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3 years ago

6

If a vehicle is traveling at constant velocity and then comes to a sudden stop, has it undergone negative acceleration or positi

ve acceleration? Explain your answer

1 answer:

sertanlavr [38]3 years ago

6 0

Answer:

Negative acceleration

Explanation:

Let us assume a vehicle is moving with a constant velocity u and then it comes to a sudden stop.

The initial velocity of the vehicle is u and finally it comes to rest, it means final velocity is 0.

As we know that,

Acceleration is equal to change in velocity divided by time taken.

$a=\dfrac{v-u}{t}\\\\\text{Put v=0}\\\\a=\dfrac{-u}{t}$

We can see that the value of acceleration is negative. Hence, it leads to negative acceleration.

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An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.60 m above

s344n2d4d5 [400]

Answer:

a) No

b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.

Explanation:

Given:

Height of the wall = 3.95m

Initial height = 1.60m

Initial speed = 5.00m/s

distance between the initial height and wall top = 3.95 - 1.60 = 2.35m

Using the formula;

v^2 = u^2 + 2as ....1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled

From equation 1

s = (v^2 - u^2)/2a ...2

Since the rock t moving up,

the acceleration = -g = -9.8m/s2

s = maximum height travelled

v = 0 (at maximum height velocity is zero)

Substituting into equation 2

s = (0 - 5^2)/(2×-9.8) = 1.28m

Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall

b) Using equation 1:

u^2 = v^2 - 2as

v = 0

a = -9.8m/s

s = 2.35m. (distance between the initial height and wall top)

u^2 = 0 - 2(-9.8 × 2.35)

u^2 = 46.06

u = √46.06

u = 6.79m/s

Therefore, the rock must have a minimum initial speed of 6.79m/s

8 0

3 years ago

A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

lara [203]

Answer:

frequency = 1475.45 Hz

Explanation:

given data

frequency f1 = 1215 Hz,

frequency f2 = 1265 Hz

police car moving vp = 25.0 m/s

solution

speed of sound u = 343 m / s

speed of the other car = v

when the police car is stationary

the frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u}$ ................1

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u}{u-v}$ ..................2

now from equation 1 and 2

$\frac{f2}{f1} = \dfrac{u+v}{u-v}$

$\frac{1275}{1240} = \frac{u+v}{u-v}$

$v =\frac{1275-1240}{1275+1240}\times 343$

v = 4.77 m/s

and

frequency the other car receives is

f2 = f1 × $\dfrac{u+v}{u-vp}$ ......................3

and

the frequency the police car receives is

f2 = f1 × $\dfrac{u+vp}{u - v}$ .......................4

now we get

f2 = f1 × $\dfrac{(u+v)(u + vp)}{(u-v)(u-vp)}$

f2 = $1240\times \frac{(343+4.77)(343+25)}{(343-4.77)(343-25)}$

f2 = 1475.45 Hz

4 0

4 years ago

Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha

mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × $10^{-26}$ N

(d) 7.37 × $10^{-29}$ N

Explanation:

(a) The minimum value of $x$ will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

$h^{2} = b^{2} + p^{2}$

$h^{2} = 5^{2} + 0.17^{2} \\h = \sqrt{} 25.03\\h= 5.002 m$

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force= $F=\frac{kq1q2}{x^{2} }$

$F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19} }{0.17^{2} }$

$F= 6.38$ × $10^{-26} N$

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F = $\frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19} }{5.002^{2} } }$

F= 7.37× $10^{-29$ N

4 0

3 years ago

I want to answer these questions about vectors

stellarik [79]

Answer:

the first one answer is no

4 0

3 years ago

A hill that has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction. At

Andreas93 [3]

The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

tanФ = vertical rise/horizontal run

tanФ = 15.5 m/100.0 m

tanФ = 0.155

<h3>Angle of incline of the hill</h3>

Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

Ф = 8.81°

So, the angle of incline of the hill above the horizontal is 8.81°.

Learn more about angle of incline of a hill here:

brainly.com/question/10056962

6 0

2 years ago