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3 years ago

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A block is on a frictionless table, on earth. The block accelerates at 7.5 m/s when a 70 N horizontal force is applied to it. Th

e block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s. The weight of the block on the moon is closest to: O 9.5 N O 13 N O 11 N O 15 N O 7.7 N

1 answer:

Liono4ka [1.6K]3 years ago

3 0

Answer:

The weight of the block on the moon is 15 kg.

Explanation:

It is given that,

The acceleration of the block, a = 7.5 m/s²

Force applied to the box, F = 70 N

The mass of the block will be, $m=\dfrac{F}{a}$

$m=\dfrac{70\ N}{7.5\ m/s^2}$

m = 9.34 kg

The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :

$W=m\times g$

$W=9.34\ kg\times 1.62\ m/s^2$

W = 15.13 N

or

W = 15 N

So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.

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A neighborhood transformer on the top of a utility pole transforms 12.0 kV 60.0 Hz alternating voltage down to 120.0 V to be use

riadik2000 [5.3K]

Answer:

<em>The primary coil has 13,400 turns</em>

Explanation:

<u>Voltage Transformers</u>

A transformer is an electrical apparatus that converts an alternating electrical voltage to another. Step-down transformers lower the voltage from higher levels (kilovolts) to consumer levels (120/240 Volts).

The ratio between both voltages can be computed as

$\displaystyle r=\frac{V_1}{V_2}$

Where V1 is the primary voltage and V2 is the secondary voltage. This ratio depends on the turns ratio of the coils wounded in a common magnetic core.

$\displaystyle r=\frac{N_1}{N_2}$

Being N1 the number of turns of the coils of the primary side and N2 the number of turns in the secondary coil. Both relations give us

$\displaystyle \frac{N_1}{N_2}=\frac{V_1}{V_2}$

Solving for N1

$\displaystyle N_1=\frac{V_1}{V_2}\cdot N_2$

We have:

$V_1=12,000\ V\\V_2=120\ V\\N_2=134$

Calculate N1

$\displaystyle N_1=\frac{12,000}{120}\cdot 134=13,400$

The primary coil has 13,400 turns

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3 years ago

A 11-inch candle is lit and burns at a constant rate of 0.9 inches per hour. Let t t represent the number of hours since the can

Serggg [28]

Answer: F(t) = 11 - 0.9(t)

Explanation:

We know the following:

The candle burns at a ratio given by:

Burning Ratio (Br) = 0.9 inches / hour

The candle is 11 inches long.

To be able to create a function that give us how much on the candle remains after turning it after a time (t). We will need to know how much of the candle have been burned after t.

Let look the following equation:

Br = Candle Inches (D) / Time for the Candle to burn (T) (1)

Where (1) is similar to the Velocity equation:

Velocity (V) = Distance (D)/Time(T)

This because is only a relation between a magnitude and time.

Let search for D on (1)

D = Br*T (2)

Where D is how much candle has been burn in a specif time

To create a function that will tell us how longer remains of the candle after be given a variable time (t) we use the total lenght minus (2):

How much candle remains? ( F(t) ) = 11 inches - Br*t

F(t) = 11 - 0.9(t)

F(t) defines the remaining length of the candle t hours after being lit

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3 years ago

Every winter i fly home to Michigan. it takes 5 hours. what is my average speed?

Alisiya [41]

It depends how far you travel.

If for instance you travel 1000 miles, then your speed is distance/time = 1000/5 = 200 mph

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2 years ago

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A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate

Mariana [72]

According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass $m=1.0 \cdot 10^{-4} kg$ . The acceleration of the insect is $a=1.0 \cdot 10^2 m/s^2$ , therefore we can calculate the force exerted by the car on the insect:
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How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.

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3 years ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

so

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

<h3>b.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

<h3>c.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

3 years ago