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Nata [24]
3 years ago
11

Students had two batteries and two different resistors. During four trials, they build four different circuits and measure the c

ircuit’s current in Amps according to the following table.
Trial Number

Voltage (V)

Resistance (Ω)

Current (A)

1

1.5

200


2

1.5

100


3

3.0

200


4

3.0

100




For which trial would the students measure the smallest current in the circuit? (AKS 10a)



A.
Trial 1

B.
Trial 2

C.
Trial 3

D.
Trial 4
Physics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

bhi jo bhi of gp oh oh gi IG 7u to uff do if goo td to yd do FP ae rt 7g hi pic vo icon

Explanation:

bh hi h bhi vc di oh x At jb jo iv hp of di of dr hi o hc x gh ki vc hi jo

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A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic fr
Ksivusya [100]

a) The initial speed of the bullet is 330.5 m/s

b) The collision is inelastic

c) The impulse is -2.97 kg m/s

Explanation:

a)

The energy lost by the block while sliding (which is equal to the work done by friction) is equal to the kinetic energy of the block after the bullet has been embedded into it, therefore we can write

KE=W

\frac{1}{2}(M+m)v^2=(\mu (M+m)g) d

where

M = 1.20 kg is the mass of the block

m = 9.00 g = 0.009 kg is the mass of the bullet

v is the combined speed of bullet+block after the collision

\mu = 0.20 is the coefficient of friction

g=9.8 m/s^2 is the acceleration of gravity

d = 0.310 m is the distance through which the block slides

Solving for v,

v=\sqrt{2gd}=\sqrt{2(9.8)(0.310)}=2.46 m/s

This is the final velocity of the block+bullet after the collision.

Now we can apply the law of conservation of momentum: in fact, the total momentum of the system before the collision must be equal to the total momentum after the collision, so we get

mu+MU = (m+M)v

where

u is the initial velocity of the bullet

U = 0 is the initial velocity of the block (initially at rest)

v = 2.46 m/s

Solving for u,

u=\frac{(m+M)v}{m}=\frac{(0.009+1.20)(2.46)}{0.009}=330.5 m/s

b)

To check whether the collision is elastic or inelastic, we just need to compare the total kinetic energy before and after the collision.

Before the collision, we have:

K_i = \frac{1}{2}mu^2 = \frac{1}{2}(0.009)(330.5)^2=491.5 J

While after the collision

K_f = \frac{1}{2}(m+M)v^2 = \frac{1}{2}(0.009+1.20)(2.46)^2=3.7 J

We see that the final kinetic energy is less than the initial kinetic energy: therefore, the collision is inelastic, since part of the energy has been converted into other forms of energy (e.g. thermal energy).

c)

The impulse of the block is equal to its change in momentum, so:

I=\Delta p =p_f - p_i = (m+M)v'-(m+M)v

where

v' = 0, since the block comes to a stop

v = 2.46 m/s is the velocity of the block just after the collision

Substituting,

I=0-(0.009+1.20)(2.46)=-2.97 kg m/s

And the impulse is negative, because its direction is opposite to the direction of motion of the block (this means that the force exerted on the block, which is the force of friction, acts in the direction opposite to the motion of the block).

Learn more about kinetic energy and momentum:

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brainly.com/question/2370982

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#LearnwithBrainly

3 0
4 years ago
Particle moves in a circle of radius 90m with a constant speed 25m/s. how many revolution does it make in 30sec​
Serggg [28]

Answer:

<em>n =1.33 revolutions</em>

Explanation:

<u>Uniform Circular Motion</u>

The angular speed can be calculated in two different ways:

\displaystyle \omega=\frac{v}{r}

Where:

v = tangential speed

r = radius of the circle described by the rotating object

Also:

\omega=2\pi f

Where:

f = frequency

Solving for f:

\displaystyle f=\frac{\omega}{2\pi}

Since the frequency is calculated when the number of revolutions n and the time t are known:

\displaystyle f=\frac{n}{t}

We can solve for n:

n=f.t

The particle moves in a circle of r=90 m with a speed v=25 m/s. Thus the angular speed is:

\displaystyle \omega=\frac{25}{90}

\displaystyle \omega=0.278\ rad/s

Now we calculate f:

\displaystyle f=\frac{0.278}{2\pi}

f=0.04421\ Hz

Calculating the number of revolutions:

n = 0.04421*30

n =1.33 revolutions

8 0
3 years ago
Q. A car is traveling
Norma-Jean [14]

Answer:

I don't even know can anybody help this person

6 0
3 years ago
A 49 kg person is being dragged in their sleeping bag to the lake by a 593 N
Alex777 [14]

Answer:

485.62 N

Explanation:

To obtain the magnitude of unbalanced forces acting on the body;

Unbalanced Force = Horizontal Component of Applied Force - Frictional Force

Frictional Force = Horizontal Component of  Applied Force - Unbalanced Force

f = frictional force  = ?

F = Applied force  = 593 N

m = mass of person = 49 kg

a = acceleration = 0.57 m/s²

θ = Angle with horizontal = 30°

Hence;

Horizontal Component of  Applied Force = (593 N)(Cos 30°)

Unbalanced Force = (49 kg)(0.57 m/s²)

f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)

f = 513.55 N - 27.93 N

f = 485.62 N

6 0
3 years ago
A ball at the end of a string is swinging in a horizontal circle of radius 0.85 m. The ball makes exactly 3 revolutions per seco
Anettt [7]

Answer:

It's centripetal acceleration is 301.7 m/s²

Explanation:

The formula to be used here is that of the centripetal acceleration which is

ac = rω²

where ac is the centripetal acceleration = ?

ω is the angular velocity = 3 revolutions per second is to be converted to radian per second: 3 × 2π  = 3 × 2 × 3.14 = 18.84 rad/s

r is the radius = 0.85 m

ac = 0.85 × 18.84²

ac = 301.7 m/s²

It's centripetal acceleration is 301.7 m/s²

8 0
3 years ago
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