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4 years ago

Help! which one has the faster average speed? vehicle 1 or vehicle 2?

Physics

1 answer:

adoni [48]4 years ago

3 0

I think it would be vehicle 2

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A person walks 1 mile every day for exercise, leaving her front porch at 9:00 am. and returning to her front porch at 9:25 am. W

Alexxx [7]

Answer: 0

Explanation:

Given that:

Distance walked per day = 1 mile

Starting position = front of house

Final position = front of house(initial position).

In other to aid understanding; Displacement simply refers to the shortest or least distance measured from an individual's initial position to his final point.

Here,

Assume, front of house = point A

Then , initial position = point A (front of house)

Final position, = point A (front of house)

Hence, the shortest distance between final and initial positions is :

Point A - Point A = 0

Since, starting point is the same as the final point, then Displacement = 0

5 0

3 years ago

Which kind of air front forms when a warm air mass meets the area of a cooler air mass? A. stationary front B. cold front C. war

zaharov [31]

This would be a cold front that forms when a warm air mass meets the area of a cooler air mass, although it should be noted that this doesn't always feel "cold".

3 0

4 years ago

Read 2 more answers

What type of evidence is needed for a hypothesis to be supported or not supported?

NISA [10]

Answer:

Hypotheses must be testable, and once tested, they can be supported by evidence. If a statement is made that cannot be tested and disproved, then it is not a hypothesis.

8 0

3 years ago

The wheel of a car has a radius of 20.0 cm. It initially rotates at 120 rpm. In the next minute it makes 90.0 revolutions. (a) W

just olya [345]

Answer:

Explanation:

Given that,

Radius of the wheel, r = 20 cm = 0.2 m

Initial speed of the wheel, $\omega_i=120\ rpm=753.98\ rad/s$

Displacement, $\theta=90\ rev=565.48\ rad$

To find,

The angular acceleration and the distance covered by the car.

Solution,

Let $\alpha$ is the angular acceleration of the car. Using equation of rotational kinematics as :

$\theta=\omega_i t+\dfrac{1}{2}\alpha t^2$

$565.48=753.98\times 60+\dfrac{1}{2}\alpha (60)^2$

$\alpha =-24.81\ rad/s^2$

Let t is the time taken by the car before coming to rest.

$t=\dfrac{\omega_f-\omega_i}{\alpha }$

$t=\dfrac{0-753.98}{-24.81}$

t = 30.39 seconds

Let v is the linear velocity of the car. So,

$v=r\times \omega_i$

$v=0.2\times 753.98$

v = 150.79 m/s

Let d is the distance covered by the car. It can be calculated as :

$d=v\times t$

$d=150.79\ m/s\times 30.39\ s$

d = 4582.5 meters

d = 4.58 km

5 0

3 years ago

A worker wants to load a 12 kg crate into a truck by sliding the crate up a straight ramp which is 2.5 m long and which makes an

olga2289 [7]

Answer:

a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Explanation:

a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:

$U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energy, measured in joules.

$W_{fr}$ - Work losses due to friction, measured in joules.

By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:

$m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta$

Where:

$m$ - Mass of the crate, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$ , $y_{2}$ - Initial and final height of the crate, measured in meters.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the crate, measured in meters per second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$\theta$ - Ramp inclination, measured in sexagesimal degrees.

The equation is now simplified and the coefficient of friction is consequently cleared:

$y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta$

$\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right]$

The final height of the crate is:

$y_{2} = (1.6\,m)\cdot \sin 30^{\circ}$

$y_{2} = 0.8\,m$

If $\theta = 30^{\circ}$ , $y_{1} = 0\,m$ , $y_{2} = 0.8\,m$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{1} = 5\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , the coefficient of friction is:

$\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}$

$\mu_{k} \approx 0.548$

Then, the magnitude of the friction force is:

$f =\mu_{k}\cdot m\cdot g \cdot \cos \theta$

If $\mu_{k} \approx 0.548$ , $m = 12\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\theta = 30^{\circ}$ , the magnitude of the force of friction is:

$f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}$

$f = 55.851\,N$

The magnitude of the force of friction is 55.851 newtons.

b) The energy equation of the situation is:

$m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta$

$y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta$

Now, the final speed is cleared:

$y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta= \frac{1}{2\cdot g}\cdot v_{2}^{2}$

$2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}$

$v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}$

Given that $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 0.8\,m$ , $y_{2} = 0\,m$ , $\mu_{k} \approx 0.548$ , $\theta = 30^{\circ}$ and $v_{1} = 0\,\frac{m}{s}$ , the speed of the crate at the bottom of the ramp is:

$v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}$

$v_{2}\approx 2.526\,\frac{m}{s}$

The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

7 0

3 years ago