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Sunny_sXe [5.5K]
2 years ago
13

Many electrical appliances use cells (batteries). there's a torch with two batteries inside it (the torch needs two cells to giv

e enough currect to light the lamp brightly) The batteries are connected end-to-end they are in series.
Question 1 What would you observe if the torch had only one cell? Explain your answer. ​
Physics
1 answer:
Shkiper50 [21]2 years ago
6 0

Answer:

torch will not light up

Explanation:

because the torch itself need two but you placed one so it will not light up

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A beaker of vegetable oil contains a beam of light that is aimed at a surface at an angle of 34 degrees as shown. If the index o
OverLord2011 [107]

Answer:

Angle of reflection of light is 34 degree

Explanation:

As per law of reflection of light we know that

angle of incidence of light = angle of reflection of light

So here we know that

angle of incidence on the surface of oil is given as

\theta_i = 34 degree

so we know that

\theta_i = \theta_r

so here we can say that reflection angle of light will be same as angle of incidence

\theta_r = 34 degree

8 0
3 years ago
You throw a ball upward from ground level with initial upward speed v0. What is the max height of the trajectory?
Inga [223]

Answer:

The max height of the ball is y = -1/2 (v0²/g).

It takes the ball t = -2 · v0/g to hit the ground.

The speed of the ball when it hits the ground is v = -v0.

Explanation:

The height and velocity of the ball is given by the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t

When the ball is at max height, the velocity is 0. So, let´s find the time at which the velocity of the ball is 0.

v = v0 + g · t

0 =  v0 + g · t

t = -v0/g

Now, replacing t =  -v0/g in the equation of height, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t²   (y0 = 0 because the origin of the frame of reference is located on the ground)

y = v0 · t + 1/2 · g · t²

Replacing t:

y = v0 · (-v0/g) + 1/2 · g ·  (-v0/g)²

y = -(v0²/g) + 1/2 · (v0²/g)

y = -1/2 (v0²/g)

The max height of the ball is y = -1/2 (v0²/g).  Remember that g is negative.

Since the acceleration of the ball is always the same, the time it takes the ball to impact the ground will be twice the time it takes to reach its max height, t = -2 v0/g.

However, let´s calculate that time knowing that at that time the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 =  v0 · t + 1/2 · g · t²

0 = t · ( v0 + 1/2 · g · t)

0 = v0 + 1/2 · g · t

-2 · v0/g = t

It takes the ball t = -2 · v0/g to hit the ground.

Let´s use the equation of velocity at final time (t = -2 · v0/g):

v = v0 + g · t

v = v0 + g · ( -2 · v0/g)

v = v0 - 2· v0

v = -v0

The speed of the ball when it hits the ground is v = -v0.

7 0
4 years ago
A mass is stretched 4 centimeters from the equilibrium position of a spring. If the displacement is doubled, what can be conclud
marta [7]
From Simple Harmonic Motion (SHM),  the restoring force is given by:

   F  =  -ω²x

We can see that restoring force is proportional to the displacement, x. So if the displacement x is doubled, then the restoring force will also be doubled.
8 0
3 years ago
Read 2 more answers
A coin dropped in the lift it takes time 0.5 s to reach the floor when lift is staionary it takes time t when lift is moving up
BARSIC [14]

Answer:

t₁ > t₂

Explanation:

A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂,  t₁ = t₂,  t₁ >> t₂ ,  t₂ > t₁

Solution:

Newton's law of motion is given by:

s = ut + (1/2)gt²;

where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.

u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.

hence:

When stationary:

s=\frac{1}{2}gt_1^2\\\\t_1^2=\frac{2s}{g}  \\\\When\ moving\ with\ acceleration(a):\\\\s=\frac{1}{2}(g+a)t_2^2\\\\t_2^2=\frac{2s}{g+a}

Hence t₂ < t₁, this means that t₁ > t₂.

4 0
3 years ago
Suppose you have two identical capacitors. You connect the first capacitor to a battery that has a voltage of 21.2 volts, and yo
HACTEHA [7]

Answer:

r=2.743

Explanation:

The energy stored on a capacitor is of type potencial, therfore depends on the capacity to "store" energy. Inthe case of the capacitor, it stores charge (Q), and the equations you use to calculate it are:

E_p=\frac{Q^2}{2C}=\frac{QV}{2}=\frac{CV^2}{2}

In this case we know V and C, therefore we use the last expression:

E_{p1}=\frac{CV_1^2}{2}

E_{p2}=\frac{CV_2^2}{2}

\frac{E_{p1}}{E_{p2}}=r=\frac{\frac{CV_1^2}{2}}{\frac{CV_2^2}{2}}  \\r=(\frac{V_1}{V_2})^2\\r=(\frac{21.2}{12.8})^2

r=2.743

3 0
3 years ago
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