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GenaCL600 [577]

4 years ago

12

A car accelerates horizontally from rest on a level road at a constant acceleration of Down the road, it passes through two phot

ocells ("electric eyes" designated by 1 for the first one and 2 for the second one) that are separated by 20.0 m. The time interval to travel this 20.0-m distance as measured by the electric eyes is 1.40 as.
(a) Calculate the speed of the car as it passes each electric eye.
(b) How far is it from the start to the first electric eye

2 answers:

Neko [114]4 years ago

7 0

Answer:A is 14.2857

B is 20M

Explanation:(A)

the velocity or speed of a car is distance/time=20/1.4=14.2857

(B)

Distance between the two eye is 20m

11Alexandr11 [23.1K]4 years ago

5 0

Answer:i totally forgot how that works

Explanation:

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A 2000 kg car moves along a horizontal road at speed vo

cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

$\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a$ (Eq. 1)

$\Sigma F_{y} = N-m\cdot g = 0$ (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

$a = -\mu_{k}\cdot g$ (Eq. 3)

Where:

$a$ - Deceleration of the car, measured in meters per square second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $\mu_{k} = 0.0735$ and $g = 9.807\,\frac{m}{s^{2}}$ , then deceleration of the car is:

$a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )$

$a = -0.721\,\frac{m}{s^{2}}$

The stopping distance of the car ( $\Delta s$ ), measured in meters, is determined from the following kinematic expression:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (Eq. 4)

Where:

$v_{o}$ - Initial speed of the car, measured in meters per second.

$v$ - Final speed of the car, measured in meters per second.

If we know that $v_{o} = 15.9\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $a = -0.721\,\frac{m}{s^{2}}$ , stopping distance of the car is:

$\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}$

$\Delta s = 175.319\,m$

The shortest possible stopping distance of the car is 175.319 meters.

8 0

4 years ago

At what altitude h above the north pole is the weight of an object reduced to 67% of its earth-surface value? Assume a spherical

lara31 [8.8K]

Answer:

The answer to the question is

At an altitude of 1413 km or 1.222·R above the north pole the weight of an object reduced to 67% of its earth-surface value

Explanation:

We make use of the gravitational formula as follows

F = G $\frac{m_{1} m_{2} }{R^{2} }$ where

m₁ = mass of the object

m₂ = mass of the earth

d = distance between the two objects and

G = gravitational constant

if at the altitude the weight is reduced to 67 % of its weight on earth then

with all other variables remaining constant, we have

67% F = G $\frac{m_{1} m_{2} }{R_{2} ^{2} }$ =0.67× G× $\frac{m_{1} m_{2} }{R_{1} ^{2} }$

cancelleing like ternss from both sides we have

1/R₂² =0.67/R₁² or r₁²/R₂² =0.67 and R₁/R₂ = 0.8185

or R₂ = R₁/0.8185 or 1.222·R = (6.37×10⁶ m)/0.8185 = 7.783×10⁶ m

Hence at an altitude = 1.413×10⁶ m the weight of the object will reduce to 67% its earth-surface value

6 0

3 years ago

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

$F=mrw^2$

$m=$ mass of the corresponding bugs

$r=$ corresponding radial distance of each bug

$w=$ angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

$F$ ∝ $mr$

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0

3 years ago

Read 2 more answers

Design an experiment to test the water temperature increases the solubility in water also increses

givi [52]

I'll solve that one second

4 0

3 years ago

A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied

Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the <em>book is</em>

F=44N

From the question we are told

A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the Force is mathematically given as

$F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\$

F=44N

Therefore

the minimum force that must be applied on the <em>book is</em>

F=44N

For more information on this visit

brainly.com/question/23379286

8 0

3 years ago