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3 years ago

13

Although the scientific method is a cyclic process, when investigating a problem for the first time, you must start somewhere. W

hich of the following should be the first step in an investigation?
draw conclusions
collect data
define the problem
analyze the data

2 answers:

Ede4ka [16]3 years ago

5 0

Answer:

Define the problem first

Explanation:

Elis [28]3 years ago

4 0

Define the problem first

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2 years ago

A 0.453 kg pendulum bob passes through the lowest part of its path at a speed of 2.58 m/s. What is the tension in the pendulum c

harkovskaia [24]

Answer with Explanation:

Mass of pendulum bob, m=0.453 kg

Speed, $v_1=$ 2.58 m/s

a.r=75.1 cm= $75.1\times 10^{-2}m$ =0.751 m

$1cm=10^{-2} m$

Tension in the pendulum cable is given by

Tension=Centripetal force+force due to gravity

$T=\frac{mv^2}{r}+mg$

Where $g=9.8 m/s^2$

Substitute the values

$T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8$

$T=8.45 N$

b.When the pendulum reaches its highest point,then

Final velocity, $v_2=0$

According to law of conservation of energy

$mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2$

$gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2$

$h_1=0$

Substitute the values

$9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2$

$3.3282=9.8h_2$

$h_2=\frac{3.3282}{9.8}=0.34 m$

The angle mad by cable with the vertical= $cos\theta=\frac{0.751-0.34}{0.751}=0.55$

$\theta=cos^{-1}(0.55)=56.6^{\circ}$

c.When the pendulum reaches at highest point then

Acceleration, a=0

Therefore, the tension in the pendulum cable

$T=mgcos\theta$

Substitute the values

$T=0.453\times 9.8cos56.6$

$T=2.4 N$

8 0

3 years ago

An automobile traveling 92.0 km/h has tires of 64.0 cm diameter. (a) What is the angular speed of the tires about their axles? (

Semenov [28]

Answer:

76.4035 m

Explanation:

r = Radius = 0.32 m

$\omega_f$ = Final angular velocity = 0

$\omega_i$ = Initial angular velocity = 92 km/h

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

Angular speed is given by

$\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{\dfrac{92}{3.6}}{0.32}\\\Rightarrow \omega=79.861\ rad/s$

The angular speed of the tires about their axles is 79.861 rad/s.

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-79.861^2}{2\times 2\pi \times 38}\\\Rightarrow \alpha=-13.355\ rad/s^2$

The magnitude of acceleration is 13.355 m/s²

Distance is given by

$d=\theta r\\\Rightarrow d=38\times 2\pi\times 0.32\\\Rightarrow d=76.4035\ m$

The distance moved while slowing down is 76.4035 m

6 0

3 years ago