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3 years ago

When a car's velocity is positive and its acceleration is negative, what is happening to the car's motion?

Physics

1 answer:

Novosadov [1.4K]3 years ago

8 0

Answer:

Assuming rightward is positive, the velocity is positive whenever the car is moving to the right, and the velocity is negative whenever the car is moving to the left. The acceleration points in the same direction as the velocity if the car is speeding up, and in the opposite direction if the car is slowing down.

Explanation:

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If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact w

Anarel [89]

Answer:

The impact force is 98000 N.

Explanation:

mass = 10 tons

The impact force is the weight of the object.

Weight =mass x gravity

W = 10 x 1000 x 9.8

W = 98000 N

The impact force is 98000 N.

5 0

3 years ago

(a) A light-rail commuter train accelerates at a rate of 1.35 m/s2 . How long does it take to reach its top speed of 80.0 km/h,

Mashcka [7]

Answer:

(a) Time t = 16.46 sec

(b) Time t =13.466 sec

Explanation:

(a) As the train starts from rest its initial velocity u = 0 m/sec

Acceleration $a=1.35m/sec^2$

Final speed v = 80 km/hr

$80km/hr=\frac{80\times 1000}{3600sec}=22.22m/sec$

From first equation of motion v =u+at

So $t=\frac{v-u}{a}=\frac{22.22-0}{1.35}=16.46 sec$

(b) Now initial speed u = 22.22 m/sec

As finally train comes to rest so final speed v=0 m/sec

Deceleration $a=1.65m/sec^2$

So $t=\frac{v-u}{a}=\frac{0-22.22}{-1.65}=13.466 sec$

Final velocity v = 0 m/sec

Time t = 8.30 sec

So acceleration is given by

$a=\frac{v-u}{t}=\frac{0-22.22}{8.3}=-2.6771m/sec^2$

As acceleration is negative so it is a deceleration

7 0

4 years ago

An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c

Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{10}=\frac{1}{v}-\frac{1}{-5}$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{5}$

v = -10 cm

The magnification factor of lens

$m=\frac{-10}{-5}$

m = 2

$m=\frac{h'}{h}$

$h'=h\times m$

$h'=3\times 2$

h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

4 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

Help with the two questions above? Correct answers?

LenKa [72]

(6) first choice: the frequency appears higher and wavelength is shorter.

The car approaches a stationary observer and so the sound will appear to have shorter wavelength. This creates an effect of its siren to sound with higher frequency than it would do if both were stationary.

(7) The Doppler formula for frequency in the case of a stationary observer and source approaching it is as follows:

$f_O = \frac{v}{v-v_s}\cdot f= \frac{343\frac{m}{s}}{(343-25)\frac{m}{s}}\cdot 400Hz \approx 431Hz$

The wavelength is then

$\lambda = \frac{343\frac{m}{s}}{431Hz}\approx 0.80 m$

The third choice "0.80m; 431Hz" is correct

7 0

3 years ago