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telo118 [61]
3 years ago
13

If an object on Earth weighs 100N what is its weight in pounds?

Physics
1 answer:
oksian1 [2.3K]3 years ago
6 0

Answer:

10.2 lbs

Explanation:

m=F/g

m=100N/9.8

m=10.2040816 lbs

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In the diagram, q1 = -3.33*10^-9 C and
Anna [14]

Answer:

The distance will be "0.319 m".

Explanation:

The diagram of the given question is not attached with the problem. Below is the attached figure.

The given values are:

q1 = -3.33\times 10^{-9}  \C

q2 = -5.22\times 10^{-9} \ C

As we know,

The electric field will be equivalent as well as opposite throughout the direction, then

⇒ \frac{R.q1}{(0.255)^2} =\frac{R.q2}{r^2}

⇒ r^2=\frac{q2}{q1}\times (0.255)^2

By putting the values, we get

⇒      =\frac{-5.22\times 10^{-9}}{-3.33\times 10^{-9}}\times (0.255)^2

⇒      =1.568\times 0.065

⇒  r^2=0.10193

⇒  r=\sqrt{0.10193}

⇒     0.319 \ m    

4 0
3 years ago
A) Find the gravitational field strength of an asteroid with the mass of 3.2 * 10^3 kg and an average radius of 30 km when at a
MrMuchimi

a) 1.96\cdot 10^{-16} m/s^2

The gravitational field strength near the surface of the asteroid is given by:

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the asteroid

R the radius of the asteroid

h is the distance from the surface

Substituting the data of the asteroid:

M=3.2\cdot 10^3 kg is the mass

R=30 km = 30000 m is the radius of the asteroid

h=3 km = 3000 m is the distance from the surface

We find

g=\frac{(6.67\cdot 10^{-11})(3.2\cdot 10^3)}{(30000+3000)^2}=1.96\cdot 10^{-16} m/s^2

b) i)  5.53\cdot 10^9 s

The acceleration of the astronaut popped out at 3 km from the surface is exactly that calculated at part a):

g=1.96\cdot 10^{-16} m/s^2

So, since its motion is at constant acceleration, we can find the time he takes to reach the surface using suvat equations:

s=ut+\frac{1}{2}gt^2

where

s = 3 km = 3000 m is his displacement to reach the surface

u = 0 is his initial velocity

t is the time

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(3000)}{1.96\cdot 10^{-16} m/s^2}}=5.53\cdot 10^9 s

b) ii) 1.08\cdot 10^{-6} m/s

Again, we can use another suvat equation:

v=u+gt

where

v is the final velocity

u is the initial velocity

g is the acceleration of gravity

t is the time

Since we have

u = 0

t=5.53\cdot 10^9 s

g=1.96\cdot 10^{-16} m/s^2

The velocity of the astronaut at the surface will be

v=0+(1.96\cdot 10^{-16} m/s^2)(5.53\cdot 10^9)=1.08\cdot 10^{-6} m/s

b) iii) 175 years

The duration of one year here is

T=3.16\cdot 10^7 s

And the time it takes for the astronaut to reach the surface of the asteroid is

t=5.53\cdot 10^9 s

Therefore, to find the number of years, we just need to divide the total time by the duration of one year:

n=\frac{t}{T}=\frac{5.53\cdot 10^9 s}{3.16\cdot 10^7}=175

So, the astronaut will take 175 years to reach the surface.

8 0
3 years ago
What would cause a beam of light to increase in wavelength and bend away from the normal? A) Passing from a round glass into a s
uranmaximum [27]

Answer:

i think it A sorry if i get it wrong but it should be A

Explanation:

8 0
2 years ago
Read 2 more answers
What temperature resistivity of tungsten is 4 times resistivity of silver?
mihalych1998 [28]

Answer:

Slightly above 20°c

Explanation:

The resistivity of silver is 1.59 x 10-8Ωm{ as discovered by experiment from scientist}

The resistivity of tungsten at 20°c is 5.6x10-8Ωm

We see the value is just an approximation of 4 times not exactly.

6 0
3 years ago
A spring does 5.0 J of work on a 0.10-kg ball bearing in a pinball machine. The ball's
lisov135 [29]

Answer:

10m/s

Explanation:

5 0
3 years ago
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