71}

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

3 0
3 years ago
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
3 years ago
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