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Tcecarenko
1 month ago
7

At the end of the school day, at exactly 2:30 pm, a group of students run out of the school building and reach the edge of the s

chool property at 2:30:45 s. Which of the following correctly describes the motion in terms of time?
Question 2 options:

A)

t1= 0, t2 = 2:30:45 s, ∆t = 45 s

B)

t1= 2:30, t2 = 45 s

C)

∆t = 2:30

D)

t2= 2:30:45 s, ∆t = 45 s
Physics
1 answer:
Yuliya22 [3.3K]1 month ago
3 0

Explanation:

The initial time is t₁ = 2:30 pm

The final time is t₂ = 2:30:45

We must analyze the students' motion concerning time. The final time exceeds the initial time by 45 seconds.

Change in time,

\Delta t=t_2-t_1\\\\\Delta t=2:30:45-2.30\\\\\Delta t=45\ s

Therefore, this is the solution we need.

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The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.
Maru [3345]

Answer:

11109.825 N

Explanation:

Provided Information:

mass = m = 1510 kg

initial acceleration (a) = 0.75g (where g = 9.81 m/s²)

Using the formula F=ma

  = (1510)*(0.75*9.81)

  = 11109.825 N

4 0
2 months ago
A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer
Keith_Richards [3271]

Response:

Clarification:

Provided

weight of disk m=12.5 kg

diameter of disc R=0.23 m

weight of ring m_r=7 kg

Force F=9.7 N

N=180 rpm

\omega =\frac{2\pi N}{60}

\omega =6\pi rad/s

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2

=13.25\times 0.23^2=0.7009 kg-m^2

At this point, Torque is T=F\times R=I\cdot \alpha

9.7\times 0.23=0.7\times \alpha

\alpha =3.18 rad/s^2

Utilizing \omega _f=\omega +\alpha t

\omega _f=0 in this scenario

0=6\pi -3.18\times t

t=\frac{6\pi }{3.18}

t=5.92 s

7 0
2 months ago
5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be
Sav [3153]

Answer:

The tension in the string when the speed increased is 134.53 N

Explanation:

Given;

Tension in the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The wave speed is expressed as;

v = \sqrt{\frac{T}{\mu} }

where;

μ represents mass per unit length

v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}

The new tension T₂ will be computed as;

T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N

Consequently, the tension in the string when the speed was increased is 134.53 N

3 0
2 months ago
A hiker walks 200m west and then walks 100m north. What is the magnitude and direction of her resulting displacement?
Maru [3345]

A hiker proceeds 200 m west and subsequently another 100 m north, resulting in a displacement of 223 m. The direction can be determined using the trigonometric function where sin(angle) = opposite/hypotenuse, yielding an angle of 26.6 degrees. Therefore, the total displacement is 223 m at an angle of 26.6 degrees north of west.

7 0
2 months ago
Read 2 more answers
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