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Karolina [17]
3 years ago
12

Need some help with this!

Physics
2 answers:
zvonat [6]3 years ago
8 0

Answer: what’s it asking

Explanation:

MariettaO [177]3 years ago
4 0

Answer:The Earth is shown as a circle. Equator is shown and labeled at 0 degrees. 30 degrees north and south latitudes are shown and labeled. 60 degrees north and south latitudes are shown and labeled. Curved arrows are shown pointing from 30 degrees north and south latitudes towards the equator.

What best describes these winds?

Polar easterlies that curve due to air above poles being relatively cooler

Jet streams that curve due to air above equator being relatively warmer

Trade winds blowing in a curved manner due to the rotation of Earth

Polar westerlies blowing in a curved manner due to Coriolis Effect.

Is this the question to it?

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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
mr Goodwill [35]

Answer:

Approximately 11.0\; \rm m \cdot s^{-1}. (Assuming that g = 9.81 \; \rm N \cdot kg^{-1}, and that the tabletop is level.)

Explanation:

Weight of the book:

W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N.

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, F(\text{normal force}) \approx 10.202\; \rm N.

As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

  • \mu_{\rm k}, the coefficient of kinetic friction, times
  • F(\text{normal force}), the normal force that's acting on it.

That is:

\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

Apply Newton's Second Law to find the acceleration of this book:

\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

6 0
2 years ago
Which picture represents 2NO2?
wariber [46]
2NO2 means that there is 2 oxygen atoms and one nitrogen with two sets of that. So its the third one

3 0
3 years ago
Read 2 more answers
Accelerating car is a _______
Pavel [41]

Acceleration is the rate of change of velocity as a function of time. For example a car traveling at 50 km/hr starts to accelerate, 10 seconds after, its speed changes to 100 km/hr then the acceleration of the car during the time can be calculated as below: initial speed = 50 km/hr.

5 0
2 years ago
Read 2 more answers
A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.
love history [14]

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

=  B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

3 0
3 years ago
A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su
Marta_Voda [28]

The net force acting on the object perpendicular to the table is

∑ F[perp] = F[normal] - mg = 0

where mg is the weight of the object. Then

F[normal] = mg = (15 kg) (9.8 m/s²) = 147 N

The maximum magnitude of static friction is then

0.40 F[normal] = 58.8 N

which means the applied 40 N force is not enough to make the object start to move. So the object has zero acceleration and does not move.

8 0
2 years ago
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