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3 years ago

Post Test: Forces and Motion

Physics

1 answer:

Len [333]3 years ago

8 0

Answer: B to C

Explanation: The line is curving inwards, practically calculating the stance that it had went down. If it went straight across, it stayed the same till a specific point, furthermore calculating the bent line bending upwards is actually a partial-raise, conclude points B to C is most likely an un-even balance, meaning it had went down; or decreasing. B to C is the decreasing segment of this equation/problem (question).

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A neon sign that requires a voltage of 11,000 v is plugged into a 120-v wall outlet. what turns ratio (secondary/primary) must a

Arisa [49]

The equation that relates the voltages and the number of turns in a transformer is
$\frac{V_s}{V_p}= \frac{N_s}{N_p}$
where $V_s = 120 V$ is the voltage in the secondary coil, $V_p=11000 V$ is the voltage in the primary coil, and $N_s$ and $N_p$ are the number of turns on the secondary and primary coils.

Using the numbers, we find the ratio between the number of turns:
$\frac{N_s}{N_p}= \frac{V_s}{V_p}= \frac{120 V}{11000 V}=0.0109$

8 0

3 years ago

Max is a network technician who just terminated the ends on a new copper cable used between two legacy switches. When he connect

Olegator [25]

Answer:

The cable used is a straight through cable.

Explanation:

The cable used here is a straight through cable, that is why it fails to establish the connection.

Cross-over cables are used to communicate over legacy switch. It is used to connect computing devices directly.

In a cross over cable, signal gets transmit from one end of the device to the other.

7 0

3 years ago

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of static friction between t

Wewaii [24]

Answer:

a) a = 6.1 m/s^2

b) a = 0.98m/s^2

Explanation:

Mass of slab = 40kg

Mass of block = 10kg

Coefficient of static friction (Us) = 0.60

Kinetic coefficient (UK) = 0.40

Horizontal force = 100N

The normal reaction from 40kg slab on 10 kg block = 10*9.81

= 98.1N

Static frictional force = Us*R

= 98.1*0.6

= 58.86N

This is less than the force applied

If 10 kg block will slide on the 40 kg slab, net force = 100 - kinetic force

Kinetic force (Uk*R) = 0.4*98.1

= 39.28N

= 39N

Net force = 100 -39

= 61N

Recall that F = ma

For 10 kg block

a = F/m

a = 61/10

a = 6.1m/s^2

b) Frictional force on 40 kg slab by 10 kg = 98.1*0.4

= 39.24

= 39N

F = ma

a = F/m

For 40kg slab

a = 39/40

a = 0.98m/s^2

3 0

3 years ago

After your patched the roof, you dropped the hammer off the house and you heard it land on your toolbox on the ground 3.3s later

Schach [20]

Answer:

53.36 m.

Explanation:

Initial velocity of the toolbox u = 0

acceleration due to gravity g = 9.8 m s⁻²

time t = 3.3 s

height of roof h = ?

h = ut + 1/2 g t²

= 0 + .5 x 9.8 x 3.3²

= 53.36 m.

3 0

4 years ago

You are driving your car, and the traffic light ahead turns red. you apply the brakes for 2.26 s, and the velocity of the car de

Lorico [155]

Let the initial velocity of the car be u.

Final velocity of the car (v) = 5.43 m/s

deceleration (a) = - 2.78 m/s^2

Time taken (t) = 2.26 s

Using the first equation of motion:

v = u + at

u = v - at

$u = 5.43 - (-2.78 \times 2.26)$

u = 5.43 + 6.28

u = 11.71 m/s

Let the car's displacement be x.

Using second equation of motion:

$x = ut + \frac{1}{2}at^2$

$x = 11.71 \times 2.26 - \frac{1}{2} \times 2.78 \times 2.26^2$

x = 26.4646 - 7.0995

x = 19.3651 meters

Hence, the displacement of the car is 19.36 meters

6 0

3 years ago