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ss7ja [257]
3 years ago
14

Select the correct answer. What was the main objective of the Discovery missions?

Physics
1 answer:
Nimfa-mama [501]3 years ago
6 0

Answer:

A

Explanation:

exploration surrounding our universe will give us a better understanding of it.

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A point charge has q=1.0×10-6
dmitriy555 [2]

Since the 2 points form a triangle with hypothenuse of √5 [ √(1²+2²)],

I guess apply the formula :

v = kq \div r

with r as √5 and q as 1x10^-6

not sure about this answer tho

draw a diagram first to understand better

8 0
3 years ago
Which of the following refers to a force acting toward the center of a circular
Masja [62]

Answer:

D. Centripetal Force :)

3 0
3 years ago
What did Rutherford’s model of the atom include that Thomson’s model did not have?
Ksivusya [100]
Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment in which he demonstrated that the atom has a tiny and heavy nucleus. Rutherford designed an experiment to use the alpha particles emitted by a radioactive element as probes to the unseen world of atomic structure.
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3 years ago
What are two ways to create a negative ion?<br> What are two ways to create a positive ion?
Iteru [2.4K]

Answer:

what that guy said

Explanation:

because he provides evidence

4 0
3 years ago
The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.
Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let \rho_{m} be the density of metal and \rho_{w} be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

W =0.10400\ N

mg=0.10400

\rho V g=0.10400

Vg=\dfrac{0.10400}{\rho_{m}}.....(I)

The weight of metal in water is

Using buoyancy force

F_{b}=0.10400-0.08400

F_{b}=0.02\ N

We know that,

F_{b}=\rho_{w} V g....(I)

Put the value of F_{b} in equation (I)

\rho_{w} Vg=0.02

Put the value of Vg in equation (II)

\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02

1000\times\dfrac{0.10400}{0.02}=\rho_{m}

\rho_{m}=5200\ kg/m^3

Hence, The density of the metal is 5200 kg/m³.

6 0
4 years ago
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