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ludmilkaskok [199]
3 years ago
12

An object moving north with an initial velocity of 14 m/s accelerates 5 m/s^2 for 20 seconds.what is the final velocity of the o

bject?
39 m/s
90 m/s
114 m/s
414m/s
Physics
2 answers:
bekas [8.4K]3 years ago
5 0

Answer:

Option C = 114 m/s

Explanation:

a=v-u/t

By substituting,

5=v-14/20

100=v-14

Thus, v=100+14

v= 114 m/s

Hope it helps :)

ELEN [110]3 years ago
5 0
The answer is 114 m/s good luck!!!
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Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations wa
OLga [1]

Answer:

<h2>15m/s</h2>

Explanation:

The equation for a traveling wave as expressed as y(x, t) = A cos(kx − \omegat) where An is the amplitude f oscillation, \omega is the angular velocity and x is the horizontal displacement and y is the vertical displacement.

From the formula; k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f where;

\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency

Before we can get the transverse speed, we need to get the frequency and the wavelength.

frequency = 1/period

Given period = 2/15 s

Frequency = \frac{1}{(2/15)}

frequency = 1 * 15/2

frequency f = 15/2 Hertz

Given wavelength \lambda = 2m

Transverse speed v = f \lambda

v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s

Hence, the transverse speed at that point is  15m/s

8 0
2 years ago
A concrete block (B-36 x10 °C-') of volume 100 mat 40°C is cooled to
ruslelena [56]
  • T1=40°C=313K
  • T_2=-10°C=263K

Applying Charles law

\\ \sf\Rrightarrow \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

\\ \sf\Rrightarrow \dfrac{100}{313}=\dfrac{V_2}{263}

\\ \sf\Rrightarrow V_2=\dfrac{26300}{313}

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6 0
2 years ago
You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height (h) from the starting
Eddi Din [679]

Answer:

h = 50.49 m

Explanation:

Data provided:

Speed of skier, u = 2.0 m/s

Maximum safe speed of the skier, v = 30.0 m/s

Mass of the skier, m = 85.0

Total work = 4000 J

Height from the starting gate = h

Now, from the law of conservation of energy

Total energy at the gate = total energy at the time maximum speed is reached

\frac{1}{2}mu^2+mgh=4000J+\frac{1}{2}mv^2

where, g is the acceleration due to the gravity

on substituting the values, we get

\frac{1}{2}\times85\times2.0^2+85\times9.81\times h=4000J+\frac{1}{2}\times85\times30^2

or

170 + 833.85 × h = 4000 + 38250

or

h = 50.49 m

7 0
3 years ago
A 75-m-long train begins uniform acceleration from rest. the the train has a speed of 23 m/s when it passes a railway worker tan
Lapatulllka [165]

Answer:

acceleration a = 1.04 m/s2

Explanation:

Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:

s = 75 + 180 = 255 m

We can use the following equation of motion to find out the distance traveled by the car:

v^2 - v_0^2 = 2aswhere v = 23 m/s is the velocity of the car when it passes the worker, v_0 = 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.

23^2 - 0^2 = 2*a*255

510a = 529

a = 529 / 510 = 1.04 m/s^2

8 0
3 years ago
Looking that following diagram of bar magnets, determine if the magnets will or will not connect (attract) and why.
vovangra [49]

Answer: They will NOT connect because like poles are facing each other, and like poles repel each other.

3 0
3 years ago
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