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Arisa [49]
2 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
Physics
1 answer:
sertanlavr [38]2 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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Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur
Len [333]

Answer:

F_c t_ c = -F_b t_b

And the forces are equal but in the opposite direction. So then we can write by general rule:

m_c \Delta V_{c} = -m_b \Delta V_b

Or equivalently:

m_c \Delta V_{c} +m_b \Delta V_b =0

Where: V_c represent the speed of the car and V_b the speed of the ball

m_c represent the mass of the car

m_b represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

F_c t_ c = -F_b t_b

And the forces are equal but in the opposite direction. So then we can write by general rule:

m_c \Delta V_{c} = -m_b \Delta V_b

Or equivalently:

m_c \Delta V_{c} +m_b \Delta V_b =0

Where: V_c represent the speed of the car and V_b the speed of the ball

m_c represent the mass of the car

m_b represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

3 0
3 years ago
What description refers to fog?
wolverine [178]
I think the answer is D
4 0
3 years ago
A mass hanged on a spring scale. what is the force exerted by gravity on 700g ?
Ipatiy [6.2K]

Answer:

6.86 N

Explanation:

Applying,

F = mg............... Equation 1

Where F = Force exerted by gravity on the mass, m = mass, g = acceleration due to gravity

Note: The Force exerted by gravity on the mass is thesame as the weight of the body.

From the question,

Given: m = 700 g = (700/1000) = 0.7 kg

Constant: g = 9.8 m/s²

Substitute these values into equation 1

F = 9.8(0.7)

F = 6.86 N

6 0
2 years ago
The driver of a car slams on the brakes, causing the car to slow down at a rate of
sdas [7]

Answer:

A. The time taken for the car to stop is 3.14 secs

B. The initial velocity is 81.64 ft/s

Explanation:

Data obtained from the question include:

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Final velocity (V) = 0

Time (t) =?

Initial velocity (U) =?

A. Determination of the time taken for the car to stop.

Let us obtain an express for time (t)

Acceleration (a) = Velocity (V)/time(t)

a = V/t

Velocity (V) = distance (s) /time (t)

V = s/t

a = s/t^2

Cross multiply

a x t^2 = s

Divide both side by a

t^2 = s/a

Take the square root of both side

t = √(s/a)

Now we can obtain the time as follow

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Time (t) =..?

t = √(s/a)

t = √(256/26)

t = 3.14 secs

Therefore, the time taken for the car to stop is 3.14 secs

B. Determination of the initial speed of the car.

V = U + at

Final velocity (V) = 0

Deceleration (a) = –26ft/s2

Time (t) = 3.14 sec

Initial velocity (U) =.?

0 = U – 26x3.14

0 = U – 81.64

Collect like terms

U = 81.64 ft/s

Therefore, the initial velocity is 81.64 ft/s

7 0
3 years ago
Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
djverab [1.8K]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

Rate of mass flow can be written as,

m = \rho A v

where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case v_{2} < v_{1} the pressure in the large cross-sectional area P_{2} will be greater than the pressure  P_{1} in the small cross sectional area, i.e.,

P_{2} > P_{1}.

6 0
3 years ago
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