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valentina_108 [34]
2 years ago
11

While driving on a rural road, your right wheels run off the pavement. You should hold the steering wheel firmly and

Physics
1 answer:
Vsevolod [243]2 years ago
3 0

Answer:

The answer is C. Steer in a straight line while gently slowing down

Explanation:

The following are advised when your cars go off the pavement while driving;

firstly, Do not panic.

ensure you hold on to your steering wheel tightly.

keep Steering straight ahead.

ensure you Stay on the shoulder.

Ease up on the accelerator and brake gently.

When you  know you can safely do so, turn back on the road at a much lower speed.

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69PTS! Which choices are also major levels of classification? Choose all answers that are correct. A. Genus B. Species C. Group
Pachacha [2.7K]
A. Genus and B. Species are also major levels of classification.
5 0
3 years ago
Read 2 more answers
Use Percentages Given that the molecular mass of magnesium
ira [324]

Answer:

27%

Explanation:

15.999 divided by 58.32 = .27433128

Move the decimal place over 2 places.

27%

5 0
3 years ago
a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic
algol13

Answer:

Approximately 281.25\; \rm m. (Assuming that the drag on this ball is negligible, and that g = 10\; \rm m \cdot s^{-2}.)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

  • Horizontal: no acceleration, velocity is constant (at v(\text{horizontal}) is constant throughout the descent.)
  • Vertical: constant downward acceleration at g = 10\; \rm m \cdot s^{-2}, starting at 0\; \rm m \cdot s^{-1}.

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: x(\text{horizontal}) = 30\; \rm m. Combine these two quantities to find the duration of this descent:

\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}.

In other words, the ball in this question start at a vertical velocity of u = 0\; \rm m \cdot s^{-1}, accelerated downwards at g = 10\; \rm m \cdot s^{-2}, and reached the ground after t = 7.5\; \rm s.

Apply the SUVAT equation \displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t to find the vertical displacement of this ball.

\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}.

In other words, the ball is 281.25\; \rm m below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 281.25\; \rm m\!.

5 0
3 years ago
An arrow is shot at an angle of 35° and a velocity of 50 m/s. How long does it take to return to its original starting height?
Ostrovityanka [42]

Answer:

4.02 s

Explanation:

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

Angle of projection (θ) = 35°

Initial velocity (u) = 50 m/s

Acceleration due to gravity (g) = 10 m/s²

Time of flight (T) =?

The time of flight of the arrow can be obtained as follow:

T = 2uSineθ / g

T = 2 × 35 × Sine 35 / 10

T = 70 × 0.5736 / 10

T = 7 × 0.5736

T = 4.02 s

Therefore, the time taken for the arrow to return is 4.02 s

4 0
3 years ago
an object starts from rest with a constant acceleration of 8.0m/s^2 along a straight line.find the average speed for 5 second in
frozen [14]

Answer:

40 m/s

Explanation:

If you multiply the acceleration with time you get the average speed.

6 0
2 years ago
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