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3 years ago

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I need help!!!! Please and thank u!

1 answer:

alexandr1967 [171]3 years ago

5 0

Answer:

False

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Gravity on the surface of the moon is only 1/6 as strong as gravity on the Earth. What is the weight of a 19 kg object on the Ea

Dahasolnce [82]

The weight of anything in any place is

(mass of the thing) x (acceleration of gravity in that place).

-- On Earth, the acceleration of gravity is about 9.807 m/s²

Weight of 19 kg of mass is (19 kg) x (9.807 m/s²) = <em>186.3 newtons</em>

-- On the Moon, the acceleration of gravity is about 1.623 m/s²

Weight of the same 19 kg of mass is (19 kg) x (1.623 m/s²) = <em>30.8 newtons</em>

7 0

3 years ago

What is the ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points 14 km and 49 k

Molodets [167]

Answer:

$\dfrac{I_1}{I_2}=12.25$

Explanation:

$r_1$ = 14 km

$r_2$ = 49 km

Intensity of a wave is inversely proportional to distance

$I\propto \dfrac{1}{r^2}$

So,

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{49^2}{14^2}\\\Rightarrow \dfrac{I_1}{I_2}=12.25$

The ratio of the intensities is $\dfrac{I_1}{I_2}=12.25$

6 0

3 years ago

PLEASE HELP!!!!!!!!Move on to electric force. Blow up the two balloons and knot them. Then tie a thread onto each balloon. Suspe

Triss [41]

Answer:

they attract

Explanation:

8 0

3 years ago

Read 2 more answers

At a point 1.2 m out from the hinge, 14.0 N force is exerted at an angle of 27 degrees to the moment arm in a plane which is per

geniusboy [140]

Answer:

$\tau = 7.63 Nm$

Explanation:

As we know that moment of force is given as

$\tau = \vec r \times \vec F$

now we have

$\vec r = 1.2 m$

$\vec F = 14 N$

now from above formula we have

$\tau = r F sin\theta$

here we know that

$\theta = 27 degree$

so we have

$\tau = (1.2)(14) sin27$

$\tau = 7.63 Nm$

3 0

3 years ago

a scale model of the solar system where 50 cm represents 1.0x10 to the fifth km is actual distance what would be the dimension o

Fofino [41]

The distance between Mars and the Sun in the scale model would be 1140 m

Explanation:

In this scale model, we have:

$x_1 = 50 cm$ represents an actual distance of

$d_1 = 1.0\cdot 10^5 km$

The actual distance between Mars and the Sun is 228 million km, therefore

$d_2=228\cdot 10^6 km$

On the scale model, this would corresponds to a distance of $x_2$ .

Therefore, we can write the following proportion:

$\frac{x_1}{d_1}=\frac{x_2}{d_2}$

And solving for $x_2$ , we find:

$x_2=\frac{x_1 d_2}{d_1}=\frac{(50)(228\cdot 10^6)}{1\cdot 10^5}=1.14\cdot 10^5 cm = 1140 m$

Learn more about distance:

brainly.com/question/3969582

#LearnwithBrainly

4 0

3 years ago