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FromTheMoon [43]
3 years ago
12

After 16.5 s, a jogger’s displacement is 200.0 m. What is her average velocity in m/s? In km/h?

Physics
1 answer:
vivado [14]3 years ago
3 0

Answer:

12.12 m/s

Explanation:

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A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10
IRINA_888 [86]

Complete Question:

A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to

Answer:

t= 16.7 sec.

Explanation:

As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:

γ = (ωf -ω₀) / t

If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:

γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².

When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:

γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec

8 0
3 years ago
Question 1
Vesna [10]

Answer:

1)  g = 4π² / m, 3) xaxis the  length of the pendulums and the y axis the period squared

Explanation:

a) students can approximate this system to a simple pendulum, in this case the angular velocity is

         w = √ g / l

angular velocity, frequency and period are related

         w = 2π f = 2π / T

we substitute

         T = 2π√ l / g

with this equation they can determine the value of the acceleration of gravity, for this they measure the period for various lengths of the pendulum and graph

        T² = 4π²  l / g

We graph T² vs l

where this is the equation of a line if the independent variable is y = T² and x = l

        y = (4π² / g)  l

so the slope is

         m = 4π² / g

clearing

         g = 4π² / m

where the slope can be found with the values ​​of the line not the experimental values.

2) to carry out the experiment, or the thread is attached to the sphere, the length of the pendulum that goes from the pivot point to the center of the sphere is measured with a tape measure and a small finished angle is turned or less than 10th is released, it is good to wait for the first oscillation to walk, the time of a determined number of oscillations is generally measured 10 or 20, the period is calculated

    T = t / n

a table of T² against the length is made and it is plotted with the length in the ax ax, we look for the slope and hence the acceleration of gravity

3) on the independent x-axis, the controlled variable must be plotted, which is the length of the pendulums, and on the y-axis, the dependent variable is the period squared

4) of the equation of the line

            m = 4pi2 / g

                 where it ends up reaching the floor

            g = 4pi2 / m

5) when the spring is cut, the sphere remains under the effect of gravity acceleration, the harmonic movement disappears and the sphere is in a vertical movement

5 0
4 years ago
Find the area of rectangle given below​
Delicious77 [7]

Answer:

6000 cm<em>²</em>

Explanation:

let width of rectangle = x

Using Pythagorean theorem,

x^{2} +<em> 50² = 130²</em>

<em />x^{2}<em> = 130² - 50² </em>

<em />x^{2}<em> = 14400</em>

<em />x <em>= 120 cm</em>

area of rectangle <em>=</em> length x width

                             <em>=</em> <em>120 x 50</em>

                             <em>= 6000 cm²</em>

4 0
3 years ago
The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b
taurus [48]

Answer:

I = 0.483 kgm^2

Explanation:

To know what is the moment of inertia I of the boxer's forearm you use the following formula:

\tau=I\alpha  (1)

τ: torque exerted by the forearm

I: moment of inertia

α: angular acceleration = 125 rad/s^2

You calculate the torque by using the information about the force (1.95*10^3 N) and the lever arm (3.1 cm = 0.031m)

\tau=Fr=(1.95*10^3N)(0.031m)=60.45J

Next, you replace this value of τ in the equation (1) and solve for I:

I=\frac{\tau}{\alpha}=\frac{60.45Kgm^2/s ^2}{125rad/s^2}=0.483 kgm^2

hence, the moment of inertia of the forearm is 0.483 kgm^2

8 0
3 years ago
A crate rests on the flatbed of a truck that is initially traveling at 15 m/s on a level road. The driver applies the brakes and
lina2011 [118]

Answer:0.3

Explanation:

Given

velocity of car=15 m/s

truck brought to halt in a distance of 38 m

We know

v^2-u^2=2as

Final velocity (v)=0

0-(15)^2=2(a)(38)

a=\frac{-225}{76}

a=-2.96 m/s^2  (deceleration)

Therefore minimum coefficient of friction \mu will be

\mu \times g=a

\mu =\frac{a}{g}

\mu =\frac{2.96}{9.8}=0.302

7 0
4 years ago
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