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3 years ago

15

A certain car is capable of accelerating at a uniform rate of 0.85 m/s^2. What is the magnitude of the car's displacement as it

accelerates uniformly from a speed of 83 km/h to one of 94km/h?

1 answer:

gogolik [260]3 years ago

7 0

Solution: 85.11 m

Given:

acceleration of the car, $a= 0.85 m/s^{2}$

initial velocity, $u=83km/h=23.05 m/s$

final velocity, $v=94 km/h=26.11 m/s$

we need to find the displacement (s) of the car.

we would use the equation of motion:

$2as=v^{2}-u^{2}\\
\Rightarrow s=\frac{v^{2}-u^{2}}{2a} \\
\Rightarrow s=\frac{26.11^{2}-23.05^{2}}{2\times0.85}\\
\Rightarrow s=85.11 m$

hence, the displacement done by the car is = 85.11 m

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3 years ago

A 7.91g bullet is fired into a 1.52-kg ballistic pendulum initially at rest and becomes embedded in it. The pendulum subsequentl

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Answer:

D

Explanation:

By law of conservation of momentum:

momentum before collision = momentum after collision

$m_{1} u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\ m_1 = mass of bullet\\u_1 = initial speed of bullet\\m_2 = mass of pendulum\\u_2 = initial speed of pendulum\\v_1 = final speed of bullet\\v_2 = final speed of pendulum$

Initial speed of bullet is unknown whereas initial speed of pendulum will be zero as it was at rest.

Final speed of bullet and pendulum will be equal as bullet is embedded in pendulum and both moves together a vertical distance of 6.89cm.

Using third equation of motion:

$2 a S = v_{f}^2 - v_{i}^2$

where:

$a = -g = acceleration due to gravity = 9.8 m/s^2 ( for vertical motion)\\S = h = 6.89cm = Vertical height or distance it covers\\v_f = final speed = 0 (after covering 6.89cm)\\v_i = initial speed (unknown)$

Thus by placing values $v_i = 1.1620 m/s$

this speed will be final speed of collision for the calculation of initial speed of bullet.

Putting values:

$(7.91) * x + (1.52*10^3) * 0 = (7.91 + 1.52*10^3) * 1.162\\7.91 x = 1775.431\\x = 224m/s$

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3 years ago

Which statements can be supported by using a position-time graph? Check all that apply. A negative slope results when an individ

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A negative slope results when an individual is moving toward the origin.

A positive slope results when an individual is moving away from the origin.

The slope of the line gives the velocity of the individual.

Explanation:

In a position-time graph, the position is represented on the y-axis (distance from the origin), while the time is represented on the x-axis. Let's analyze each statement:

A negative slope results when an individual is moving toward the origin. --> TRUE. In fact, a negative slope means that the value on the y is decreasing as the time passes, so the distance of the individual is decreasing as the time increases, therefore the person is moving towards the origin.

A horizontal line on the graph means the individual is moving at the same velocity. --> FALSE. A horizontal line means that the position of the individual is constant, so the person is not moving.

A positive slope results when an individual is moving away from the origin. --> TRUE. In this case, the value on the y is increasing with time, so the distance of the person from the origin is increasing with time.

The slope of the line gives the velocity of the individual. --> TRUE. In a position-time graph, the velocity of the individual is equal to the ratio between the displacement and the time taken. But the displacement corresponds to the increase in the y-value, while the time corresponds to the increase in the x-value, so the velocity corresponds to the slope of the graph. In order to know velocity, we must also know the direction of the motion.

The speed of an individual cannot be determined from this type of graph. --> FALSE. The speed is equal to the slope of the position-time graph (in fact, the speed is the magnitude of the velocity)

8 0

4 years ago

Read 2 more answers

an object is thrown horizontally off a cliff with an initial velocity of 5.0 meters per second. the object strikes the ground 3.

Olegator [25]

You know that the formula for finding horizontal displacement is;

s=uₓt
s=(5 m/s)(3s)
s=15 m

Therefore, the ball travelled 15 m.

Hope I helped :)

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3 years ago

Read 2 more answers

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

8 0

4 years ago