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dalvyx [7]
4 years ago
9

Please help me with this ​

Physics
1 answer:
xxTIMURxx [149]4 years ago
7 0

Answer:

12 Ω

Explanation:

Data obtained from the question include:

Resistor 1 (R1) = 50 Ω.

Resistor 2 (R1) = 30 Ω.

Resistor 3 (R3) = 20 Ω.

Resistor 4 (R4) = R

Galvanometer reading = Zero deflection.

The Resistor 4 (R4) in the Wheatstone bridge can be obtained as follow:

Since the galvanometer gives zero deflection, it means the bridge is balanced. Therefore,

R1/R2 = R3/R4

50/30 = 20/R

Cross multiply

50 x R = 30 x 20

Divide both side by 50

R = (30 x 20)/50

R = 12 Ω

Therefore, the value of R in the wheatstone bridge is 12 Ω.

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Answer:

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Explanation:

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A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position
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The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

v_x= \frac{dx}{dt}

Write an equation for x.

dx=v_xdt\\ x=\int v_xdt

Substitute the equation for vₓ =2t² in the integral.

x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C

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When t =0, x = 1. 1m

x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m

Substitute 2.0s for t.

x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m

The position of the particle at t =2.0 s is <u>6.4m</u>




5 0
3 years ago
How many amps are there in the current of a circuit if the voltage source is 140v and the resistance is 2 ohms?
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A pickup truck, initially stationary at a stop light, accelerates at a rate of 1.60m/s2 for 14.0 s. The truck then cruises at co
Alex Ar [27]

Answer:

The total distance traveled by the truck is 1797 m

Explanation:

Hi there!

The equation of position and velocity of an object moving in a straight line with constant acceleration are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the truck at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time

v = velocity at time t.

Let's calculate the position of the truck after the first 14.0 s:

x = x0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference at the first stop light, the initial position, x0, is zero. Since the truck starts from rest, v0 = 0. So, the equation of position will be:

x = 1/2 · a · t²

x = 1/2 · 1.60 m/s² · (14.0 s)²

x = 157 m

Then, the truck travels with constant speed (a = 0) for 70.0 s. The equation of position will be:

x = x0 + v · t

In this case, let's consider the initial position as the the position where the car is after 14.0 s (157 m from the stop light). The velocity is the velocity reached after the 14.0 s of acceleration. Let's calculate it with the equation of velocity:

v = v0 + a · t  (v0 = 0)

v = 1.60 m/s² · 14.0 s

v = 22.4 m/s

So, the position will be:

x = 157 m + 22.4 m/s · 70.0 s

x = 1725 m

Now, the truck slows down with an acceleration of 3.50 m/s² until it stops (until its velocity is zero). Let's calculate the time at which the velocity of the truck is zero:

v = v0 + a · t

0 = 22.4 m/s - 3.50 m/s² · t

-22.4 m/s / -3.50 m/s² = t

t = 6.4 s

Now let's calculate the position of the truck after that time considering the initial position as the position at which the truck was after the 70.0 s traveling at constant speed (1725 m from the stop light):

x = x0 + v0 · t + 1/2 · a · t²

x = 1725 m + 22.4 m/s · 6.4 s + 1/2 · (-3.50 m/s²) · (6.4 s)²

x = 1797 m

The total distance traveled by the truck is 1797 m

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Mass of the dart = 0.01 kg
Speed at which the dart is thrown = 20 m/s
Kinetic Energy = (1/2) * mass * speed * speed
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So the kinetic energy of the dart is 2 Joules. I hope this is the correct answer and it has helped you.
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3 years ago
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