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1 year ago

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A car is travelling to the right with a speed of 42\,\dfrac{\text m}{\text s}42 s m 42, space, start fraction, m, divided by,

s, end fraction when the driver slams on the brakes. The car skids for 4.0\,\text s4.0s4, point, 0, space, s with constant acceleration before it comes to a stop. How many meters did the car skid before coming to a stop? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.

2 answers:

Effectus [21]1 year ago

7 0

Answer:

d = 84 m

Explanation:

As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object

here we know that

initial speed $v_i = 42 m/s$

final speed $v_f = 0$

time taken by the car to stop

$t = 4s$

now the distance moved by the car before it stop is given as

$d = \frac{v_f + v_i}{2} \times t$

now we have

$d = \frac{42 + 0}{2} \times 4$

$d = 84 m$

wolverine [178]1 year ago

5 0

Answer:

84

Explanation:

khan academy

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The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un

Elza [17]

Answer:

The answer to your question is: 15 m/s2

Explanation:

Equation x = at3 - bt2 + ct

a = 4.1 m/s3

b = 2.2 m/s2

c = 1.7 m/s

First we find x at t = 4.1 s

x = 4.1(4.1)3 - 2.2(4.1)2 + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we find speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s2

6 0

1 year ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

3 0

1 year ago

A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle wi

Kamila [148]

Answer:

A.)1.52cm

B.)1.18cm

Explanation:

angular speed of 120 rev/min.

cross sectional area=0.14cm²

mass=12kg

F=120±12ω²r

=120±12(120×2π/60)^2 ×0.50

=828N or 1068N

To calculate the elongation of the wire for lowest and highest point

δ=F/A

= 1068/0.5

δ=2136MPa

'E' which is the modulus of elasticity for alluminium is 70000MPa

δ=ξl=φl/E =2136×50/70000=1.52cm

δ=F/A=828/0.5

=1656MPa

δ=ξl=φl/E

=1656×50/70000=1.18cm

$δ=1.18cm$

6 0

2 years ago

Read 2 more answers

A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b

mars1129 [50]

Answer:

ω2 = 216.47 rad/s

Explanation:

given data

radius r1 = 460 mm

radius r2 = 46 mm

ω = 32k rad/s

solution

we know here that power generated by roller that is

power = T. ω ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2

so

ω2 = $\frac{460\times 32}{34\times 2}$

ω2 = 216.47

8 0

1 year ago

13. An aircraft heads North at 320 km/h rel:

AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

$v' = v + v_a$

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

$v_a$ is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

$v_a = -80 km/h$ (since the air is moving from North)

Therefore, we find

$v'=+320 + (-80) = +240 km/h$ (north)

Learn more about vector addition:

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2 years ago