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kherson [118]
1 year ago
10

A transformer is to be used to provide power for a computer drive. The number of turns in the primary is 1000, and it delivers a

secondary current of 600 milliamperes (mA) at a secondary voltage of 8.0 V for the drive. A voltage of 400 volts is applied across the primary. What kind of transformer is this?a. step upb. Loadc. Step-Downd. AC
Physics
1 answer:
Yuri [45]1 year ago
3 0

Answer:

C.Step-Down

Explanation:

Given that

Number of turn in the primary N₁= 1000

Secondary current I₂= 600 mA

Secondary voltage V₂= 8 V

Primary voltage V₁= 400 V

From the above information we can say that output voltage is low compare to input voltage.

V₂ < V₁

We know that if

1 . V₂ < V₁  - Then transformer will be step down .

2. V₂ > V₁  - Then transfer will be step up .

Here

V₂ < V₁  ( 8 V < 400 V)

So the transformer is step down transformer.

C.Step-Down

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To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch
sergij07 [2.7K]

Answer:

man upward acceleration is 0.14m/s^2

Explanation:

given data:

mass of man = 72 kg

downward force = 360 N

The mass of man of weight 72 kg is hang from two sections of rope, one section pf rope ties around man waist and other section is ties in man hands. when he pulls down the rope  with 360 N force then each section of  rope pulls with 360 N

we know that

Weight= mass × gravity= 72kg × 9.8 = 705.6N

Force = mass× acceleration

Force= -705.6 + (2 × 358) = 10.4 N

acceleration = \frac{10.4}{72} = 0.14m/s^2

4 0
1 year ago
A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn
EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, C=5.00 \mu F, the charged stored on the capacitor at the end of the process is

Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

V=\frac{Q}{C}

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

C=\frac{\epsilon_0 \epsilon_r A}{d}

where:

\epsilon_0 is the permittivity of free space

\epsilon_r is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: C'=\frac{C}{2}. The new voltage across the plate is given by

V'=\frac{Q}{C'}

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V

So, the new voltage is

V'=2 (12.0 V)=24.0 V

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

A=\pi r^2

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

A'=\pi (2r)^2 = 4 \pi r^2 = 4A

While the separation between the plate was unchanged (d); so, the new capacitance will be

C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C

So, the capacitance has increased by a factor 4; therefore, the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}

which means

V'=\frac{12.0 V}{4}=3.0 V

3 0
1 year ago
A disk of mass m and radius r is initially held at rest just above a larger disk of mass M and radius R that is rotating at angu
Leto [7]

Answer:

Explanation:

Moment of inertia of larger disk   I₁ = 1/2 MR²

Moment of inertia of smaller  disk   I₂ = 1/2 m r ²

Initial angular velocity

We shall apply law of conservation of angular momentum .

initial total momentum = final angular momentum

I₁ X ωi  = ( I₁ + I₂ )ωf

1/2 MR² x ωi = 1/2 ( m r² + MR² ) ωf

ωf =  ωi   / ( 1 + m r²/MR² )

6 0
1 year ago
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
2 years ago
Here are the positions at three different times for a bee in flight (a bee's top speed is about 7 m/s). Time 6.6 s 6.9 s 7.2 s P
Ber [7]

Answer:

(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)

Explanation:

Given the following :

Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s

Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)

(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?

Vavg = Distance / time

[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6

Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3

Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3

Vavg = (- 4.33, 6.33 , 0)

b) Between 6.6 s and 7.2 s, what was the bee's average velocity?

Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6

Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6

Vavg = (- 3.66, 7.5, 0)

c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s

D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?

Displacement = Velocity * time

Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s

= (- 4.33, 6.33 , 0) * 0.05

= (- 0.2165, 0.3165, 0)

5 0
2 years ago
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