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3 years ago

13

You are running at a speed of 10km/h and hit a patch of mud. Two seconds later you speed is 8km/h. What is your acceleration in

units of m/s^2

1 answer:

Vlad1618 [11]3 years ago

5 0

Answer:

$0.28 m/s^2$

Explanation:

Acceleration is given by

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time interval

In this problem:

$u = 10 km/h \cdot \frac{1000 m/km}{3600 s/h}=2.78 m/s$ is the initial velocity

$v = 8 km/h \cdot \frac{1000 m/km}{3600 s/h}= 2.22 m/s$ is the final velocity

t = 2 s is the time

Substituting, we find the acceleration:

$a=\frac{2.78-2.22}{2}=0.28 m/s^2$

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a car is moving 8.80 m/s when it begins to accelerate at 2.45 m/s^2. how much time does it take to trav 138m. please help me (':

Ulleksa [173]

Answer:

7.6 s

Explanation:

Considering kinematics formula for final velocity as

$v^{2}=u^{2}+2as$

Where v and u are final and initial velocities, a is acceleration and s is distance moved.

Making v the subject then

$v=\sqrt{u^{2}+2as}$

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

$v=\sqrt{8.8^{2}+2*2.45*138}\\v=27.45 m/s$

Also, v=u+at and making t the subject of the formula

$t=\frac {v-u}{a}$

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

$t=\frac {27.45-8.8}{2.45}=7.6122448979591\approx 7.6s$

Therefore, it needs 7.6 seconds to travel

7 0

2 years ago

A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of 0.11 m and a mass of 0.01

AnnZ [28]

Answer:

Explanation:

The magnetic force acting horizontally will deflect the wire by angle φ from the vertical

Let T be the tension

T cosφ = mg

Tsinφ = Magnetic force

Tsinφ = BiL , where B is magnetic field , i is current and L is length of wire

Dividing

Tanφ = BiL / mg

= .055 x 29 x .11 / .010 x 9.8

= 1.79

φ = 61° .

Tension T = mg / cosφ

= .01 x 9.8 / cos61

= .2 N .

5 0

2 years ago

You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b

Radda [10]

Answer:

$\theta=34 \textdegree$

Explanation:

From the question we are told that:

Mass $m=55kg$

Angle $\theta =28.0$

Coefficient of static friction $\alpha =0.680$

Generally, the equation for Newtons second Law is mathematically given by

For

$\sum_y=0$

$N=mgcos \theta$

for

$\sum_x=0$

$F_{s}=mgsin\theta$

Where

$F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta$

$F_{s}=0.68*55*9.8*cos 28$

$F_{s}=323.62N$

Therefore

$\alpha mgcos \theta=mg sin \theta$

$\theta=tan^{-1}(0.68)$

$\theta=34 \textdegree$

6 0

2 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.6 m from the center of the ride. then the operator turns on the rid

ioda

<span>Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116 So coefficient of static friction = 0.6116</span>

4 0

2 years ago

ou are out stargazing with your 13.4-cm telescope. You point your telescope at an interesting formation in the sky, which you th

Alinara [238K]

Answer:

θ = 4.716 10⁻⁶ rad

Explanation:

In order for the releases to be considered separate, they must meet the Rayleigh criterion that establishes that the maximum diffraction of one star must coincide with the first minimum of the diffraction pattern of the second star.

We use the diffraction equation for a slit

a sin θ = m λ

The minimum occurs at m = 1

sin θ = λ / a

Since the angles in these systems are very small, we can approximate the sine to its angle in radians

θ = λ / a

The telescope has a circular aperture whereby polar cords should be used, which introduces a constant number

θ = 1.22 λ / a

Let's calculate

θ = 1.22 518 10⁻⁹ / 13.4 10⁻²

θ = 4.716 10⁻⁶ rad

8 0

2 years ago