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RoseWind [281]
3 years ago
5

70 POINTS HELP PHYSICS THE PICTURE HAS THE QUESTION AND POSSIBLE CHOICES

Physics
1 answer:
snow_tiger [21]3 years ago
6 0

Answer:

0.3 m/s

Explanation:

0.5 - 0.2 = .3

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3 years ago
How is grounding a positive object different from grounding a negative object?
Natali5045456 [20]

Answer:

Grounding a Positively Charged Object

Electrons were transferred from the electroscope to the ground. As in the case of grounding a negatively charged electroscope, the grounding of a positively charged electroscope involves charge sharing. The excess positive charge is shared between the electroscope and the ground.

Explanation:

8 0
3 years ago
Evaluate x and y in the equation: E=Cm^xV^y , where E is kinetic energy , m is mass , V is velocity and C is a dimension less co
Korvikt [17]

Answer: Do I look like Einstein

8 0
2 years ago
Considerando que você comece a caminhar em velocidade constante, inicialmente a 350 m de um ponto referencial escolhido. Você ca
ExtremeBDS [4]

Answer:

a. S(t)=350−1t

Explanation:

To determine the equation of motion you take into account the general form of motion with constant velocity:

S(t)=S_o+vt    ( 1 )

So is the initial position from a specific reference frame. In this case is 350 m.

v is the speed of the motion, in this case is 1m/s. However, the motion is forward the zero point of the reference frame, hence, the speed is - 1m/s.

You replace the values of So and v in the equation ( 1 ) and you obtain:

S(t)=350-(1m/s)t

Hence, the answer is:

a. S(t)=350−1t

- - - - - - - - - - - - - - - - - - - - -

Para determinar a equação do movimento, você leva em consideração a forma geral do movimento com velocidade constante:

             (1)

Assim é a posição inicial de um quadro de referência específico. Neste caso, é de 350 m.

v é a velocidade do movimento, neste caso é de 1m / s. No entanto, o movimento é avançar o ponto zero do quadro de referência, portanto, a velocidade é de - 1m / s.

Você substitui os valores de So ev na equação (1) e obtém:

Portanto, a resposta é:

uma. S (t) = 350-1t, movimento retrógrado

4 0
3 years ago
This time particle A starts from rest and accelerates to the right at 65.5 cm/s
FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

d = v_i t + \frac{1}{2}at^2

d = 0 + \frac{1}{2}(65.5)t^2

d_1 = 32.75 t^2 cm

Now we know that B moves with constant speed so in the same time B will move to another distance

d_2 = 44 \times t

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

d_1 = 349 + d_2

32.75 t^2 = 349 + 44 t

on solving above kinematics equation we have

t = 4 s

4 0
3 years ago
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