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3 years ago

Match each statement with the characteristic of scientists.

Physics

2 answers:

Anarel [89]3 years ago

8 0

Answer: B, A, D, C,

Explanation: i got it correct :))

Dimas [21]3 years ago

5 0

Answer:

Options B, A, D, C

Explanation:

When a scientists, let's say Roberto wonders if the presence of other elements also affects the color of a flame, he can decide to prove this through a study. Therefore, in chemistry class, Roberto sees that traces of lithium makes a flame appear bright red. Subsequently, Roberto designs an experiment to test flame color in the presence of different elements and finally Roberto's friend tells him the color of a flame cannot be changed, but Roberto is still unsure.

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Why do people cover their flowers on a cool night. Think about radiation absorption conduction and temperature

Blababa [14]

When the temperature lowers the plants can freeze and die. One way to prevent that is to cover the plants so they dont freeze

3 0

3 years ago

A car drives 100 km north, 50 km east, then 10 km south. What is the car's displavement?

Over [174]

Answer:

$\sqrt{ ({40}^{2} } + {50}^{2} ) = \sqrt{(1600 + 2500)} = \sqrt{4100} = 64.03$

6 0

3 years ago

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

3 years ago

The amount of friction divided by the weight of an object forms a unit less number called the

Romashka [77]

Answer:

Coefficient of friction.

Explanation:

The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :

$F=\mu N$

N is normal force.

$\mu$ = coefficient of friction

$\mu=\dfrac{F}{N}$

3 0

3 years ago