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2 years ago

Before scientists come up with a scientific question, what do they have to do first?

1 answer:

8 0

Scientist would check what other scientists have said before they come up with their own scientific question. That way they can record the results from the others and compare them with their own findings when they conduct their own experiments and record the findings.

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Solved, can someone check it over!

kakasveta [241]

Answer:

its good no need to change anything :))

4 0

2 years ago

A ball is thrown at 20 m/s from the ground upwards at an angle of elevation of 30°. How far away does it land? 35.35 m

mestny [16]

Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m

8 0

3 years ago

Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net

kotykmax [81]

Answer:

0.29D

Explanation:

Given that

F = G M m / r2

F = GM(6m) / (D-r)2

G Mm/r2 = GM(6m) / (D-r)2

1/r2 = 6 / (D-r)2

r = D / (Ö6 + 1)

r = 0.29 D

See diagram in attached file

7 0

2 years ago

A positively charged particle Q1 = +35nC is held fixed at the origin. A second charge of mass m = 3.5ug is floating a distance d

cupoosta [38]

Answer:

Explanation:

Q1 = 35 nC = 35 x 10^-9 C

m = 3.5 micro gram = 3.5 x 10^-9 Kg

d = 35 cm = 0.35 m

(a) The electrostatic force between the two charges is balanced by the weight of another charge.

F = m g

$\frac{1}{4\pi \epsilon _{0}}\frac{Q_{1}Q_{2}}{d^{2}}=mg$

$Q_{2}=\frac{4\pi \epsilon _{0}mgd^{2}}{Q_{1}}$

(b) By substituting the values

$Q_{2}=\frac{3.5\times 10^{-9}\times 9.8\times 0.35\timees 0.35}{9\times 10^{9}\times 35\times 10^{9}}$

Q2 = 13.34 x 10^-12 C

Q2 = 0.0134 nC

4 0

3 years ago

A 5.20 kg chunk of ice is sliding at 13.5 m/s on the floor of an ice-covered valley when it collides with and sticks to another

Stella [2.4K]

Answer:

The chunk went as high as

2.32m above the valley floor

Explanation:

This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.

Applying the principle of energy conservation for the two ice we have based on the scenery

Momentum before impact = momentum after impact

M1U1+M2U2=(M1+M2)V

Given data

Mass of ice 1 M1= 5.20kg

Mass of ice 2 M2= 5.20kg

velocity of ice 1 before impact U1= 13.5 m/s

velocity of ice 2 before impact U2= 0m/s

Velocity of both ice after impact V=?

Inputting our data into the energy conservation formula to solve for V

5.2*13.5+5.2*0=(10.4)V

70.2+0=10.4V

V=70.2/10.4

V=6.75m/s

Therefore the common velocity of both ice is 6.75m/s

Now after impact the chunk slide up a hill to solve for the height it climbs

Let us use the equation of motion

v²=u²-2gh

The negative sign indicates that the chunk moved against gravity

And assuming g=9.81m/s

Initial velocity of the chunk u=0m/s

Substituting we have

6.75²= 0²-2*9.81*h

45.56=19.62h

h=45.56/19.62

h=2.32m

5 0

2 years ago