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kykrilka [37]
3 years ago
6

Which graph best represents the relationship between the electric current and the rate at which a magnet is turning inside an el

ectric generator?
Physics
2 answers:
Ludmilka [50]3 years ago
5 0

Answer:

The second one.

Explanation:

Yuri [45]3 years ago
5 0

Answer:

B. Just took it on USA Test Prep

Explanation:

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A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the
Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.  

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

-F_{f} + F = 0      

-F_{f} + ma = 0      

\mu mg = ma

\mu = \frac{a}{g}

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

v_{f}^{2} = v_{0}^{2} + 2ad

a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}

Where:  

d: is the distance traveled = 46.1 m

v_{f}: is the final speed of the truck = 0 (it stops)      

v_{0}: is the initial speed of the truck = 17.9 m/s

a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}        

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.  

\mu = \frac{a}{g}  

\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}

\mu = 0.35

Therefore, the minimum coefficient of friction required is 0.35.  

I hope it helps you!

4 0
3 years ago
Acording to Newton’s second law of motion if a rigid body of unchanging mass is observed accelerating what must be happening
blondinia [14]

A force is being applied to the mass.

6 0
3 years ago
Read 2 more answers
While traveling along a highway a driver slows from 34 m/s to 17 m/s in 6 seconds. What is the automobiles acceleration?​
xxTIMURxx [149]

Answer:

-2.83 m/s²

Explanation:

  • Initial velocity (u) = 34 m/s
  • Final velocity (v) = 17 m/s
  • Time taken (t) = 6 seconds

❖ Acceleration is defined as the rate of change in velocity with time.

→ a = (v - u)/t

  • v denotes final velocity
  • a denotes acceleration
  • u denotes initial velocity
  • t denotes time

→ a = (17 - 34)/6 m/s²

→ a = -17/6 m/s²

<h3>→ Acceleration = -2.83 m/s²</h3>

(Minus sign implies that the velocity is decreasing.)

5 0
3 years ago
The components of vector A are Ax = +2.2 and Ay = -6.9 , and the components of vector B are given are Bx = -6.1 and By = -2.2. W
Zina [86]
For simplicity, let's call vector B-A  vector C  Then C is
Cx = (-6.1 - 2.2)  
Cy = (-2.2 - (-6.9))  Or,
Cx = -8.3  Cy = 4.7
The magnitude is found with the Pythagorean theorem
||C|| = √(-8.3² + 4.7²) = 9.538
3 0
3 years ago
A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant
mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

3 0
3 years ago
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