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3 years ago

1. A bicycle initially moving with a velocity

Physics

1 answer:

ki77a [65]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf 15 \ m/s \ or \ 15 \ m*s^{-1}}}$

Explanation:

We are asked to find the final velocity. We are given the acceleration, time, and initial velocity, so we can use the following kinematics formula.

$v_f= v_i+ at$

In this formula, $v_f$ is the final velocity, $v_i$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

The bicycle has an initial velocity of 5.0 m *s⁻¹ or m/s, acceleration of 2 m/s², and a time of 5 seconds.

$\bullet \ v_i = 5.0 \ m/s \\\bullet \ a= 2\ m/s^2\\\bullet \ t= 5 \ s$

Substitute the values into the formula.

$v_f=5.0 \ m/s + ( 2\ m/s^2 * 5 \ s)$

Solve inside the parentheses.

$\frac {2 \ m}{s^2}* 5 \ s = \frac{ 2 \ m}{s} * 5 = \frac{ 10 \ m}{s} = 10 \ m/s$

$v_f= 5.0 \ m/s + (10 \ m/s)$

Add.

$v_f= 15 \ m/s$

The units can also be written as:

$v_f= 15 \ m*s^{-1}$

The bicycle's final velocity is 15 meters per second.

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You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauc

pickupchik [31]

a) $15.4^{\circ}$

b) 5.2 m/s

c) 151.2 N

Explanation:

When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by

$mg sin \theta$

where

m =16 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

$\theta$ is the angle of the ramp

According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:

$F=ma\\mgsin \theta = ma$

where a is the acceleration.

From the equation above we get

$a=g sin \theta$

And we are told that the acceleration must not exceed

$a=2.6 m/s^2$

Substituting this value and solving for $\theta$ , we find the maximum angle of the ramp:

$\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}$

Here we are told that the vertical distance of the ramp is

$h=1.4 m$

Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.

So we have:

$GPE=KE\\mgh=\frac{1}{2}mv^2$

where:

m = 16 kg is the mass of the box

$g=9.8 m/s^2$

h = 1.4 m height of the ramp

v = final speed of the box at the bottom of the ramp

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s$

There are two forces acting on the box in the direction perpendicular to the ramp:

- The normal force, N, upward

- The component of the weight perpendicular to the ramp, downward, of magnitude

$mg cos \theta$

Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:

$N-mg cos \theta$

and by substituting:

m = 16 kg

$g=9.8 m/s^2$

$\theta=15.4^{\circ}$

We can find the normal force:

$N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N$

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Explanation:

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4 years ago

horizontal force pushes block up a 20.0 incline with an initial speed 12.0 m//s. a) how high up the plane does slide before comi

Annette [7]

Answer:

Explanation:

We shall apply conservation of mechanical energy .

initial kinetic energy = 1/2 m v²

= .5 x m x 12 x 12

= 72 m

This energy will be spent to store potential energy . if h be the height attained

potential energy = mgh , h is vertical height attined by block

= mg l sin20 where l is length up the inclined plane

for conservation of mechanical energy

initial kinetic energy = potential energy

72 m = mg l sin20

l = 72 / g sin20

= 21.5 m

deceleration on inclined plane = g sin20

= 3.35 m /s²

v = u - at

t = v - u / a

= (12 - 0) / 3.35

= 3.58 s

it will take the same time to come back . total time taken to reach original point = 2 x 3.58

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3 years ago