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3 years ago

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A 64.9 kg sprinter starts a race with an acceleration of 3.89 m/s2. She keeps this acceleration for 17 m and then maintains the

velocity for the remainder of the 100-m dash, what will be her time for the race?

1 answer:

xxMikexx [17]3 years ago

3 0

Answer:

Time of race = 10.18 s

Explanation:

She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash

Time to travel 17 m can be calculated

s = ut + 0.5at²

17 = 0 x t + 0.5 x 3.89 x t²

t = 2.96 s

Velocity after 2.96 seconds

v = 3.89 x 2.96 = 11.50 m/s

Remaining distance = 100 - 17 = 83 m

Time required to cover 83 m with a speed of 11.50 m/s

$t=\frac{83}{11.50}=7.22s$

Time of race = 2.96 + 7.22 = 10.18 s

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marishachu [46]

Answer:

Yes, that is correct.

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2 years ago

Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her

leva [86]

To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

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$\tau_l = F*d$

And in angular movement like

$\tau_a = I*\alpha$

Where,

F= Force

d= Distance

I = Inertia

$\alpha =$ Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

$\tau= F*cos(19)*d$

On the other hand we have the speed data expressed in RPM, as well

$\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 1.0471rad/s$

Acceleration can be calculated by

$\alpha = \frac{\omega_f}{t}$

$\alpha = \frac{1.0471}{9}$

$\alpha = 0.11rad/s^2$

In the case of Inertia we know that it is equivalent to

$I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2$

$I = 1983.75kg.m^2$

Matching the two types of torque we have to,

$\tau_l=\tau_a$

$Fd=I\alpha$

$Fcos(19)*2.3=1983.75(0.11)$

$F=100.34N$

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

$W = \frac{1}{2}I\omega_f^2$

$W= \frac{1}{2}(1983.75)(1.0471)^2$

$W=1087.51J$

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2 years ago

Use the accompanying seismogram to answer which of the three types of seismic waves reached the seismograph first.

UkoKoshka [18]

Answer:

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Explanation:

Due to excess of the energy inside the earth when the tectonic plates begin to slide or fracture then the energy is released in the form of seismic waves, this causes the earthquake.

<u>Two types of seismic waves are generally responsible for the earth quakes:</u>

body waves
surface waves

Body waves are of two types:

Primary waves (P-waves)

These are the fastest of all the waves involved in the earth-quake which travel at a speed of 1.6 km to 8 km per second.

They can pass trough solids, liquids and gases. They arrive at the surface as an instant thud.

Secondary waves (S-waves)

They can only pass through the solids and they move slower than the P-waves.

As S-waves move, they displace the rock particles, pushing them outwards perpendicular to the wave-path that leads to the earthquake-related first rolling period.

Surface waves (L-waves/ long waves)

These waves move along the surface of the earth. They are responsible for the earthquake's carnage.

They move up and down the Earth's surface, rocking the foundations of man-made structures.

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3 years ago

EASY BRAINLIEST PLEASE HELP!!

Soloha48 [4]

Answer:

Solution given:

frequency[f]=x

velocity[V]=15000m/s

wave length=59m

we have

wave length= $\frac{V}{f}$

59m= $\frac{15000}{x}$

x= $\frac{15000}{59}$ =254.Hz

frequency=254Hz

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2 years ago

Read 2 more answers

A straight trail with a uniform inclination of 15 degrees leads from a lodge at an elevation of 600 feet to a mountain lake at a

9966 [12]

Answer:

The length of the trail = 22796 ft

Explanation:

From the ΔABC

AC = length of the trail = x

AB = 6100 - 600 = 5500 ft

Angle of inclination $\theta$ = 15°

$\sin \theta = \frac{AB}{AC}$

$\sin 15 = \frac{5900}{x}$

$x = \frac{5900}{0.2588}$

x = 22796 ft

Since x = AC = Length of the trail.

Therefore the length of the trail = 22796 ft

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3 years ago