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2 years ago

What is the weight of a 4.5 kg mass on Earth?

Physics

1 answer:

swat322 years ago

4 0

Answer:

7.535×10^25 earth mass

Explanation:

for an approximate result,divide the mass value by 9.223e+18

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When plugging in metric facts, always remember that 1 big unit = # small units. Fill in these facts: 1.________ s= ________μs 2.

balandron [24]

Answer:

1. 1 s = 1 x 10⁶ μs

2. 1 g = 0.001 kg

3. 1 km = 1000 m

4. 1 mm = 1 x 10⁻³ m

5. 1 mL = 1 x 10⁻³ L

6. 1 g = 100 dg

7. 1 cm = 1 x 10⁻² m

8. 1 ms = 1 x 10⁻³ s

Explanation:

1 x 10⁻⁶ s = 1 μs

(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs

1 s = 1 x 10⁶ μs

1000 g = 1 kg

1 g = 1/1000 kg

1 g = 0.001 kg

1 km = 1000 m

1 mm = 1 x 10⁻³ m

1 mL = 1 x 10⁻³ L

1 x 10⁻² g = 1 dg

(1 x 10⁻² x 10²) g = 1 x 10² dg

1 g = 100 dg

1 cm = 1 x 10⁻² m

1 ms = 1 x 10⁻³ s

4 0

3 years ago

A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

We know the distance between the two Cartesian points is given as:

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

For the electric field effect to be zero at point D we need equal and opposite field at the point.

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$

and

$y=2\times 8=16\ nm$

3 0

3 years ago

What does the person in back of the ambulance experience? A) a lower frequency of the siren B) a lower amplitude of the siren C)

hoa [83]

The answer to your question is "A. a lower frequency of the siren.

Because the person in back of the ambulance will hear a lower frequency of the siren. This is because the waves are stretched out. A longer wavelength results in a lower frequency.

5 0

3 years ago

Read 2 more answers

What holds the moon in place, orbiting around the Earth

AnnZ [28]

The Earth's gravity keeps the Moon orbiting us. It keeps changing the direction of the Moon's velocity. This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

4 0

2 years ago

Which of the following statements explains how total time spent in the air is affected as a projectile's angle of launch is incr

charle [14.2K]

Answer:

Therefore letter C is the correct answer.

Explanation:

In a projectile motion the total time in the air can be calculated using the following equation:

We analyze the y-component motion.

$v_{fy}=v_{iy}-gt$

When the final velocity (v(f)) is equal to zero we calculate the upward time and multiplying it by 2 we find the total time in the air. So we will have:

$t_{tot}=2\frac{v_{iy}}{g}$

$t_{tot}=2\frac{v_{i}sin(\theta)}{g}$

We can see that the total time is directly proportional to the angle, then when θ increase t increase.

Therefore letter C is the correct answer.

I hope it helps you!

3 0

3 years ago