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3 years ago

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Dois automóveis A e B percorreram uma trajetória retilínea conforme as equações horárias Sa = 30 + 20t e Sb = 90 – 10t, sendo a

posição S em metros e o tempo t em segundos. No instante t = 0s, a distância, em metros, entre os automóveis era de:

1 answer:

Ray Of Light [21]3 years ago

4 0

Answer:

a yeHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

Explanation:

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calculate the value of net force acting on given body?5N acting towards right & 15N force acting towards left on the brick.(

lys-0071 [83]

Answer:

The net force acting on the body is 10N directed to the left.

Explanation:

Magnitude of force to the right = 5N

Magnitude of force to the left = 15N

Net force acting on the object and in what direction;

Solution:

It is the vector sum of all forces acting on a body. This net force is the single force that will replace the forces acting on a body;

For the problem;

Net force = Force to the left + Force to the right

Let us take left to be negative and right to be positive;

Force to the left = -15N

Net force = -15N + 5N = -10N

The net force acting on the body is 10N directed to the left.

4 0

3 years ago

When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c

e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = $e^{kt}$

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = $e^{-0.01t}$

V = V₀ $e^{-0.01t}$

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀ $e^{-0.01t}$

0.1V₀ = V₀ $e^{-0.01t}$

0.1V₀/V₀ = $e^{-0.01t}$

0.1 = $e^{-0.01t}$

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0

3 years ago

Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work?

kifflom [539]

Based on the law of conservation of energy, we know that we can't create energy, machines can only convert one type of energy into another. So, if we want to improve a machines's ability then we need to reduce it's loss energy (part of energy which is useless). Out of all the options only Option C fits best with it.

In short, Your Answer would be Option C

Hope this helps!

6 0

3 years ago

Read 2 more answers

A hockey puck is pushed by a stick with a force of 750 newton’s. the puck tracked 2.0 meters in 0.30 seconds. how powerful is th

Anit [1.1K]

Work= force x distance
work= 750 x 2
work =1500
power =work/time
power= 1500/ 0.3
power= 5000W

answer: b. 5000W

7 0

3 years ago

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h

vodka [1.7K]

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

$R_A=2R_B$

Also Potential of both sphere is same

$V_A=V_B$

$V=\frac{kQ}{R}$

thus

$k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}$

$\frac{Q_A}{Q_B}=\frac{R_A}{R_B}$

$\frac{Q_A}{Q_B}=\frac{2}{1}=2$

$\frac{Q_B}{Q_A}=\frac{1}{2}$

(b)Ratio of $\frac{E_B}{E_A}$

Electric Field is given by $E=\frac{kQ}{R^2}$

thus $E_A=\frac{kQ_A}{R_A^2}$ ----1

$E_B=\frac{kQ_B}{R_B^2}$ ----2

Divide 2 by 1

$\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}$

$\frac{E_B}{E_A}=\frac{1}{2}\times 4=2$

8 0

3 years ago