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olganol [36]
3 years ago
14

During which interval is the object not moving

Physics
1 answer:
Bezzdna [24]3 years ago
6 0

Answer:

Between 2.0 s and 4.0 s (B and C)

Between 5.0 s and 8.0 s (D and E)

Between 10.0 s and 11.0 s (F and G)

Explanation:

The graph shown in the figure is a velocity-time graph, which means that:

- On the x-axis, the time is plotted

- On the y-axis, the velocity is plotted

Therefore, this means that the object is not moving when the line is horizontal (because at that moment, the velocity is constant, so the object is not moving). This occurs in the following intervals:

Between 2.0 s and 4.0 s (B and C)

Between 5.0 s and 8.0 s (D and E)

Between 10.0 s and 11.0 s (F and G)

From the graph, it would be possible to infer additional information. In particular:

- The area under the graph represents the total distance covered by the object

- The slope of the graph represents the acceleration of the object

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A man walking on a tightrope carries a long a pole which has heavy items attached to the two ends. If he were to walk the tight-
katen-ka-za [31]

Answer:

 I_weight = M L²

this value is much larger and with it it is easier to restore balance.I

Explanation:

When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by

            v = w r

For man to maintain equilibrium needs the total moment to be zero

             ∑τ = I α

              S  τ = 0

The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.

Therefore the moment of the masses and the open is the one that must be zero.

If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope

              I = ⅓ m L² / 4

As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.

If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is

             I_weight = M L²

this value is much larger and with it it is easier to restore balance.

5 0
3 years ago
An over-caffeinated student stands on a table which weighs 600 newtons. The student has a mass of 50 kg. What is the weight of t
liberstina [14]

Answer:50kg

Explanation:

6 0
3 years ago
The alpha line in the balmer series of the hydrogen spectrum consists of light having a wavelength of 6.56. calculate the freque
guajiro [1.7K]
The alpha line in the Balmer series is the transition from n=3 to n=2 and with the wavelength of λ=656 nm = 6.56*10^-7 m. To get the frequency we need the formula: v=λ*f where v is the speed of light, λ is the wavelength and f is the frequency, or c=λ*f. c=3*10^8 m/s. To get the frequency: f=c/λ. Now we input the numbers: f=(3*10^8)/(6.56*10^-7)=4.57*10^14 Hz. So the frequency of the light from alpha line is f= 4.57*10^14 Hz. 
5 0
3 years ago
A motorist is driving at 20m/s when she sees that a traffic light 200m ahead has just turned red. She knows that this light stay
yuradex [85]

Answer:

5.71428571422 m/s

Explanation:

u = Initial velocity = 20 m/s

v = Final velocity

s = Displacement

a = Acceleration

Time taken = 15-1 = 14 s

Distance traveled in 1 second = 20\times 1=20\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2

v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s

The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s

4 0
4 years ago
Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp
grandymaker [24]

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

5 0
3 years ago
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