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3 years ago

Q 36 - URGENT HELP PLS WILL BE MARKED AS BRAINLIEST.

Physics

1 answer:

Sliva [168]3 years ago

5 0

The answer is B.

This is because you add up all of the times (1.44s+1.70s+1.58s+1.76s) and you get 6.48 then you divide 6.48 by 4 to get the average of the times. Now you get the distance (200m) and because speed=distance/time you divide 200m/1.62s to get 123m/s. I hope this made sense :)

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On earth, you swing a simple pendulum in simple harmonic motion with a period of 1.6 seconds. what is the period of this same pe

polet [3.4K]

The period of a simple pendulum is given by
$T= 2 \pi \sqrt{ \frac{L}{g} }$
where
L is the pendulum length
g is the acceleration of gravity

If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
$T_e= 2 \pi \sqrt{ \frac{L}{g_e} }$
where $g_e$ is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
$T_m= 2 \pi \sqrt{ \frac{L}{g_m} }$
where $g_m$ is the acceleration of gravity on the Moon.

If we do the ratio of the two periods, we get
$\frac{T_m}{T_e} = \sqrt{ \frac{g_e}{g_m} }$
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write $g_e = 6 g_m$ and we can rewrite the previous ratio as
$\frac{T_m}{T_e} = \sqrt{ \frac{6 g_m}{g_m} }= \sqrt{6}$

so the period of the pendulum on the Moon is
$T_m = \sqrt{6} T_e = \sqrt{6} (1.6 s)=3.9 s$

8 0

4 years ago

Pls help its about baseball and energy (one question) '^'

avanturin [10]

The answer would be A. As the bat is swung, it gains kinetic energy. But once it hits the ball, it loses, or transfers, it’s kinetic energy to the ball.

5 0

3 years ago

The law of combining volumes only applies under these conditions EXCEPT :_________.

kenny6666 [7]

Answer:

measured at constant temperature and pressure

Hope this helps :)

3 0

2 years ago

Expain how a hurricane develops

Firlakuza [10]

Answer:

A hurricane starts to form when warm air rises and then that warm air turns into cooler air. After a while of this happening large clouds and thunderstorms are made. The clouds and thunderstorms keep growing and rotating from earth's Coriolis Effect. Then a hurricane forms.

6 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction tim

miss Akunina [59]

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
We can find this distance simply applying the definition of average velocity, as follows:

$\Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)$

The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

$v_{1f} ^{2} - v_{1o} ^{2} = 2* a* \Delta x (2)$

where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
Solving for Δx, we get:

$\Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)$

So, the total distance traveled was the sum of (1) and (3):
Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

$\Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)$

Δx₂, is the distance traveled while decelerating, and can be obtained using (2):

$v_{omax} ^{2} = 2* a* \Delta x_{2} (7)$

Solving for Δx₂, we get:

$\Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a} = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)$

Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
Taking the positive root in the quadratic formula, we get the following value for vomax:
v₀max = 24.3 m/s.

6 0

3 years ago