Ask question

Ask question

All categories

2 years ago

11

The three notches in the graph occur where the driver changed gears. Describe the changes in velocity and acceleration of the ca

r while in first gear. Is the acceleration just before a gear change larger or smaller than the acceleration just after the change?
Explain your answer

1 answer:

kaheart [24]2 years ago

6 0

<h3>Answer : </h3><h3 /><h3>A ) The larger gear can be moved by applying a relatively small force on the smaller gear.</h3>

<h3>B ) The force applied on the smaller gear is transmitted without any loss to the larger gear .</h3><h3 /><h3>C ) the direction of motion can be changed without changing the direction of the applied force .</h3>

D ) the system would continue to move without any further, after and initial force has set in motion.

You might be interested in

The Air Force is F 22 raptor fighter jets mass is 21,000 KG the F 22 is flying at a height of 26,000 miles what is its gravitati

leva [86]

Answer:

5460000000 J OR 5460000 KJ

Explanation:

GPE = mgh

21000*10*26000

=5460000000J OR 5460000KJ

5 0

1 year ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

3 0

2 years ago

Read 2 more answers

A 56 kg astronaut stands on a bathroom scale inside a rotating circular space station. The radius of the space station is 250 m.

Zielflug [23.3K]

Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,

Mass of astronaut = 56 kg

Radius = 250 m

We need to calculate the speed of space station floor

Using centripetal force and newton's second law

$F=mg$

$\dfrac{mv^2}{r}=mg$

$\dfrac{v^2}{r}=g$

$v=\sqrt{rg}$

Where, v = speed of space station floor

r = radius

g = acceleration due to gravity

Put the value into the formula

$v=\sqrt{250\times9.8}$

$v=49.49\ m/s$

Hence, The speed of space station floor is 49.49 m/s.

6 0

2 years ago

Jared is reading an article about the heavy rains that have fallen recently in his town. But he can’t separate the facts from th

Crank

<span>FACTS:
|
|
It’s that time of year again when the days are wet and cool.
   The rainy season is the best season.
Rain makes up part of Earth’s water cycle.
Water evaporates from streams, lakes, and oceans, then condensation
   and precipitation occur in the form of rain.
   Precipitation in the form of rain is better than snow.
   Snow this time of year makes people gloomy.
Rain is a great boon to local farmers.
It helps their crops grow.

</span>OPINIONS:
|
<span>|   It’s that time of year again when the days are wet and cool.
The rainy season is the best season.
   Rain makes up part of Earth’s water cycle.
   Water evaporates from streams, lakes, and oceans, then condensation
and precipitation occur in the form of rain.
Precipitation in the form of rain is better than snow.
Snow this time of year makes people gloomy.
   Rain is a great boon to local farmers.
   It helps their crops grow.</span>

6 0

2 years ago

Read 2 more answers

How many amps are there in the current of a circuit if the voltage source is 140v and the resistance is 2 ohms?

VLD [36.1K]

Voltage (V)= Current (I) * Resistance (R)
I=V/R=140/2=70A

Hope this helps!

7 0

2 years ago