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Citrus2011 [14]

3 years ago

9

Later, while taking your groceries back to the car, you accidentally let go of your cart. It rolls straight down a grassy slope

at a constant speed of 10 m/s. What is the net force on the cart during this motion

1 answer:

VladimirAG [237]3 years ago

6 0

Car is moving on the glassy slope with constant speed

Now we know that

$a = \frac{dv}{dt}$

so acceleration is rate of change in velocity

as we know that velocity is constant here so acceleration is zero

so here

$a = 0$

now as we know by Newton's II law

$F = ma$

since a = 0

$F = 0$

so net force will be ZERO on it during this motion

You might be interested in

A car goes from rest to a velocity of 108 km/h north in 10s what is the car's acceleration in m/s2

fomenos

initial velocity of the car given as

$v_i = 0$

final velocity is given as

$v_f = 108 km/h$

as we know that

$1 km/h = 0.277 m/s$

now we can convert final speed into m/s

$v_f = 108 * 0.277 = 30 m/s$

now acceleration is rate of change in velocity

$a = \frac{v_f - v_i}{t}$

$a = \frac{30 - 0}{10}$

$a = 3 m/s^2$

so the acceleration of the car is 3 m/s^2

7 0

3 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

If you traveled one mile at a speed of 100 miles per hour and another mile at a speed if 1 mile per hour, your average speed wou

Semenov [28]

Answer:

v = 1.98 mph

Explanation:

Given that,

Speed to travel one mile is 100 mph

Speed to travel another mile is 1 mph

The formula used to find your average speed is given by :

$v=\dfrac{2v_1v_2}{v_1+v_2}$

Putting the values, we get :

$v=\dfrac{2\times 100\times 1}{100+1}$

v = 1.98 mph

So, yours average speed is 1.98 mph.

7 0

3 years ago

A train is travelling towards the station on a straight track. It is a certain distance from the station when the engineer appli

AleksandrR [38]

Answer:

500 m

Explanation:

t = Time taken

u = Initial velocity = 50 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -2.5 m/s²

Equation of motion

$v=u+at\\\Rightarrow 0=50-2.5\times t\\\Rightarrow \frac{-50}{-2.5}=t\\\Rightarrow t=20\ s$

Time taken by the train to stop is 20 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=50\times 20+\frac{1}{2}\times -2.5\times 20^2\\\Rightarrow s=500\ m$

∴ The engineer applied the brakes 500 m from the station

4 0

3 years ago

¿Qué ocurre como consecuencia de las corrientes de convección en la astenosfera?

Vladimir [108]

Respuesta:En la astenosfera existen lentos movimientos de convección que explican la deriva continental. Además, el basalto de la astenosfera fluye por extrusión a lo largo de las dorsales oceánicas, lo cual hace que se renueve y expanda constantemente el fondo oceánico. :D

6 0

2 years ago